In the previous set of notes we saw that functions that were holomorphic on an open set
enjoyed a large number of useful properties, particularly if the domain
was simply connected. In many situations, though, we need to consider functions
that are only holomorphic (or even well-defined) on most of a domain
, thus they are actually functions
outside of some small singular set
inside
. (In this set of notes we only consider interior singularities; one can also discuss singular behaviour at the boundary of
, but this is a whole separate topic and will not be pursued here.) Since we have only defined the notion of holomorphicity on open sets, we will require the singular sets
to be closed, so that the domain
on which
remains holomorphic is still open. A typical class of examples are the functions of the form
that were already encountered in the Cauchy integral formula; if
is holomorphic and
, such a function would be holomorphic save for a singularity at
. Another basic class of examples are the rational functions
, which are holomorphic outside of the zeroes of the denominator
.
Singularities come in varying levels of “badness” in complex analysis. The least harmful type of singularity is the removable singularity – a point which is an isolated singularity (i.e., an isolated point of the singular set
) where the function
is undefined, but for which one can extend the function across the singularity in such a fashion that the function becomes holomorphic in a neighbourhood of the singularity. A typical example is that of the complex sinc function
, which has a removable singularity at the origin
, which can be removed by declaring the sinc function to equal
at
. The detection of isolated removable singularities can be accomplished by Riemann’s theorem on removable singularities (Exercise 35 from Notes 3): if a holomorphic function
is bounded near an isolated singularity
, then the singularity at
may be removed.
After removable singularities, the mildest form of singularity one can encounter is that of a pole – an isolated singularity such that
can be factored as
for some
(known as the order of the pole), where
has a removable singularity at
(and is non-zero at
once the singularity is removed). Such functions have already made a frequent appearance in previous notes, particularly the case of simple poles when
. The behaviour near
of function
with a pole of order
is well understood: for instance,
goes to infinity as
approaches
(at a rate comparable to
). These singularities are not, strictly speaking, removable; but if one compactifies the range
of the holomorphic function
to a slightly larger space
known as the Riemann sphere, then the singularity can be removed. In particular, functions
which only have isolated singularities that are either poles or removable can be extended to holomorphic functions
to the Riemann sphere. Such functions are known as meromorphic functions, and are nearly as well-behaved as holomorphic functions in many ways. In fact, in one key respect, the family of meromorphic functions is better: the meromorphic functions on
turn out to form a field, in particular the quotient of two meromorphic functions is again meromorphic (if the denominator is not identically zero).
Unfortunately, there are isolated singularities that are neither removable or poles, and are known as essential singularities. A typical example is the function , which turns out to have an essential singularity at
. The behaviour of such essential singularities is quite wild; we will show here the Casorati-Weierstrass theorem, which shows that the image of
near the essential singularity is dense in the complex plane, as well as the more difficult great Picard theorem which asserts that in fact the image can omit at most one point in the complex plane. Nevertheless, around any isolated singularity (even the essential ones)
, it is possible to expand
as a variant of a Taylor series known as a Laurent series
. The
coefficient
of this series is particularly important for contour integration purposes, and is known as the residue of
at the isolated singularity
. These residues play a central role in a common generalisation of Cauchy’s theorem and the Cauchy integral formula known as the residue theorem, which is a particularly useful tool for computing (or at least transforming) contour integrals of meromorphic functions, and has proven to be a particularly popular technique to use in analytic number theory. Within complex analysis, one important consequence of the residue theorem is the argument principle, which gives a topological (and analytical) way to control the zeroes and poles of a meromorphic function.
Finally, there are the non-isolated singularities. Little can be said about these singularities in general (for instance, the residue theorem does not directly apply in the presence of such singularities), but certain types of non-isolated singularities are still relatively easy to understand. One particularly common example of such non-isolated singularity arises when trying to invert a non-injective function, such as the complex exponential or a power function
, leading to branches of multivalued functions such as the complex logarithm
or the
root function
respectively. Such branches will typically have a non-isolated singularity along a branch cut; this branch cut can be moved around the complex domain by switching from one branch to another, but usually cannot be eliminated entirely, unless one is willing to lift up the domain
to a more general type of domain known as a Riemann surface. As such, one can view branch cuts as being an “artificial” form of singularity, being an artefact of a choice of local coordinates of a Riemann surface, rather than reflecting any intrinsic singularity of the function itself. The further study of Riemann surfaces is an important topic in complex analysis (as well as the related fields of complex geometry and algebraic geometry), but unfortunately this topic will probably be postponed to the next course in this sequence (which I will not be teaching).
ā 1. Laurent series ā
Suppose we are given a holomorphic function and a point
in
. For a sufficiently small radius
, the circle
and its interior both lie in
, and the Cauchy integral formula tells us that
Lemma 1 (Cauchy integral formula decomposition in annular regions) Letbe a holomorphic function. Let
,
be simple closed anticlockwise curves in
such that
is contained in the interior
of
(or equivalently, by Exercise 49 of Notes 3, that
is contained in the exterior
of
). Suppose also that the “annular region”
is contained in
. Then there exists a decomposition
on
, where
is holomorphic on the union of
and the interior of
, and
is holomorphic on the union of
and the exterior of
, with
as
. Furthermore, this decomposition is unique.
In addition, we have the Cauchy integral type formulae
for
in the interior of
, and
for
in the exterior of
. In particular, we have
for
in
.
Proof: We begin with uniqueness. Suppose we have two decompositions
Now for existence. Suppose that we can establish the identity (1) for in
. Then we can define
on
by
Thus it remains to establish (1). This follows from the homology form of the Cauchy integral formula (Exercise 63(v) of Notes 3), but we can also avoid explicit use of homology by the following “keyhole contour” argument. For , we have from the definition of winding number that
By perturbing using Cauchy’s theorem we may assume that these curves are simple closed polygonal paths (if one wishes, one can also restrict the edges to be horizontal and vertical, although this is not strictly necessary for the argument). By connecting a point in
to a point in
by a polygonal path in the interior of
, and removing loops, self-intersections, or excursions into the interior (or image) of
, we can find a simple polygonal path
from a point
in
to a point
in
that lies entirely in
except at the endpoints. By rearranging
and
we may assume that
is the initial and terminal point of
, and
is the initial and terminal point of
. Then the closed polygonal path
has vanishing winding number in the interior of
or exterior of
, thus
contains all the points where the winding number is non-zero. This path is not simple, but we can approximate it to arbitrary accuracy
by a simple closed polygonal path
by shifting the simple polygonal paths
and
slightly; for
small enough, the interior of
will then lie in
. Applying Cauchy’s theorem (Theorem 52 of Notes 3) we conclude that
Exercise 2 Letbe a simple closed anticlockwise curve, and let
be simple closed anticlockwise curves in the interior
of
whose images are disjoint, and such that the interiors
are also disjoint. Let
be an open set containing
and the region
Let
be holomorphic. Show that for any
, one has
(Hint: induct on
using Lemma 1.)
Exercise 3 (PainlevĆ©’s theorem on removable singularities) Letbe an open subset of
. Let
be a compact subset of
which has zero length in the following sense: for any
, one can cover
by a countable number of disks
such that
. Let
be a bounded holomorphic function. Show that the singularities in
are removable in the sense that there is an extension
of
to
which remains holomorphic. (Hint: one can work locally in some disk in
that contains a portion of
. Cover this portion by a finite number of small disks, group them into connected components, use the previous exercise, and take an appropriate limit.) Note that this result generalises Riemann’s theorem on removable singularities, see Exercise 35 from Notes 3. The situation when
has positive length is considerably more subtle, and leads to the theory of analytic capacity, which we will not discuss further here.
Now suppose that is holomorphic for some annulus of the form
Exercise 4 (Fourier inversion formula) Letbe holomorphic on some open set
that contains an annulus of the form (3), and let
be the coefficients of the Laurent expansion (4) in this annulus. Show that the coefficients
are uniquely determined by
and
, and are given by the formula
for all integers
, whenever
is a simple closed curve in the annulus with
. Also establish the bounds
and
The following modification of the above exercise may help explain the terminology “Fourier inversion formula”.
Exercise 5 (Fourier inversion formula, again) Let.
- (i) Show that if
is holomorphic on the annulus
, then we have the Fourier expansion
for all
, where the Fourier coefficients
are given by the formula
Furthermore, show that the Fourier series in (7) is absolutely convergent, and the coefficients
obey the asymptotic bounds (5), (6).
- (ii) Conversely, if
are complex numbers obeying the asymptotic bounds (5), (6), show that there exists a function
holomorphic on the annulus
obeying the Fourier expansion (7) and the inversion formula (8).
The Laurent series for a given function can vary as one varies the annulus. Consider for instance the function . In the annulus
, the Laurent expansion coincides with the Taylor expansion:
Exercise 6 Find the Laurent expansions for the functionin the regions
,
, and
. (Hint: use partial fractions.)
We can use Laurent series to analyse an isolated singularity. Suppose that is holomorphic on a punctured disk
. By the above discussion, we have a Laurent series expansion (4) in this punctured disk. If the singularity is removable, then the Laurent series must coincide with the Taylor series (by the uniqueness component of Exercise 4), so in partcular
for all negative
; conversely, if
vanishes for all negative
, then the Laurent series matches up with a convergent Taylor series and so the singularity is removable. We then adopt the following classification:
- (i)
has a removable singularity at
if one has
for all negative
. If furthermore there is an
such that
and
for
, we say that
has a zero of order
at
(after removing the singularity). Zeroes of order
are known as simple zeroes, zeroes of order
are known as double zeroes, and so forth.
- (ii)
has a pole of order
at
for some
if one has
, and
for all
. Poles of order
are known as simple poles, poles of order
are double poles, and so forth.
- (iii)
has an essential singularity if
for infinitely many negative
.
It is clear that any holomorphic function will be of exactly one of the above three categories. Also, from the uniqueness of Laurent series, shrinking
does not affect which of the three categories
will lie in (or what order of pole
will have, in the second category). Thus, we can classify any isolated singularity
of a holomorphic function
with singularities as being either removable, a pole of some finite order, or an essential singularity by restricting
to a small punctured disk
and inspecting the Laurent coefficients
for negative
.
Example 7 The functionhas a Laurent expansion
and thus has an essential singularity at
.
It is clear from the definition (and the holomorphicity of Taylor series) that (as discussed in the introduction), a holomorphic function has a pole of order
at an isolated singularity
if and only if it is of the form
for some holomorphic
with
. Similarly, a holomorphic function would have a zero of order
at
if and only if
for some
with
.
We can now define a class of functions that only have “nice” singularities:
Definition 8 (Meromorphic functions) Letbe an open subset of
. A function
defined on
outside of a singular set
is said to be meromorphic on
if
- (i)
is closed in
and discrete (i.e., all points in
are isolated);
- (ii)
is holomorphic on
; and
- (iii) Every
is either a removable singularity or a pole of finite order.
Two meromorphic functions ,
are said to be equivalent if they agree on their common domain of definition
. It is easy to see that this is an equivalence relation. It is common to identify meromorphic functions up to equivalence, similarly to how in measure theory it is common to identify functions which agree almost everywhere.
Exercise 9 (Meromorphic functions form a field) Letdenote the space of meromorphic functions on a connected open set
, up to equivalence. Show that
is a field (with the obvious field operations). What happens if
is not connected?
Exercise 10 (Order is a valuation) Ifis a meromorphic function, and
, define the order
of
at
as follows:
Establish the following facts:
- (a) If
has a removable singularity at
, and has a zero of order
at
once the singularity is removed, then
.
- (b) If
is holomorphic at
, and has a zero of order
at
, then
.
- (c) If
has a pole of order
at
, then
.
- (d) If
is identically zero, then
.
In the language of abstract algebra, the above facts are asserting that
- (i) If
and
are equivalent meromorphic functions, then
for all
. In particular, one can meaningfully define the order of an element of
at any point
in
, where
is as in the preceding exercise.
- (ii) If
and
, show that
. If
is not zero, show that
.
- (iii) If
and
, show that
. Furthermore, show if
, then the above inequality is in fact an equality.
is a valuation on the field
.
The behaviour of a holomorphic function near an isolated singularity depends on the type of singularity.
Theorem 11 Letbe holomorphic on an open set
outside of a singular set
, and let
be an isolated singularity in
.
- (i) If
is a removable singularity of
, then
converges to a finite limit as
.
- (ii) If
is a pole of
, then
as
.
- (iii) (Casorati-Weierstrass theorem) If
is an essential singularity of
, then every point
of
is a limit point of
as
, that is to say there exists a sequence
converging to
such that
converges to
(where we adopt the convention that
converges to
if
converges to
).
Proof: Part (i) is obvious. Part (ii) is immediate from the factorisation and noting that
converges to the non-zero value
as
. The
case of (iii) follows from Riemann’s theorem on removable singularities (Exercise 35 from Notes 3). Now suppose
is finite. If (iii) failed, then there exist
such that
avoids the disk
on the domain
. In particular, the function
is bounded and holomorphic on
, and thus extends holomorphically to
by Riemann’s theorem. This function cannot vanish identically, so we must have
on
for some
and some holomorphic
that does not vanish at
. Rearranging this as
, we see that
has a pole or removable singularity at
, a contradiction.
In Theorem 56 below we will establish a significant strengthening of the Casorati-Weierstrass theorem known as the Great Picard Theorem.
Exercise 12 Letbe holomorphic in
outside of a discrete set
of singularities. Let
. Show that the radius of convergence of the Taylor series of
around
is equal to the distance from
to the nearest non-removable singularity in
, or
if no such non-removable singularity exists. (This fact provides a neat way to understand the rate of growth of a sequence
: form its generating function
, locate the singularities of that function, and find out how close they get to the origin. This is a simple example of the methods of analytic combinatorics in action.)
A curious feature of the singularities in complex analysis is that the order of singularity is “quantised”: one can have a pole of order ,
, or
(for instance), but not a pole of order
or
. This quantisation can be exploited: if for instance one somehow knows that the order of the pole is less than
for some integer
and real number
, then the singularity must be removable or a pole of order at most
. The following exercise formalises this assertion:
Exercise 13 Letbe holomorphic on a disk
except for a singularity at
. Let
be an integer, and suppose that there exist
,
such that one has the upper bound
for all
. Show that the singularity of
at
is either removable, or a pole of order at most
(the latter option is only possible for
positive, of course). (Hint: use Exercise 4 and a limiting argument to evaluate the Laurent coefficients
for
.) In particular, if one has
for all
, then the singularity is removable.
As mentioned in the introduction, the theory of meromorphic functions becomes cleaner if one replaces the complex plane with the Riemann sphere. This sphere is a model example of a Riemann surface, and we will now digress to briefly introduce this more general concept (though we will not develop the general theory of Riemann surfaces in any depth here). To motivate the definition, let us first recall from differential geometry the notion of a smooth
-dimensional manifold
(over the reals).
Definition 14 (Smooth manifold) Let, and let
be a topological space. An (
-dimensional real) atlas for
is an open cover
of
together with a family of homeomorphisms
(known as coordinate charts) from each
to an open subset
of
. Furthermore, the atlas is said to be smooth if for any
, the transition map
, which maps one open subset of
to another, is required to be smooth (i.e., infinitely differentiable). A map
from one topological space
(equipped with a smooth atlas of coordinate charts
for
) to another
(equipped with a smooth atlas of coordinate charts
for some
) is said to be smooth if it is continuous, for any
and
, the maps
are smooth; if
is invertible and
and
are both smooth, we say that
is a diffeomorphism, and that
and
are diffeomorphic. Two smooth atlases on
are said to be equivalent if the identity map from
(equipped with one of the two atlases) to
(equipped with the other atlas) is a diffeomorphism; this is easily seen to be an equivalence relation, and an equivalence class of such atlases is called a smooth structure on
. A smooth
-dimensional real manifold is a Hausdorff topological space
equipped with a smooth structure. (In some texts the mild additional condition of second countability on
is also imposed.) A map
between two smooth manifolds is said to be smooth, if the map
from
(equipped with one of the atlases in the smooth structure on
) to
(equipped with one of the atlases in the smooth structure on
) is smooth; it is easy to see that this definition is independent of the choices of atlas. We may similarly define the notion of a diffeomorphism between two smooth manifolds.
This definition may seem excessively complicated, but it captures the modern geometric philosophy that one should strive as much as possible to work with objects that are coordinate-independent in that they do not depend on which atlas of coordinate charts one picks within the equivalence class of the given smooth structure in order to perform computations or to define foundational concepts. One can also define smooth manifolds more abstractly, without explicit reference to atlases, by working instead with the structure sheaf of the rings of smooth real-valued functions on open subsets
of the manifold
, but we will not need to do so here.
Example 15 A simple example of a smooth-dimensional manifold is the unit circle
; there are many equivalent atlases one could place on this circle to define the smooth structure, but one example would be the atlas consisting of the two charts
,
, defined by setting
,
,
,
,
for
, and
for
. Another smooth manifold, which turns out to be diffeomorphic to the unit circle
, is the one-point compactification
of the real numbers, with the two charts
,
defined by setting
,
,
,
to be the identity map, and
defined by setting
for
and
.
Exercise 16 Verify that the unit circleis indeed diffeomorphic to the one-point compactification
.
A Riemann surface is defined similarly to a smooth manifold, except that the dimension is restricted to be one, the reals
are replaced with the complex numbers, and the requirement of smoothness is replaced with holomorphicity (thus Riemann surfaces are to the complex numbers as smooth curves are to the real numbers). More precisely:
Definition 17 (Riemann surface) Letbe a Hausdorff topological space. A holomorphic atlas on
is an open cover
of
together with a family of homeomorphisms
(known as coordinate charts) from each
to an open subset
of
, such that, for any
, the transition map
, which maps one open subset of
to another, is required to be holomorphic. A map
from one space
(equipped with coordinate charts
for
) to another
(equipped with coordinate charts
for some
) is said to be holomorphic if it is continuous and, for any
and
, the maps
are holomorphic; if
is invertible and
and
are both holomorphic, we say that
is a complex diffeomorphism, and that
and
are complex diffeomorphic. Two holomorphic atlases on
are said to be equivalent if the identity map from
(equipped with one of the atlases) to
(equipped with the other atlas) is a complex diffeomorphism; this is easily seem to be an equivalence relation, and we refer to such an equivalence class as a (one-dimensional) complex structure on
. A Riemann surface is a Hausdorff topological space
, equipped with a one-dimensional complex structure. (Again, in some texts the hypothesis of second countability is imposed. This makes little difference in practice, as most Riemann surfaces one actually encounters will be second countable.)
By considering dimensions greater than one, one can arrive at the more general notion of a complex manifold, the study of which is the focus of complex geometry (and also plays a central role in the closely related fields of several complex variables and complex algebraic geometry). However, we will not need to deal with higher-dimensional complex manifolds in this course. The notion of a Riemann surface should not be confused with that of a Riemannian manifold, which is the topic of study of Riemannian geometry rather than complex geometry.
Clearly any open subset of the complex numbers
is a Riemann surface, in which one can use the atlas that only consists of one “tautological” chart, the identity map
. More generally, any open subset of a Riemann surface is again a Riemann surface. If
are open subsets of the complex numbers, and
is a map, then by unpacking all the definitions we see that
is holomorphic in the sense of Definition 17 if and only if it is holomorphic in the usual sense.
Now we come to the Riemann sphere , which is to the complex numbers as
is to the real numbers. As a set, this is the complex numbers
with one additional point (the point at infinity)
attached. Topologically, this is the one-point compactification of the complex numbers
: the open sets of
are either subsets of
that were already open, or complements
of compact subsets
of
. As a Riemann surface, the complex structure can be described by the atlas of coordinate charts
,
, where
,
,
,
is the identity map, and
equals
for
with
. It is not difficult to verify that this is indeed a Riemann surface (basically because the map
is holomorphic on
). One can identify the Riemann sphere with a geometric sphere, and specifically the sphere
, through the device of stereographic projection through the north pole
, identifying a point
in
with the point
on
collinear with that point, and the point at infinity
identified with the north pole
. This geometric perspective is especially helpful when thinking about Mƶbius transformations, as is for instance exemplified by this excellent video. (We may cover Mƶbius transformations in a subsequent set of notes.)
By unpacking the definitions, we can now work out what it means for a function to be holomorphic to or from the Riemann sphere. For instance, if is a map from an open subset
of
to the Riemann sphere
, then
is holomorphic if and only if
- (i)
is continuous;
- (ii)
is holomorphic on the set
(which is open thanks to (i)); and
- (iii)
is holomorphic on the set
(which is open thanks to (i)), where we adopt the convention
.
Similarly, if a function is a map from an open subset
of the Riemann sphere
to the Riemann sphere, then
is holomorphic if and only if
- (i)
is holomorphic on
; and
- (ii)
is holomorphic on
, where we adopt the convention
.
We can then identify meromorphic functions with holomorphic functions on the Riemann sphere:
Exercise 18 Letbe open connected, let
be a discrete subset of
, and let
be a function. Show that the following are equivalent:
Furthermore, if (ii) holds, show that
- (i)
is meromorphic on
.
- (ii)
is the restriction of a holomorphic function
to the Riemann sphere, that is not identically equal to
.
is uniquely determined by
, and is unaffected if one replaces
with an equivalent meromorphic function.
Among other things, this exercise implies that the composition of two meromorphic functions is again meromorphic (outside of where the composition is undefined, of course).
Exercise 19 Letbe a holomorphic map from the Riemann sphere to itself. Show that
is either identically
, or is a rational function in the sense that there exist polynomials
of one complex variable, with
not identically zero, such that
for all
with
. (Hint: show that
has finitely many poles, and eliminate them by multiplying
by appropriate linear factors. Then use Exercise 29 from Notes 3.)
Exercise 20 (Partial fractions) Letbe a polynomial of one complex variable, which by the fundamental theorem of algebra we may write as
for some distinct roots
, some non-zero
, and some positive integers
. Let
be another polynomial of one complex variable. Show that there exist unique polynomials
, with each
having degree less than
for
, such that one has the partial fraction decomposition
for all
. Furthermore, show that
vanishes if the degree
of
is less than the degree
of
, and has degree
otherwise.
ā 2. The residue theorem ā
Now we can prove a significant generalisation of the Cauchy theorem and Cauchy integral formula, known as the residue theorem.
Suppose one has a function holomorphic on an open set
outside of a singular set
. If
is an isolated singularity of
, then we have a Laurent expansion
We then have
Theorem 21 (Residue theorem) Letbe a simply connected open set, and let
be holomorphic outside of a closed discrete singular set
(thus all singularities in
are isolated singularities). Let
be a closed curve in
. Then
where only finitely many of the terms on the right-hand side are non-zero.
Proof: Being simply connected, can be contracted to a point inside
. This homotopy takes values inside some compact subset
of
, and thus only contains finitely many of the singularities in
. By Rouche’s theorem, the winding number
then vanishes for any singularity
in
(since
can be contracted to a point without touching
). Hence on the right-hand side we may replace
by
without loss of generality.
Next, we exploit the linearity of the residue theorem in to reduce to the case where all the residues
vanish. We introduce the rational function
defined by
Now we repeat the proof of Cauchy’s theorem (Theorem 4 of Notes 3), discretising the homotopy into short closed polygonal paths
(each of diameter less than
) around which the integral of
is zero, to conclude (9). The argument is completely analogous, save for the technicality that the paths
may occasionally pass through one of the points in
. But this can be easily rectified by perturbing each of the paths
by adding a short detour around any point of
that is passed through; we leave the details to the interested reader.
Combining the residue theorem with the Jordan curve theorem, we obtain the following special case, which is already enough for many applications:
Corollary 22 (Residue theorem for simple closed curves) Letbe a simple closed anticlockwise curve in
. Suppose that
is holomorphic on an open set
containing the image and interior of
, outside of a closed discrete
that does not intersect the image of
. Then we have
If
is oriented clockwise instead of anticlockwise, then we instead have
Exercise 23 (Homology version of residue theorem) Show that the residue theorem continues to hold when the closed curveis replaced by a
-cycle (as in Exercise 63 of Notes 3) that avoids all the singularities in
, and the requirement that
be simply connected is replaced by the requirement that
contains all the points
outside of the image of
where
.
Exercise 24 (Exterior version of residue theorem) Letbe a simple closed anticlockwise curve in
. Suppose that
is holomorphic on an open set
containing the image and exterior of
, outside of a finite
that does not intersect the image of
. Suppose also that
converges to a finite limit
in the limit
. Show that
If
is oriented clockwise instead of anticlockwise, show instead that
In order to use the residue theorem effectively, one of course needs some tools to compute the residue at a given point. The Fourier inversion formula (4) expresses such residues as a contour integral, but this is not so useful in practice as often the best way to compute such integrals is via the residue theorem, leaving one back where one started! But if the singularity is not an essential one, we have some useful formulae:
Exercise 25 Letbe holomorphic on an open set
outside of a singular set
, and let
be an isolated point of
.
Using these facts, show that Cauchy’s theorem (Theorem 14 from Notes 3), the Cauchy integral formula (Theorem 39 from Notes 3), and the higher order Cauchy integral formula (Exercise 40 from Notes 3) can be derived from the residue theorem. (Of course, this is not an independent proof of these theorems, as they were used in the proof of the residue theorem!)
- (i) If
has a removable singularity at
, show that
.
- (ii) If
has a simple pole at
, show that
.
- (iii) If
has a pole of order at most
at
for some
, show that
In particular, if
near
for some
that is holomorphic at
, then
The residue theorem can be applied in countless ways; we give only a small sample of them below.
Exercise 26 Use the residue theorem to give an alternate proof of the fundamental theorem of algebra, by considering the integralfor a polynomial
of degree
and some large radius
.
Exercise 27 Letbe a Dirichlet polynomial of the form
for some sequence
of complex numbers, with only finitely many of the
non-zero. Establish Perron’s formula
for any real numbers
with
not an integer. What happens if
is an integer? Generalisations and variants of this formula, particularly with the Dirichlet polynomial replaced by more general Dirichlet series in which infinitely many of the
are allowed to be non-zero, are of particular use in analytic number theory; see for instance this previous blog post.
Exercise 28 (Spectral theorem for matrices) This exercise presumes some familiarity with linear algebra. Letbe a positive integer, and let
denote the ring of
complex matrices. Let
be a matrix in
. The characteristic polynomial
, where
is the
identity matrix, is a polynomial of degree
in
with leading coefficient
; we let
be the distinct zeroes of this polynomial, and let
be the multiplicities; thus by the fundamental theorem of algebra we have
We refer to the set
as the spectrum of
. Let
be any closed anticlockwise curve that contains the spectrum of
in its interior, and let
be an open subset of
that contains
and its interior.
- (i) Show that the resolvent
is a meromorphic function on
with poles at the spectrum of
, where we call a matrix-valued function meromorphic if each of its
components are meromorphic. (Hint: use the adjugate matrix.)
- (ii) For any holomorphic
, we define the matrix
by the formula
(cf. the Cauchy integral formula). We refer to
as the holomorphic functional calculus for
applied to
. Show that the matrix
does not depend on the choice of
, depends linearly on
, and equals the identity matrix when
is the constant function
. Furthermore, if
is the function
, show that
Conclude in particular that if
is a polynomial
with complex coefficients
, then the function
(as defined by the holomorphic functional calculus) matches how one would define
algebraically, in the sense that
- (iii) Prove the Cayley-Hamilton theorem
. (Note from (ii) that it does not matter whether one interprets
algebraically, or via the holomorphic functional calculus.)
- (iv) If
is holomorphic, show that the matrix-valued function
has only removable singularities in
.
- (v) If
are holomorphic, establish the identity
- (vi) Show that there exist matrices
that are idempotent (thus
for all
), commute with each other and with
, sum to the identity (thus
), annihilate each other (thus
for all distinct
) and are such that for each
, one has the nilpotency property
In particular, we have the spectral decomposition
where each
is a nilpotent matrix with
. Finally, show that the range of
(viewed as a linear operator from
to itself) has dimension
. Find a way to interpret each
as the (negative of the) “residue” of the resolvent operator
at
.
Under some additional hypotheses, it is possible to extend the analysis in the above exercise to infinite-dimensional matrices or other linear operators, but we will not do so here.
ā 3. The argument principle ā
We have not yet defined the complex logarithm of a complex number
, but one of the properties we would expect of this logarithm is that its derivative should be the reciprocal function:
. In particular, by the chain rule we would expect the formula
A general rule of thumb in complex analysis is that holomorphic functions behave like generalisations of polynomials, and meromorphic functions behave like generalisations of rational functions. In view of this rule of thumb and the above calculation, the following lemma should thus not be surprising:
Lemma 29 Letbe a holomorphic function on an open set
outside of a singular set
, and let
be either an element of
or an isolated point of
.
- (i) If
is holomorphic and non-zero at
, then the log-derivative
is also holomorphic at
.
- (ii) If
is holomorphic at
with a zero of order
, then the log-derivative
has a simple pole at
with residue
.
- (iii) If
has a removable singularity at
, and is non-zero once the singularity is removed, then the log-derivative
has a removable singularity at
.
- (iv) If
has a removable singularity at
, and has a zero of order
once the singularity is removed, then the log-derivative
has a simple pole at
with residue
.
- (v) If
has a pole of order
at
, then the log-derivative
has a simple pole at
with residue
.
Proof: The claim (i) is obvious. For (ii), we use Taylor expansion to factor for some
holomorphic and non-zero near
, and then from (11) we have
Remark 30 Note that the lemma does not cover all possible singularity and zero scenarios. For instance,could be identically zero, in which case the log-derivative is nowhere defined. If
has an essential singularity then the log-derivative can be a pole (as seen for instance by the example
for some
) or another essential singularity (as can be seen for instance by the example
). Finally, if
has a non-isolated singularity, then the log-derivative could exhibit a wide range of behaviour (but probably will be quite wild as one approaches the singular set).
By combining the above lemma with the residue theorem, we obtain the argument principle:
Theorem 31 (Argument principle) Letbe a simple closed anticlockwise curve. Let
be an open set containing
and its interior. Let
be a meromorphic function on
that is holomorphic and non-zero on the image of
. Suppose that after removing all the removable singularities of
,
has zeroes
in the interior of
(of orders
respectively), and poles
in the interior of
(of orders
respectively). (
is also allowed to have zeroes and poles in the exterior of
.) Then we have
where
is the closed curve
.
Proof: The first equality of (13) follows from the residue theorem and Lemma 29. From the change of variables formula (Exercise 16(ix) of Notes 2) we have
We isolate the special case of the argument principle when there are no poles for special mention:
Corollary 32 (Special case of argument principle) Letbe a simple closed anticlockwise curve, let
be an open set containing the image of
and its interior, and let
be holomorphic. Suppose that
has no zeroes on the image of
. Then the number of zeroes of
(counting multiplicity) in the interior of
is equal to the winding number
of
around the origin.
Recalling that the winding number is a homotopy invariant (Lemma 41 of Notes 3), we conclude that the number of zeroes of a holomorphic function in the interior of a simple closed anticlockwise curve is also invariant with respect to continuous perturbations, so long as zeroes never cross the curve itself. More precisely:
Corollary 33 (Stability of number of zeroes) Letbe an open set. Let
,
be simple closed anticlockwise curves that are homotopic as closed curves via some homotopy
; suppose also that
contains the interiors of
and
. Let
be holomorphic, and let
be a continuous function such that
and
for all
. Suppose that
for all
and
(i.e., at time
, the curve
never encounters any zeroes of
). Then the number of zeroes (counting multiplicity) of
in the interior of
equals the number of zeroes of
in the interior of
(counting multiplicity).
Proof: By Corollary 32, it suffices to show that
Informally, the above corollary asserts that zeroes of holomorphic functions cannot be created or destroyed, as long as they are confined within a closed curve.
Example 34 Letbe the unit circle
. The polynomial
has a double zero at
, so (counting multiplicity) has two zeroes in the interior of
. If we consider instead the perturbation
for some
, this has simple zeroes at
and
respectively, so as long as
, the holomorphic function
also has two zeroes in the interior of
; but as
crosses
, the zeroes of
pass through
, and one no longer has any zeroes of
in the interior of
. The situation can be contrasted with the real case: the function
has a double zero at the origin when
, but as soon as
becomes positive, the zeroes immediately disappear from the real line. Note that the stability of zeroes fails if we do not count zeroes with multiplicity; thus, as a general rule of thumb, one should always try to count zeroes with multiplicity when doing complex analysis. (Heuristically, one can think of a zero of order
as
simple zeroes that are “infinitesimally close together”.)
Example 35 When one considers meromorphic functions instead of holomorphic ones, then the number of zeroes inside a region need not be stable any more, but the number of zeroes minus the number of poles will be stable. Consider for instance the meromorphic function, which has a removable singularity at
but no zeroes or poles. If we perturb it to
for some
, then we suddenly have a double pole at
, but this is balanced by two simple zeroes at
and
; in the limit as
we see that the two zeroes “collide” with the double pole, annihilating both the zeroes and the poles.
A particularly useful special case of the stability of zeroes is Rouche’s theorem:
Theorem 36 (Rouche’s theorem) Letbe a simple closed curve, and let
be an open set containing the image of
and its interior. Let
be holomorphic. If one has
for all
in the image of
, then
and
have the same number of zeroes (counting multiplicity) in the interior of
.
Proof: We may assume without loss of generality that is anticlockwise. By hypothesis,
and
cannot have zeroes on the image of
. The claim then follows from Corollary 33 with
,
,
,
, and
.
Rouche’s theorem has many consequences for complex analysis. One basic consequence is the open mapping theorem:
Theorem 37 (Open mapping theorem) Letbe an open connected non-empty subset of
, and let
be holomorphic and not constant. Then
is also open.
Proof: Let . As
is not constant, the zeroes of
are isolated (Corollary 24 of Notes 3). Thus, for
sufficiently small,
is nonvanishing on the image of the circle
. Clearly
has at least one zero in the interior of this circle. Thus, by Rouche’s theorem, if
is sufficiently close to
, then
will also have at least one zero in the interior of this circle. In particular,
contains a neighbourhood of
, and the claim follows.
Exercise 38 Use Rouche’s theorem to obtain another proof of the fundamental theorem of algebra, by showing that a polynomialwith
and
has exactly
zeroes (counting multiplicity) in the complex plane. (Hint: compare
with
inside some large circle
.)
Exercise 39 (Inverse function theorem) Letbe an open subset of
, let
, and let
be a holomorphic function such that
. Show that there exists a neighbourhood
of
in
such that the map
is a complex diffeomorphism; that is to say, it is holomorphic, invertible, and the inverse is also holomorphic. Finally, show that
for all
. (Hint: one can either mimic the real-variable proof of the inverse function theorem using the contraction mapping theorem, or one can use Rouche’s theorem and the open mapping theorem to construct the inverse.)
Exercise 40 Letbe an open subset of
, and
be a map. Show that the following are equivalent:
- (i)
is a local complex diffeomorphism. That is to say, for every
there is a neighbourhood
of
in
such that
is open and
is a complex diffeomorphism (as defined in the preceding exercise).
- (ii)
is holomorphic on
and is a local homeomorphism. That is to say, for every
there is a neighbourhood
of
in
such that
is open and
is a homeomorphism.
- (iii)
is holomorphic on
and is locally injective. That is to say, for every
there is a neighbourhood
of
in
such that
is injective.
- (iv)
is holomorphic on
, and the derivative
is nowhere vanishing.
Exercise 41 (Hurwitz’s theorem) Letbe an open connected non-empty subset of
, and let
be a sequence of holomorphic functions that converge uniformly on compact sets to a limit
(which is then necessarily also holomorphic, thanks to Theorem 34 of Notes 3). Prove the following two versions of Hurwitz’s theorem:
- (i) If none of the
have any zeroes in
, show that either
also has no zeroes in
, or is identically zero.
- (ii) If all of the
are univalent (that is to say, they are injective holomorphic functions), show that either
is also univalent, or is constant.
Exercise 42 (Bloch’s theorem) The purpose of this exercise is to establish a more quantitative variant of the open mapping theorem, due to Bloch; this will be useful later in this notes for proving the Picard and Montel theorems. Letbe a holomorphic function on a disk
, and suppose that
is non-zero
- (i) Suppose that
for all
. Show that there is an absolute constant
such that
contains the disk
. (Hint: one can normalise
,
,
. Use the higher order Cauchy integral formula to get some bound on
for
near the origin, and use this to approximate
by
near the origin. Then apply Rouche’s theorem.)
- (ii) Without the hypothesis in (i), show that there is an absolute constant
such that
contains a disk of radius
. (Hint: if one has
for all
, then we can apply (i) with
replaced by
. If not, pick
with
, and start over with
replaced by
and
replaced by
. One cannot iterate this process indefinitely as it will create a singularity of
in
.)
ā 4. Branches of the complex logarithm ā
We have refrained until now from discussing one of the most basic transcendental functions in complex analysis, the complex logarithm. In real analysis, the real logarithm can be defined as the inverse of the exponential function
; it can also be equivalently defined as the antiderivative of the function
, with the initial condition
. (We use
here for the real logarithm in order to distinguish it from the complex logarithm below.)
Let’s see what happens when one tries to extend these definitions to the complex domain. We begin with the inversion of the complex exponential. From Euler’s formula we have that ; more generally, we have
whenever
for some integer
. In particular, the exponential function
is not injective. Indeed, for any non-zero
, we have a multi-valued logarithm
Of course, one also encounters multi-valued functions in real analysis, starting when one tries to invert the squaring function , as any given positive number
has two square roots. In the real case, one can eliminate this multi-valuedness by picking a branch of the square root function – a function which selects one of the multiple choices for that function at each point in the domain. In particular, we have the positive branch
of the square root function on
, as well as the negative branch
. One could also create more discontinuous branches of the square root function, for instance the function that sends
to
for
, and
to
for
.
Suppose now that we have a branch of the logarithm function, thus
On the other hand, if is a simply connected open subset of
, then from Cauchy’s theorem the function
is conservative on
. If we pick a point
in
and arbitrarily select a logarithm
of
, we can then use the fundamental theorem of calculus to find an antiderivative
of
on
with
. By definition,
is holomorphic, and from the chain rule we have for all
that
Thus, for instance, the region formed by excluding the negative real axis from the complex plane is simply connected (it is star-shaped around
), and so must admit a holomorphic branch of the complex logarithm. One such branch is the standard branch
of the complex logarithm, defined as
Exercise 43 Letbe a connected non-empty open subset of
.
- (i) If
and
are continuous branches of the complex logarithm, show that there exists a natural number
such that
for all
.
- (ii) Show that any continuous branch
of the complex logarithm is holomorphic.
- (iii) Show that there is a continuous branch
of the logarithm if and only if
and
lie in the same connected component of
. (Hint: for the “if” direction, use a continuity argument to show that the winding number of any closed curve in
around
vanishes. For the “only if”, you may use without proof the fact that if two points
in a compact Hausdorff space
do not lie in the same connected component, then there exists a clopen subset
of
that contains
but not
(sketch of proof: show that the intersection of all the clopen neighbourhoods of
is connected and also contains all the connected sets containing
). Use this fact to encircle
by a simple polygonal path in
.)
Exercise 44 Letbe a continuous injective map with
and
as
.
- (i) Show that
is not all of
. (Hint: modify the construction in Section 4 of Notes 3 that showed that a simple closed curve admitted at least one point with non-zero winding number.)
- (ii) Show that the complement
is simply connected. (Hint: modify the remaining arguments in Section 4 of Notes 3). In particular, by the preceding discussion, there is a branch of the complex logarithm that is holomorphic outside of
.
It is instructive to view the identity
, through the lens of branches of the complex logarithm such as the standard branch
One can use branches of the complex logarithm to create branches of the root functions
for natural numbers
. As with the complex exponential, the function
is not injective, and so
is multivalued (see Exercise 15 of Notes 0). One cannot form a continuous branch of this function on
for any
, as the corresponding branch of
would then contradict the quantisation of order of singularities (Exercise 13). However, on any domain
where there is a holomorphic branch
of the complex logarithm, one can define a holomorphic branch
of the
root function by the formula
The presence of branch cuts can prevent one from directly applying the residue theorem to calculate integrals involving branches of multi-valued functions. But in some cases, the presence of the branch cut can actually be exploited to compute an integral. The following exercise provides an example:
Exercise 45 Compute the improper integralby applying the residue theorem to the function
for some branch
of
with branch cut on the positive real axis, and using a “keyhole” contour that is a perturbation of
the key point is that the branch cut makes the contribution of (the perturbations) of
and
fail to cancel each other.
The construction of holomorphic branches of can be extended to other logarithms:
Exercise 46 Letbe a simply connected subset of
, and let
be a holomorphic function with no zeroes on
.
- (i) Show that there exists a holomorphic branch
of the complex logarithm
, thus
. Furthermore, this branch is unique up to the addition of an integer multiple of
(i.e., if
are two such branches, then
for some integer
).
- (ii) Show that for any natural number
, there exists a holomorphic branch
of the root function
, thus
.
Actually, one can invert other non-injective holomorphic functions than the complex exponential, provided that these functions are a covering map. We recall this topological concept:
Definition 47 (Covering map) Letbe a continuous map between two connected topological spaces
. We say that
is a covering map if, for each
, there exists an open neighbourhood
of
in
such that the preimage
is the disjoint union of open subsets
of
, such that for each
, the map
is a homeomorphism. In this situation, we call
a covering space of
.
In complex analysis, one specialises to the situation in which are Riemann surfaces (e.g. they could be open subsets of
), and
is a holomorphic map. In that case, the homeomorphisms
are in fact complex diffeomorphisms, thanks to Exercise 40.
Example 48 The exponential mapis a covering map, because for any element of
written in polar form as
, one can pick (say) the neighbourhood
of
, and observe that the preimage
of
is the disjoint union of the open sets
for
, and that the exponential map
is a diffeomorphism. A similar calculation shows that for any natural number
, the map
is a covering map from
to
. However, the map
is not a covering map from
to
, because it fails to be a local diffeomorphism at zero due to the vanishing derivative (here we use Exercise 40). One final (non-)example: the map
is not a covering map from the upper half-plane
to
, because the preimage of any small disk
around
splits into two disconnected regions, and only one of them is homeomorphic to
via the map
.
From topology we have the following lifting property:
Lemma 49 (Lifting lemma) Letbe a continuous covering map between two path-connected and locally path-connected topological spaces
. Let
be a simply connected and path connected topological space, and let
be continuous. Let
, and let
be such that
. Then there exists a unique continuous map
such that
and
, which we call a lift of
by
.
Proof: We first verify uniqueness. If we have two continuous functions with
and
, then the set
is clearly closed in
and contains
. From the covering map property we also see that
is open, and hence by connectedness we have
on all of
, giving the claim.
To verify existence of the lift, we first prove the existence of monodromy. More precisely, given any curve with
we show that there exists a unique curve
such that
and
(the reader is encouraged to draw a picture to describe this situation). Uniqueness follows from the connectedness argument used to prove uniqueness of the lift
, so we turn to existence. As in previous notes, we rely on a continuity argument. Let
be the set of all
for which there exists a curve
such that
and
, where
is the restriction of
to
. Clearly
contains
; using the covering map property (and the uniqueness of lifts) it is not difficult to show that
is also open and closed in
. Thus
is all of
, giving the claim.
Now let ,
be homotopic curves with fixed endpoints, with initial point
and some terminal point
, and let
be a homotopy. For each
, we have a curve
given by
, and by the preceding paragraph we can associate a curve
such that
and
. Another application of the continuity method shows that for all
, the map
is continuous; in particular, the map
is continuous. On the other hand,
lies in
, which is a discrete set thanks to the covering map property. We conclude that
is constant in
, and in particular that
.
Since is simply connected, any two curves
with fixed endpoints are homotopic. We can thus define a function
by declaring
for any
to be the point
, where
is any curve from
to
, and
is constructed as before. By construction we have
, and from the local path connectedness of
and the covering map property of
we can check that
is continuous. The claim follows.
We can specialise this to the complex case and obtain
Corollary 50 (Holomorphic lifting lemma) Letbe a holomorphic covering map between two connected Riemann surfaces
. Let
be a simply connected and path connected Riemann surface, and let
be holomorphic. Let
, and let
be such that
. Then there exists a unique holomorphic map
such that
and
, which we call a lift of
by
.
Proof: A Riemann surface is automatically locally path-connected, and a connected Riemann surface is automatically path connected (observe that the set of all points on the surface that can be path-connected to a reference point is open, closed, and non-empty). Applying Lemma 49, we obtain all the required claims, except that the lift
produced is only known to be continuous rather than holomorphic. But then we can locally express
as the composition of one of the local inverses of
with
. Applying Exercise 40, these local inverses are holomorphic, and so
is holomorphic also.
Remark 51 It is also possible to establish the above corollary using the monodromy theorem and analytic continuation.
Exercise 52 Establish Exercise 46 using Corollary 50.
Exercise 53 Letbe simply connected, and let
be holomorphic and avoid taking the values
. Show that there exists a holomorphic function
such that
. (This can be proven either through Corollary 50, or by using the quadratic formula to solve for
and then applying Exercise 46.)
In some cases it is also possible to obtain lifts in non-simply connected domains:
Exercise 54 Show that there exists a holomorphic functionsuch that
for all
. (Hint: use the Schwarz reflection principle, see Exercise 37 of Notes 3.)
As an illustration of what one can do with all this machinery, let us now prove the Picard theorems. We begin with the easier “little” Picard theorem.
Theorem 55 (Little Picard theorem) Letbe entire and non-constant. Then
omits at most one point of
.
The example of the exponential function , whose range omits the origin, shows that one cannot make any stronger conclusion about
.
Proof: Suppose for contradiction that we have an entire non-constant function such that
omits at least two points. After applying a linear transformation, we may assume that
avoids
and
, thus
takes values in
.
At this point, the most natural thing to do from a Riemann surface point of view would be to cover by a bounded region, so that Liouville’s theorem may be applied. This can be done easily once one has the machinery of elliptic functions; but as we do not have this machinery yet, we will instead use a more ad hoc covering of
using the exponential and trigonometric functions to achieve a passable substitute for this strategy.
We turn to the details. Since avoids
, we may apply Exercise 46 to write
for some entire
. As
avoids
,
must avoid the integers
.
Next, we apply Exercise 53 to write for some entire
. The set
must now avoid all complex numbers of the form
for natural numbers
and integers
. In particular, if
is large enough, we see that
does not contain any disk of the form
. Applying Bloch’s theorem (Exercise 42(ii)) in the contrapositive, we conclude that for any disk
in
, one has
for some absolute constant
. Sending
to infinity and using the fundamental theorem of calculus, we conclude that
is constant, hence
and
are also constant, a contradiction.
Now we prove the more difficult “great” Picard theorem.
Theorem 56 (Great Picard theorem) Letbe holomorphic on a disk
outside of a singularity at
. If this singularity is essential, then
omits at most one point of
.
Note that if one only has a pole at , e.g. if
for some natural number
, then the conclusion of the great Picard theorem fails. This result easily implies both the little Picard theorem (because if
is entire and non-polynomial, then
has an essential singularity at the origin) and the Casorati-Weierstrass theorem (Theorem 11(iii)). By repeatedly passing to smaller neighbourhoods, one in fact sees that with at most one exception, every complex number
is attained infinitely often by a function holomorphic in a punctured disk around an essential singularity.
Proof: This will be a variant of the proof of the little Picard theorem; it would again be more natural to use elliptic functions, but we will use some passable substitutes for such functions concocted in an ad hoc fashion out of exponential and trigonometric functions.
Assume for contradiction that has an essential singularity at
and avoids at least two points in
. Applying linear transformations to both the domain and range of
, we may normalise
,
, and assume that
avoids
and
, thus we have a holomorphic map
with an essential singularity at
.
The domain is not simply connected, so we work instead with the function
On the other hand, from (16) one has
We now upgrade this bound on (18) by exploiting the quantisation of pole orders (Exercise 13). As the function is periodic with period
on
, we may write
Exercise 57 (Montel’s theorem) Letbe an open subset of the complex plane. Define a holomorphic normal family on
to be a collection
of holomorphic functions
with the following property: given any sequence
in
, there exists a subsequence
which is uniformly convergent on compact sets (i.e., for every compact subset
of
, the sequence
converges uniformly on
to some limit). Similarly, define a meromorphic normal family to be a collection
of meromorphic functions
such that for any sequence
in
, there exists a subsequence
that are uniformly convergent on compact sets, using the metric on the Riemann sphere induced by the identification with the geometric sphere
. (More succinctly, normal families are those families of holomorphic or meromorphic functions that are precompact in the locally uniform topology.)
- (i) (Little Montel theorem) Suppose that
is a collection of holomorphic functions
that are uniformly bounded on compact sets (i.e., for each compact
there exists a constant
such that
for all
and
). Show that
is a holomorphic normal family. (Hint: use the higher order Cauchy integral formula to establish some equicontinuity on this family on compact sets, then use the ArzelĆ”-Ascoli theorem.
- (ii) (Great Montel theorem) Let
be three distinct elements of the Riemann sphere, and suppose that
is a family of meromorphic functions
which avoid the three points
. Show that
is a meromorphic normal family. (Hint: use some elementary transformations to reduce to the case
. Then, as in the proof of the Picard theorems, express each element
of
locally in the form
and use Bloch’s theorem to get some uniform bounds on
.)
Exercise 58 (Harnack principle) Letbe an open connected subset of
, and let
be a sequence of harmonic functions which is pointwise nondecreasing (thus
for all
and
). Show that
is either infinite everywhere on
, or is harmonic. (Hint: work locally in a disk. Write each
on this disk as the real part of a holomorphic function
, and apply Montel’s theorem followed by the Hurwitz theorem to
.) This result is known as Harnack’s principle.
Exercise 59
- (i) Show that the function
is harmonic on
but has no harmonic conjugate.
- (ii) Let
, and let
be a harmonic function obeying the bounds
for all
and some constants
. Show that there exists a real number
and a harmonic function
such that
for all
. (Hint: one can find a conjugate of
outside of some branch cut, say the negative real axis restricted to
. Adjust
by a multiple of
until the conjugate becomes continuous on this branch cut.)
Exercise 60 (Local description of holomorphic maps) Letbe a holomorphic function on an open subset
of
, let
be a point in
, and suppose that
has a zero of order
at
for some
. Show that there exists a neighbourhood
of
in
on which one has the factorisation
, where
is holomorphic with a simple zero at
(and hence a complex diffeomorphism from a sufficiently small neighbourhood of
to a neighbourhood of
). Use this to give an alternate proof of the open mapping theorem (Theorem 37).
Exercise 61 (Winding number and lifting) Let, let
be a closed curve avoiding
, and let
be an integer. Show that the following are equivalent:
- (i)
.
- (ii) There exists a complex number
and a curve
from
to
such that
for all
.
- (iii)
is homotopic up to reparameterisation as closed curves in
to the curve
that maps
to
for some
.
142 comments
Comments feed for this article
11 October, 2016 at 12:52 pm
Anonymous
Is it possible for the set of singularities to have Hausdorff dimension greater than 1 ? (it is known that dimension 1 is attainable by a natural boundary.)
11 October, 2016 at 1:39 pm
Terence Tao
Certainly; for instance the interior of the Koch snowflake is conformally equivalent to the unit disk thanks to the Riemann mapping theorem, so one can take your favourite example of a function with singularities on a significant portion on the unit circle and translate it to one with singularities on the snowflake (or many other fractals).
11 October, 2016 at 1:27 pm
Lior Silberman
In the beginning when you discuss compactifying the range of a holomorphic function to be the Riemann sphere, you say that
can be extended to a function … on the Riemann sphere, but I think you instead mean that
can be extended to a function $f\colon U\to S\cup\{\infty\}$ to the Riemann sphere.
[Corrected, thanks – T.]
11 October, 2016 at 2:52 pm
Anonymous
It seems that the concept of analytic continuation is necessary for a precise definition and classification of all types of singularities (also on the boundary of the domain
).
[Note added that we are not discussing boundary singularities in this set of notes. -T.]
11 October, 2016 at 3:17 pm
Anonymous
Since any non-polynomial entire function has (on the Riemann sphere) essential singularity at
, Picard’s theorem applies also for these functions.
13 October, 2016 at 1:49 am
Anonymous
It is easy to verify that theorem 56 is equivalent to the (apparently stronger) formulation: Let
be an essential singularity of
, then, with at most one exception, any complex number is attained by
infinitely many times in any neighborhood of
.
[Fair enough; remark to this effect added. -T.]
13 October, 2016 at 5:03 am
M. Scorsese
You say in the beginning that there is a need to consider functions that are holomorphic only on a certain subset of $U$ to motivate the following discussion of singularities of homorphic functions resp. of meromorphic functions.
But where do such functions actually/naturally come up ? I understood the study of such function always as being intrinsically motivated, as they have nice properties, and not as bring motivated by some extrinsic need to understand them deeply.
13 October, 2016 at 7:57 am
Terence Tao
Well, one is always dividing one holomorphic function by another to obtain a quotient
which is meromorphic (and thus only holomorphic away from the zeroes of
). There are also some important transcendental functions like the Gamma function
and the zeta function
that have poles. Finally, with the Riemann sphere perspective, even functions that are entire (but are not polynomials) can be viewed as having an essential singularity at infinity, which can then be studied by these techniques.
13 October, 2016 at 5:26 am
Anonymous
In the proof of theorem 55,
should avoid (all branches of)
(not
) for any integer
. This “excluded set” has real part
(for
or
(for
) – which might affect the latter application of Bloch’s theorem (as the distance of any
from the “excluded set” is not bounded – growing with
).
[Corrected, thanks – T.]
13 October, 2016 at 6:04 am
Anonymous
Correction: Since
where
is any integer, it follows that the above “excluded set” corresponds only to nonnegative(!) integers
, and the “excluded set” corresponding to the remaining negative(!) integers
is generated by (real!) translations of odd multiples of
of the previous “excluded set”. Therefore the distance of any
from the full “excluded set” is bounded (as desired for the application of Bloch’s theorem).
13 October, 2016 at 12:00 pm
Anonymous
In the proof of theorem 55, the expression “
” should be corrected to “
” (as in the proof of theorem 56).
[Corrected, thanks – T.]
14 October, 2016 at 9:43 pm
Anonymous
There is a typo in the
term.
[Corrected, thanks – T.]
14 October, 2016 at 1:32 am
Anonymous
Every rational function is meromorphic on the Riemann sphere, but it seems that the converse is also true (since any meromorphic function
on the Riemann sphere has only finitely many poles as its singularities, so it can be written as
for some polynomial
and entire function
with polynomial growth – implying that
is also polynomial and therefore
is rational).
[This is Exercise 19. -T]
14 October, 2016 at 5:04 am
Anonymous
“such a function would be holomorphic save for a singularity at {z_0}.”
[This is what is written in the text – I am not sure what point you are making here. -T.]
14 October, 2016 at 7:00 pm
Anonymous
Ah, I just don’t understand what that sentence means… Sorry for my bad English. I thought the word “save” was a typo.
14 October, 2016 at 10:25 pm
Anonymous
Let
be holomorphic on some open disk
in the complex plane. By analytic continuation it may be viewed as a restriction of its continuation to some (connected) Riemann surface
. If such
is maximal (i.e. no further analytic continuation of
to a larger Riemann surface is possible), is it uniquely determined? (i.e. is its topology uniquely determined and somehow encoded in the Taylor coefficients of
at
?)
15 October, 2016 at 11:36 pm
Terence Tao
Not as stated. Suppose for instance we have a holomorphic function
with natural boundary on the unit circle, and with
. Let
be the restriction of
to a small ball
, then of course we have a maximal extension of
to
. But
on
can also be identified with the identity function on
(if
is small enough), which has a maximal extension to the identity function on
. So the analytic continuation is not unique if we allow
to be an arbitrary Riemann surface with an embedded copy of
.
On the other hand, if we place the disk
inside a fixed Riemann surface
, then there is a unique maximal continuation of
to some Riemann surface
that projects down to
and contains a copy of
on which the projection is the identity. Basically,
consists of those paths from
to a point in
along which one has an analytic continuation, up to homotopy.
18 November, 2016 at 1:28 am
Anonymous
Is it possible to extend the above unique maximal continuation for each real analytic function in
variables defined over some ball in
to be analytic over some unique maximal
-dimensional complex manifold ?
16 October, 2016 at 5:02 am
Anonymous
Every algebraic function has only finitely many branches on the Riemann sphere. Is the converse also true?
16 October, 2016 at 8:34 am
Terence Tao
I *think* this follows from Chow’s theorem, but am not 100% sure of this (maybe one also needs the Riemann existence theorem?).
16 October, 2016 at 12:10 pm
Anonymous
A more elementary (heuristic) approach is to assume
meromorphic branches
for a function
on the Riemann sphere, and observing that
is a polynomial in
whose coefficients are symmetrical functions of
which are meromorphic outside the branch cuts and are the same (because of symmetry) at both sides of each branch cut. Therefore (using e.g. Morera’s theorem) the branch cuts are removable singularities for these coefficients of
which are meromorphic on the Riemann sphere and thereby (exercise 19) rational. Since (by its definition)
on the Riemann sphere, it follows that
is algebraic.
19 October, 2016 at 9:24 am
Terence Tao
Ah, yes, you’re right, Chow’s theorem is definitely overkill here. By the way, I found a reference for this result in Theorem 8.3 of Forster’s “Lectures on Riemann surfaces”.
16 October, 2016 at 7:04 am
Steven Gubkin
Another approach to the open mapping theorem (which may be a bit more memorable geometrically) is that one can always find local holomorphic coordinates about a point and its image so that the map is equal to z^(n+1) in those coordinates (where n is the order of the derivative at that point). Since these maps are all obviously open (even at the origin) the claim follows. Since you can prove this just with technology you have already developed, this might make a nice additional exercise for your students.
Really enjoying these notes!
[Exercise to this effect added to the end of the notes. Thanks for the suggestion -T.]
16 October, 2016 at 7:08 pm
Anonymous
Few minor typos
at the end should be removed.
should be
.
should be
instead. Actually, I am not sure where the vanishing of winding numbers is used here.
.
should be
.
.
1. In the formula preceding (2) in the proof of Lemma 1,
2. Last paragraph in the proof of Lemma 1,
3. Same sentence, “vanishing winding number in the .. exterior of
4. In the first formula of Exercise 2 there should be intersection of n sets, e.g.
5. In the inequality statement of Exercise 13, replace m with -m (may also add that m is a positive integer).
6. In Definition 17,
7. The Exercise 24, the second formula is not true without an extra residue term at infinity (properly defined), or some assumption of behavior at infinity. A simple counterexample is
[Corrected, thanks -T.]
16 October, 2016 at 8:23 pm
Anonymous
In definition 14, a formula does not parse.
[Corrected, thanks – T.]
16 October, 2016 at 10:10 pm
Anonymous
I exercise 43, it may be added that a continuous branch of the complex logarithm exists on
if and only if
does not separate
and
on the Riemann sphere (i.e.
and
should belong to some connected component of the complement of
on the Riemann sphere – to allow a connecting branch cut between them.)
[Exercise expanded, thanks – T.]
18 October, 2016 at 2:14 am
Anonymous
As an interesting application, with the assumptions of theorem 31, the function
has a holomorphic logarithm on (an open neighborhood of) the image of
if and only if the image of
does not separate
and
on the Riemann sphere – which is equivalent to the vanishing of the RHS of (13), and thereby equivalent to the vanishing of the LHS of (13).
inside
equals(!) the sum of the orders of its poles) for the existence of holomorphic logarithm of
on the image of
may be (heuristically) explained by assuming that inside(!)
the logarithm of
has branch cuts connecting each zero of
to a corresponding(!) pole of
.
This simple necessary and sufficient condition (that the sum of the orders of the zeros of
17 October, 2016 at 1:33 am
Just watched Donald Duck in Mathmagic Land – vamavamavama
[…] https://terrytao.wordpress.com/2016/10/11/math-246a-notes-4-singularities-of-holomorphic-functions/#… […]
17 October, 2016 at 9:10 am
Anonymous
Is it possible to extend exercise 2 to the case of infinitely many contours?
17 October, 2016 at 1:05 pm
Terence Tao
Yes, but only for the degenerate reason that only finitely many of the
can encircle a point outside of
(if not, one can use the Bolzano-Weierstrass theorem to locate an element of
that is adherent to the complement of $latex $U$). I don’t know if there is a non-degenerate version of this exercise that usefully allows for an infinite number of singularities inside the big closed curve.
17 October, 2016 at 12:07 pm
Anonymous
If I understand correctly, a function is defined to be holomorphic only(!) on open sets, so its singular set is automatically closed. Therefore there seems to be no need for the remark in lines 9-10 (that we always deal with closed singular sets and open sets on which the function is holomorphic).
[Paragraph reworded to emphasise this – T.]
17 October, 2016 at 5:49 pm
Anonymous
Third paragraph of section 2,
should be 
Proof of Corollary 50, first sentence is redundant (appears twice).
[Corrected, thanks – T.]
18 October, 2016 at 12:28 am
Anonymous
In theorem 31, the summands in the LHS of (13) should have running index
.
[Corrected, thanks – T.]
18 October, 2016 at 3:08 am
Anonymous
In the first line of theorem 31,
should be
.
[Corrected, thanks – T.]
18 October, 2016 at 9:31 pm
246A, Notes 5: conformal mapping | What's new
[…] Anonymous on Math 246A, Notes 4: singularit… […]
21 October, 2016 at 12:23 pm
Georges Elencwajg
Dear Terence, Exercise 9 is correct only under the assumption that
is connected : else
has non trivial zero divisors.
[Corrected, thanks – T.]
26 October, 2016 at 7:18 pm
Dan Asimov
A little exercise (or puzzle): Let f: C ā> C be holomorphic and onto with nowhere vanishing derivative. Is f necessarily invertible?
27 October, 2016 at 4:33 am
Anonymous
No! (since
is a non-vanishing entire function, it has the form
for some entire function
. Now, to get a counterexample, let
be any even(!) non-constant entire function, and assume the initial condition
. This determines
as the odd(!) non-polynomial entire function
. Since
has essential singularity at the origin, it follows from theorem 56 that
omits at most one complex number
, say. But since
is an odd function, if it omits
it must omit also
which is possible only if
i.e.
, but this contradicts the initial condition
. Hence,
is onto, and by theorem 56,
attains every(!) complex number infinitely many times!)
which (according to the above discussion) is an entire function with nowhere vanishing derivative, that attains every complex number infinitely many times.
one of simplest counter-examples seems to be the error function
27 October, 2016 at 12:56 am
Anonymous
No! (e.g. the exponential function).
27 October, 2016 at 1:02 am
Anonymous
My error! (the exponential function omits
).
30 October, 2016 at 7:54 pm
Anonymous
It seems that in equation (18), it should be |k|x instead of |k|. Also, the display equation after (19), 1/z should be 1/|z|, and the above typo of (18) would carry over, hence |k| should be |k| \log(1/|z|).
[Corrected, thanks – T.]
1 November, 2016 at 10:34 am
Anonymous
Dear professor, I can’t see why we need U to be connected in the proof of uniqueness of the decomposition in the very first lemma. By the definition given,
are defined on
,
are defined on
. I think the overlap is U, not just
.
[Fair enough – proof adjusted accordingly. -T.]
2 November, 2016 at 9:41 am
Anonymous
I am wondering if the condition that S is compact is necessary in Exercise 3. It seems we want to prove integral of f along a simple closed polygonal contour $\latex \gamma$is zero, $\latex int \gamma \cap \gamma \cap S$ is always compact, whether S is or not.
[Yes, one can extend Painleve’s theorem to non-compact closed singular sets of zero length also. -T.]
3 November, 2016 at 12:12 am
Anonymous
Typo, Exercise 20: “polynomial of one complex polynomial”
3 November, 2016 at 2:34 am
Anonymous
In the beginning of line 7.
3 November, 2016 at 11:23 am
Anonymous
In the third line there is a latex typo.
[Corrected, thanks – T.]
3 November, 2016 at 3:10 pm
Anonymous
The typo is still there.
[Hopefully fixed now – T.]
10 November, 2016 at 7:42 am
Anonymous
In the last part of lemma 1, it seems the path
does not have winding number zero – the third term should be
instead. Cheers, L.
[Corrected, thanks – T.]
25 November, 2016 at 3:30 pm
JD
In Exercise 25 (iii), shouldn’t you divide by
instead of
? (in order to cancel for the
that appears from the $m-1$th derivative)
[Corrected, thanks – T.]
6 March, 2017 at 12:17 pm
Anonymous
Small typo in Exercise 6 (the very last set). [Corrected, thanks – T.]
āāāāāāā
1. In the discussion right above Example 7, do we have any assumption regarding the āsingular setā S? If S is a branch cut for some multivalued function, then z_0\in S would be in none of the three categories? [The discussion above Example 7 applies only to isolated singularities; elements of branch cuts are not isolated. -T.]
2. In Definition 8, what would go wrong if one drops the condition (i) in the definition? Is there an example such that (ii) is satisfied but not meromorphic? [Property (i) is mainly for emphasis, as the isolated nature of the elements of
is already implicit in (ii). -T.]
10 March, 2017 at 6:51 am
Anonymous
“For instance, if
is a map from an open subset
of
to the Riemann sphere
, then
is holomorphic if and only if…”
Typo in “to the Riemann sphere
“?
[Typo corrected. -T]
11 March, 2017 at 8:18 am
Anonymous
I saw in many textbooks (e.g. in Gameline) the following calculation using (10) is made. For a closed curve
I think the first equality can be made rigorous by using similar argument in the discussion of (15) in this post. But by the Cauchy-Riemann equations, the map
would not necessarily be holomorphic anywhere (unless
is a constant?). It would not even make sense to talk about
.
Is there any way to make this rigorous?
11 March, 2017 at 11:34 am
Anonymous
One can also see in Ahlfors that

15 March, 2017 at 9:34 pm
Terence Tao
One can interpret integrals such as
using integration of differential forms (or line integrals) rather than contour integrals.
20 March, 2017 at 4:09 pm
Anonymous
In the proof of Lemma 1, “… are hence restrictions of a single entire function
“: Why does such entire function
exists?
[If two functions
and
agree on their common domain
, then they are restrictions of a single function
. As holomorphicity is a local condition, if one glues together two holomorphic functions in this fashion, the resulting function is also holomorphic. -T.]
20 March, 2017 at 4:28 pm
Anonymous
In the uniqueness part of the proof of Lemma 1, why does one need the analytic continuation in the last sentence? Once one shows that
by Liouville’s theorem, doesn’t one have
on
and similarly
on
?
[Fair enough; I have shortened this part of the argument accordingly. -T.]
20 March, 2017 at 4:31 pm
Anonymous
In the proof of Lemma 1, might be better to use
instead of
, which might be confused with the derivatives of
.
[Notation changed – T.]
20 March, 2017 at 5:05 pm
Anonymous
(Sorry for the separate comments.)
in the definition of
and
? How is Exercise 36 of Notes 3 relevant.
In the proof of Lemma 1 again, why do we need to put the extra term
[Without the terms of
, the two definitions of
would not agree on their common domain, and similarly for
. If it were not for Exercise 36, how would one show that
and
were holomorphic in
? -T]
In the statement of Lemma 1, if one replaces
with
as the domain of
and uses
as the domain of
, would the theorem be changed fundamentally?
[I think the existence component of this theorem might more inconvenient to use in some applications, as one may wish all the functions involved to be holomorphic on the original domain
. -T.]
21 March, 2017 at 12:23 pm
Anonymous
In the proof of Lemma 1:
1. How is the factor theorem needed for proving the identity above (2)? Isn’t it true that
is holomorphic on
?
[The function
is not defined at
(but the factor theorem ensures that the singularity here is removable). -T]
2. Why does (2) not directly follow from Theorem 4 (Cauchy’s theorem) in Notes 3?
[How does one know that
and
are homotopic in
? This is a true statement, but to prove it one basically needs the same sort of methods used in the proof provided for (2) (insertion of an auxiliary path
and analysis of the winding number). -T.]
3. What estimate does one need to justify rigorously the passage to limit in the very last sentence?
[Cauchy’s theorem ensures that the integral does not depend on
once
is small enough. -T]
22 March, 2017 at 7:21 am
Anonymous
Thank you for the comments!
For 1,
, and
is either on
or
in establishing (2), so
… ?
For 3, I understand that
How should one show that
31 March, 2017 at 9:23 am
jura05
Dear Professor Tao!
In the definition of meromorphic function, I think we do not need that S is closed. If S is discrete, then
is guaranteed to be open.
For example, if
, and $S=\{1/n\colon n=1,2,\ldots\}$ is discrete, but not closed. And there are meromophic functions on U with singular set S.
31 March, 2017 at 9:27 am
jura05
Sorry, it is not guaraneed, of course. I guess S should be “closed in U”, not just “closed”.
5 June, 2017 at 10:04 am
Five-Value Theorem of Nevanlinna – Elmar Klausmeier's Weblog
[…] Math 246A, Notes 4: singularities of holomorphic functions […]
7 June, 2017 at 3:20 pm
David Hruska
Hi, I’m a little bit stuck at the Exercise 28 devoted to the spectral theorem for matrices. In particular I don’t see how to prove that the contour integral from the resolvent equals identity matrix. Of course there is a standard way of proving that the holomorphic calculus “preserves” polynomials in general, from which this follows, but the structure of the execrise suggests that another (hopefully more elementary) approach can be used for matrices. Can somebody help? Thanks in advance.
[Move the contour to infinity – T.]
22 September, 2017 at 12:31 pm
Anonymous
In Stein-Shakarchi,
If
is a nowhere vanishing holomorphic function in a simply connected region
, then there exists a holomorphic function
on
such that
. The function
in the theorem can be denoted by
, and determines a “branch” of that logarithm.
Is this alluded somewhere in this note? How should one understand “
determines a “branch” of that logarithm”?
[See Exercise 46. -T]
26 October, 2017 at 4:07 pm
Anonymous
It is said in Stein-Shakarchi that
“There is a general principle in the theory, already implicit in Riemann’s work, which states that analytic functions are in an essential way characterized by their singularities. That is to say, globally analytic functions are “effectively” determined by their zeros, and meromorphic functions by their zeros and poles.”
I don’t quite understand this. Can one find any theorems in this note that can serve as instances of the mentioned principle?
26 October, 2017 at 7:25 pm
Terence Tao
Partial fractions (Exercise 20) is an instance of this in the category of rational functions. The Residue theorem (Section 2) is an instance in the context of contour integration. But probably Stein and Shakarchi are referring most directly to factorisation theorems such as the Weierstrass factorisation theorem, which was not covered in this course (typically at UCLA we cover it in the sequel to this course, 246B).
3 November, 2017 at 5:14 am
Anonymous
Can meromorphic functions have an essential singularity at infinity?
7 November, 2017 at 4:00 pm
Terence Tao
If
is not part of the domain, then sure. For instance the exponential function is meromorphic on the complex plane but has an essential singularity at infinity (and as such, will not be meromorphic on the Riemann sphere).
8 November, 2017 at 2:30 pm
Anonymous
Stein-Shakarchi shows the following chain of implications:
Cauchy integral formula, -> residue theorem, -> argument principle, -> Rouche theorem, -> open mapping theorem, -> maximum principle
Can this chain be closed as a circle so that everything mentioned is actually equivalent to the Cauchy integral formula?
27 April, 2018 at 2:00 am
Anonymous
In the definition of holomorphic function on open subset of CU{ā} (before the exercise 18) , it seems that the set described in (ii) should be {zāC:1/zāU} ļ¼the open cover is U ā© C and U ā© C\{0}U{ā}ļ¼so 1/z ā U ā© C\{0}U{ā}ļ¼for 1/z ā C\{0}U{ā}ļ¼it should be zāC. And then (ii) can be full definition based on the definition of holomorphic function on open subset of C before.
[Corrected, thanks – T.]
2 May, 2018 at 6:06 am
Anonymous
Pro Taoļ¼I find that the definition of meromorphic in your Notes is a little different from it in Wiki and some other books. And there isn’t a inclusion relation between them. Because your definition requires that S is closed which isn’t needed in Wiki’s definition(poles can converge to a point doesn’t belong to U, so in Wiki’s definition closed is not confirm), while Wiki’s definition require S can’t include removable singularities which yours don’t require that. So, which one is better? In subsequent study, which definition should we choose?
2 May, 2018 at 11:32 am
Terence Tao
The two definitions are equivalent after removing any singularities that are removable. Note that the set of singularities is necessarily closed, because one can only define the notion of being holomorphic at a point if there is a neighbourhood of that point that is free of singularities.
If one calls the two notions of meromorphic functions “meromorphic functions with removable singularities”, and “meromorphic functions with all removable singularities removed”, the former is analogous to non-reduced fractions such as
, and the latter to reduced fractions such as
. The advantage of the latter is that there is a canonical description: every fraction has a unique reduced form, and every meromorphic function has a unique representation after removing all singularities. In the former, one does not have unique forms any more, but it can be faster to manipulate the fractions and meromorphic functions because one does not have to keep checking that the numerators and denominators remain coprime, or that all singularities are genuine poles and not merely removable.
2 May, 2018 at 7:16 pm
Anonymous
Thanks for your exhaustive reply. But I still have a small problem. We know U is open in C, U\S is open in C and SāU, with these condition I can just deduct that S is closed in U rather than in C. And the example is that a series of singularities converge to a point that is not in U, so S is not closed in C but U\S can still be open.
2 May, 2018 at 7:42 pm
Terence Tao
This was a typo, now corrected:
is required to be closed in
, but need not be closed in
.
3 May, 2018 at 1:13 am
Anonymous
Thanks for your patient reply. So in theorem 21(Residue theorem), if S is closed in U rather than in C, it seems that we can’t ensure the singularities in the large ball which contains γ are finite. Because without closed in C, we can’t ensure S restricted to the large ball is compact. Then the function g can’t be defined successfully without the assumption S is finite.
if we restrict S to the image of homotopic function γ^~, then S will be finite, g can be defined with z0 which belongs to the S, but if we do this and resume the left step in the proofļ¼finally we just can derive the equality where S in the right-hand side is a restriction to image of homotopic function γ^~. we can’t ensure the point not in the image of homotopic function has 0 winding number, can we expand the restriction of S to the whole S?
3 May, 2018 at 1:02 pm
Terence Tao
Fair enough. I’ve changed this part of the argument to resolve this.
5 May, 2018 at 6:36 am
Anonymous
I’ve seen thatļ¼but I have some new question about your reviseļ¼just about how we define “sufficiently fine grid”, “outer boundary” and “perturbing”. For “sufficiently fine grid”, my interpretation is that decompose a sufficiently big grid including the whole curve γ uniformly and repeatedly until every small grid which contains part of γ is in U. Then the union of the finite grid will contains all the γ. I don’t know whether I am right about “sufficiently fine grid”. And How to define “outer boundary” to keep it closed , polygonal and γ contained in? is the “outer boundary” self-intersected? if yes, how do we “perturbing”?
5 May, 2018 at 7:39 pm
Terence Tao
The complement of the union of the collection
of squares that touch
partition the remainder of the complex plane into connected components, one of which is unbounded. The boundary of that component is the outer boundary; it is the union of a finite number of edges of the squares. It can be traversed as a piecewise linear closed contour (right-hand rule), but it may touch itself due to the possibility of two of the squares in
touching each other at a vertex without sharing a common edge, and with neither of the two other squares at that vertex being in
. But then one can perturb the contour away from this vertex, for instance by replacing the both corners where the contour passes through that vertex with a diagonal “shortcut”. (This should be visually obvious once one draws a picture.)
6 May, 2018 at 4:06 am
Anonymous
thank you very muchļ¼
8 May, 2018 at 1:11 am
Anonymous
In the definition 14 and 17ļ¼if a function f is said to be smooth or holomorphic, then should every f^{-1}(U’_β)ā©U_α be open? if f can’t ensure every f^{-1}(U’_β)ā©U_α to be open , then f can’t be said to be smooth or holomorphic ?
8 May, 2018 at 7:33 am
Terence Tao
As
is a homeomorphism, the sets
and hence
will be open.
9 May, 2018 at 5:14 am
Anonymous
In the definition 14 and 17, when you define a smooth or holomorphic function f:M->M’ from one topological space M to another M’, you don’t require f should be homemorphism or even continuous on M. I try proving that a smooth or holomorphic function as you defined in definition 14 and 17 is continuous on M, and find this proof requires every f^{-1}(Uā_β)ā©U_α to be open. So I have that trivial problem.(i.e. if f canāt ensure every f^{-1}(Uā_β)ā©U_α to be open , then f canāt be said to be smooth or holomorphic ?) , if f can be holomorphic or smooth without ensuring every f^{-1}(Uā_β)ā©U_α to be open, then f can be holomorphic or smooth but not continuous on M.
10 May, 2018 at 4:49 pm
Terence Tao
Thanks for pointing this out. I’ve added the hypothesis of continuity to both definitions.
11 May, 2018 at 6:34 am
Weibo Shu
Thanks for your reply, It’s my pleasure to do this.
15 May, 2018 at 6:02 am
Anonymous
In chapter 4 of Wolff’s Harmonic analysis notes, in calculating the Fourier transform of
where
, one ends up with showing that the map
is holomorphic. How can one estimate the integral
where
so that one can take complex differentiation under the integral sign? Does one have
15 May, 2018 at 6:41 pm
Terence Tao
Generally speaking, to show that an integral expression is holomorphic, it is easiest to proceed using Morera’s theorem, rather than differentiation under the integral sign, as it is easier to interchange integrals than to interchange a derivative and an integral.
16 May, 2018 at 5:52 am
Anonymous
Thanks for the comment. I think I see your point of using Moreara’s theorem: one can use Fubini’s theorem and the fact that the map
is analytic when
.
Is it still possible to exploit the differentiation approach? In this case,
Since
, one can expect a bound (on
) of the map
But this seems to be rather useless for bounding the difference quotient
since mean value theorem is not necessarily true for complex-valued functions.
18 May, 2018 at 2:24 pm
Anonymous
In Wolff’s notes (chapter 4), he pointed out that the analyticity of
“may be done by using the dominated convergence theorem to justify complex differentiation under the integral sign.” Hardly can I guess how he did it.
This seems to be a bit different from the “compilation errors” mentioned in the advice (https://plus.google.com/u/0/+TerenceTao27/posts/TGjjJPUdJjk)
since it is very unlikely (in the notes) that the author would come back to discuss this little step.
One can take this for granted (or give an alternative proof) though, it is quite “frustrating” to read ahead without really knowing what the author intended to say in such a small step. And it seems that this is a type of “runtime exception” that is not mentioned in the advice (https://plus.google.com/u/0/+TerenceTao27/posts/TGjjJPUdJjk).
5 November, 2020 at 2:29 pm
Anonymous
While “contour integrals” is understood as complex integrals for continuous curves in notes 2, the term “contours” refers to the piecewise smooth curves. Does “contour” simply mean “continuous curves” in Lemma 1 (and various other places in this set of notes)?
[Yes; wording has now been changed to reflect this – T.]
6 November, 2020 at 12:53 pm
Anonymous
Yes, it doesnāt need to be rectifiable.
8 November, 2020 at 4:26 am
Image crop error
I think that the a portion (towards the right) of first image in the section on Laurent series gets cropped out. Also pulling the image using cursor works.
[Now corrected – T.]
21 November, 2020 at 10:12 am
Anonymous
Is the topology on
the same as the subspace topology of the unit sphere in
? Is there any advantage of using the definition of one point compactification in this note?
21 November, 2020 at 1:34 pm
Terence Tao
Yes, once one sets up the appropriate stereographic projection, one can show that the Riemann sphere
is homeomorphic to a round sphere in
with the subspace topology. This is certainly one very useful model for the Riemann sphere, as it allows one to use one’s existing geometric intuition about round spheres to understand the Riemann sphere. However, it makes the coordinate charts on the sphere moderately complicated (one has to invert the stereographic projection). The model
of the Riemann sphere based on the one-point compactification of
is more convenient for many algebraic calculations due to the simplicity of the coordinate charts (they are just given by
and
).
22 November, 2020 at 8:23 am
adityaguharoy
Last day prof. Tao posed the problem of finding an open map that is not continuous.
We notice that if we can construct a map
such that
is not continuous, but for any non-empty open interval
(with
) we have
then we are done (since any open set in
can be written as a union of open intervals).
One way to construct such functions is to play around with the expressions of reals in base 10 or other bases.
I wonder if we can find an explicit expression for such a function which is open but not continuous.
23 November, 2020 at 8:59 am
Anonymous
In Definition 17,
when defining the holomorphicity of a map between manifolds:
… the maps
are holomorphic
Is there a typo there? I think the map should be
[Corrected, thanks – T.]
23 November, 2020 at 9:44 am
Anonymous
and
should be
?
[Corrected, thanks – T.]
23 November, 2020 at 10:48 am
Anonymous
In Exercise 45, how does one interpret
? Is it
,
, or something else?
23 November, 2020 at 1:48 pm
Ben Johnsrude
It should be understood as a joint limit: saying that
is defined as saying
such that
we have
. – Ben
24 November, 2020 at 2:35 pm
Anonymous
Can the discrete set
in Exercise 18 be infinite? (Exercise 19 seems to suggest that it should be finite.)
24 November, 2020 at 3:36 pm
Ben Johnsrude
In the setting of Exercise 19,
will always be finite, because the (sequential) compactness of
implies that all infinite subsets have an accumulation point. – Ben
24 November, 2020 at 4:14 pm
Anonymous
In the “Laurent expansion” in the definition of residue of
at
, there are several interpretations of the double-sided sum
. Does one have the absolute convergence there so that all the interpretations are equivalent or the double-sided sum refers to a particular limit?
[Yes, one has absolute convergence of the doubly infinite series near the singularity; see the comments after (4). -T.]
25 November, 2020 at 9:05 am
Ben Johnsrude
The series does converge absolutely on the punctured disk
. – Ben
25 November, 2020 at 12:10 pm
Anonymous
The absolute convergence region for the series is (generally) for an annular region whose inner radius may be positive (not necessarily zero as for the particular case of the punctured disk example.)
25 November, 2020 at 6:28 am
Anonymous
Can one define the contour integral of functions on the Riemann sphere so that the interior and exterior residue theorems are unified as one theorem?
25 November, 2020 at 8:56 am
Ben Johnsrude
For integrating over Riemann surfaces, one has to consider instead the integral of the one-form
instead of the integral of the function
. One can reformulate the notion of residue to work for holomorphic one-forms (whereby our definition of residue is understood as the residue of
, obtained by multiplying
by the canonical choice of one-form in
,
). Reformulating the discussion as such, the interior and exterior theorems are equivalent. – Ben
25 November, 2020 at 11:10 am
chaikens
Typo in the uniqueness proof of the lifting lemma:
is clearly closed in
(not $latex N)?
25 November, 2020 at 3:10 pm
Anonymous
In Corollary 33, is the statement still true if
is only separately continuous in
and
?
26 November, 2020 at 12:00 pm
Terence Tao
I think it is false. A near-example is provided by
,
where
is a curve in
that approaches a boundary point non-tangentially, so that the pointwise limit is
, and the zero at
inside the disk disappears in the limit. These functions have poles just outside the unit disk and so this isn't directly a counterexample, but I believe if one replaces each
with a Taylor series approximation (where the order of approximation goes to infinity as
) one should be able to build a suitable counterexample.
27 November, 2020 at 9:00 am
Anonymous
In the Proof of Lemma 49, when proving the monodromy is path-independent, should
in particular, the map
be “…
“?
In the definition of the lifted map:
for which there exists a curve
is the condition
missing?
[Corrected, thanks – T.]
30 November, 2020 at 11:12 am
Anonymous
This may just be a pedantic comment.
The symbol
is used for the multivalued complex logarithm while
is used for the real logarithm. But throughout the proof of Theorem 56,
is used when it really means
.
(Maybe it is just a (confusing) “convention” that when the input
is a positive real number, the expression “
” is considered the same as
.)
[Corrected, thanks – T.]
1 December, 2020 at 6:56 pm
Anonymous
In the proof of Great Picard,
… the function
for some function
,
I see why periodicity is necessary for the factorization in (19), why is it sufficient?
Also, why is
the domain of
?
[The domain has been corrected to
. One can use (19) as a definition of
, namely
whenever
, or equivalently
; the periodicity of the RHS counteracts the multiplicity of the logarithm. Using branches of the logarithm one can verify that
is holomorphic. -T]
1 December, 2020 at 6:58 pm
Anonymous
And should “on the right half-plane” be
?
[Corrected, thanks, -T.]
1 December, 2020 at 7:36 pm
Anonymous
In the last sentence of the proof of Great Picard,
this implies that
[Here I am viewing a pole of negative order as a zero of the opposite order – T.]
5 December, 2020 at 8:25 am
Iota
Prof. Tao
In reference to the sentence “In this set of notes we shall only consider interior singularities…….”, can you mention some places where other types of singularities are studied or appear.
[See e.g., https://mathworld.wolfram.com/NaturalBoundary.html -T]
7 December, 2020 at 6:38 am
Anonymous
In the proof of Lemma 1,
is holomorphic.
… From Exercise 36 of Notes 3 we see that
In the case when
(similarly for
) is smooth (or piecewise smooth), one can write the integral in the definition of
on
as (by Ex16(iv) of Notes 2)
so that one can then apply Exercise 36 of Notes 3.
In the case when
is only rectifiable, can one still use Exercise 36 of Notes 3?
If one wants to directly use the “Cauchy+Fubini+Morera” strategy in the proof of Exercise 36 of Notes 3, then one needs a version of “Fubini” that allows something like
In 245A Notes 6, it takes quite an amount of time to establish the Fubini theorem. Does one have a convenient way to convert the integral above in the setting of Lebesgue integrals so that one can indeed use the Fubini theorem?
7 December, 2020 at 11:27 am
Terence Tao
The simplest thing to do here is to first use Cauchy’s theorem to temporarily deform
to a polygonal path so that Exercise 36 can be applied; I’ve adjusted the notes accordingly.
7 December, 2020 at 4:48 pm
Anonymous
In Exercise 13:
… Hint: use Lemma 4 and a limiting argument to evaluate the Laurent coefficients
There is no Lemma 4 in the notes. It probably refers to something else?
[Corrected to Exercise 4 – T.]
8 December, 2020 at 6:48 am
Anonymous
In Definition 14,
are said to be equivalent if the identity map from
(equipped with one of the two atlases) to
(equipped with the other atlas) is a diffeomorphism;
Two smooth atlases on
Is it sufficient to define the notion of equivalent atlases by requiring only that the identity above is smooth (instead of a diffeomorphism)?
(In other words, is it sufficient to say that
is smooth for all
?)
8 December, 2020 at 9:50 am
Terence Tao
No. For instance, consider the line
with two different smooth structures: one coming from the standard coordinate chart
given by
and the other coming from an exotic coordinate chart
given by
. Then the identity map from
with the first structure to
with the second is smooth, but the inverse is not smooth, so the two smooth structures are not equivalent.
8 December, 2020 at 12:17 pm
Anonymous
I’m a bit confused with the notions of “atlases” and “smooth structures” in Definition 14. There seem to be different levels of equivalences.
Given a topological space
, one can have different “atlases”, say,
and
.
If the *identity map*
and
are equivalent “atlases”.
is a diffeomorphism, then we say that
On the other hand, if one puts all the atlases that are equivalent to
together, and denote the equivalent class as
, then
is one smooth “structure” on
.
If we have two smooth “structures” on
, we can then ask if the two structures are “equivalent”.
If there exists a diffeomorphism
, then the two smooth structures,
and
are equivalent.
So it is possible that
and
are not equivalent “atlases”, but the two smooth structures,
and
can still be equivalent. Correct?
It seems that in Definition 14 (and similarly Definition 17), only “equivalent atlases” are defined, but not “equivalent smooth structures” (at least not explicitly). The former uses the identity map, while the latter needs only the existence of a diffeomorphism.
8 December, 2020 at 7:19 pm
Terence Tao
There is no notion of “equivalent smooth structures”. There are “equal smooth structures” and “diffeomorphic smooth structures”.
Two smooth manifolds
,
are equal as smooth manifolds iff the underlying base manifold
is the same and two representative atlases
are equivalent.
Two smooth manifolds
,
are diffeomorphic if there is an invertible map
such that
are smooth maps between the smooth manifolds (which is equivalent to
,
being smooth maps for two representative atlases
.
In the example I provided in a different comment,
and
are not the same smooth manifold, but they are diffeomorphic (using a non-identity map such as
or
to provide the diffeomorphism). On the other hand it does not make sense to ask whether the two smooth manifolds are “equivalent”.
9 December, 2020 at 5:31 am
Anonymous
Thanks for the clarification.
There is no notion of āequivalent smooth structuresā.
Well, you said in the previous comment that “… so the two smooth structures are not equivalent”
(now I see you may just mean to say that they are not the same smooth manifold.)
Moreover, it seems that people do use the notion of āequivalent smooth structuresā sometime: https://en.wikipedia.org/wiki/Smooth_structure#Equivalence_of_smooth_structures
The Wikipedia page phrases the John Milnor result as “the 7-dimensional sphere admits a smooth structure that is not equivalent to the standard smooth structure.”
8 December, 2020 at 12:21 pm
Anonymous
I just submitted a long comment. But I don’t know why it disappears.
8 December, 2020 at 12:31 pm
Anonymous
So in your example,
and
are not equivalent atlases since the identity map
is not a diffeomorphism (the inverse is not smooth).
But
and
are two equivalent “smooth structures” since
is a diffeomorphism (between manifolds). Correct?
8 December, 2020 at 9:51 am
Anonymous
In the definition of the topology on the Riemann sphere, what is the motivation of using
(with
compact in
) as open sets?
8 December, 2020 at 9:55 am
adityaguharoy
I think that helps us to avoid any ambiguity arising out of bounded-unboundedness.
8 December, 2020 at 7:11 pm
Terence Tao
This is the smallest possible enlargement of the space
that preserves the topology of the subspace
and which turns the space into a compact Hausdorff space; it is a special case of the one-point compactification (or Alexandroff compactification).
9 December, 2020 at 12:28 am
adityaguharoy
Ah yes ! This subtle point was touched upon during the lectures. How bad that I forgot it !
9 December, 2020 at 12:18 pm
Anonymous
The open interval
has a compactification
, which gives another compactification for
than
. Similarly one can consider the closed unit disk for
as another compactification, or the closed upper-half complex plane. Are other compactifications less useful than the Riemann sphere when studying meromorphic functions?
9 December, 2020 at 12:43 pm
Terence Tao
These are topological compactifications of the plane, but they are not Riemann surface compactifications (note that the disk and half-plane are merely homeomorphic to the plane, rather than being complex diffeomorphic to the plane). The Riemann sphere is the only compact Riemann surface that contains a copy of the complex plane as an open dense subset, though I don’t know of an elementary way to prove this (not using, for instance Riemann-Roch).
8 December, 2020 at 11:38 am
Anonymous
In Def 8, the singular set
is required to be closed in
while the assumptions in Exercise 18 do not explicitly state so. (So one implicitly assumes that
is closed in
if one says
is meromorphic?)
[Yes, holomorphic and meromorphic functions are only defined on open domains in this series of notes. -T]
12 December, 2020 at 9:05 am
Anonymous
In the proof of the Residue theorem (Theorem 21),
The absolute convergence of the series on the right-hand side in
…
Also, for any
Should both
above be
?
[The first one yes; the second one no (note that when
that all the negative coefficients in the Laurent series vanish) -T.]
12 December, 2020 at 12:10 pm
Anonymous
What “compactness argument” is the proof of Theorem 21 referring to?
In the proof of Corollary 13 (local Cauchy theorem), it is also said that “From compactness, there must exist a radius
such that
for all
and
.”
(The image of the homotopy is compact, so it can always be covered by finitely many balls and one can then take the smallest radius
and cover the bigger balls with finitely many balls of radius
as well??)
How does one get the uniform radius
and why one needs it?
12 December, 2020 at 5:17 pm
Terence Tao
For each
, one can find a radius
such that
. Now the disks
form an open cover of
, so one can find a finite subcover. Let
be the smallest radius in this subcover, then one can check from the triangle inequality that
for all
. The uniformity in
is helpful in the proof of Corollary 13 of Notes 3 in order to ensure that all the quadrilateral curves
are contained in a disk inside of
; if one only had the non-uniform disks
it is not clear that one could ensure that every such
was inside one of the disks. The situation is similar in Theorem 21.
16 December, 2020 at 2:10 am
Anonymous
Hello Sir,
In the Ex 26 (proof of fundamental thm of algebra using residue theorem) I have some problem.
In order to use residue theorem at first I assume that p(z) has no zero in large disk of radius R. Then by residue thm integral of z^(n-1)/p(z) is zero. But I want to show that integral of that function on large disk is non-zero then by contradiction we have our result. Now in large disk p(z) can be dominated by some leading term z^n but using this we can show only that modulus of given function is non zero but not for the given function.
There is any other way to recover this problem ?
Thanking you.
[Parameterise a large circle by
for
, express the integral as an integral over
, and apply the dominated convergence theorem. -T]
23 December, 2020 at 5:54 pm
246B, Notes 1: Zeroes, poles, and factorisation of meromorphic functions | What's new
[…] again for avoiding the zeroes of both and . Thus we see that the zeroes and poles of a rational function describe the behaviour of that rational function, as well as close relatives of that function such as the log-magnitude and log-derivative . We have already seen these sorts of formulae arise in our treatment of the argument principle in 246A Notes 4. […]
27 December, 2020 at 7:01 am
Anonymous
The last paragraph of establishing (1) in the proof of Lemma 1 is sketchy and informal; it is not difficult to draw a “picture” according to the informal narrative, but a “picture”, while illustrating ideas, is not quite a substitute of a proof. You once mentioned that in practice, one uses the less formal
(but *still* formalisible) modes of mathematical reasoning when doing analysis.
But unlike the “hard analysis” that uses explicit
estimates in lots of other proofs of this set of notes, or “soft analysis” (like “by compactness …”) that conceals the epsilon-delta, I find it very difficult to formalize the mentioned sketchy proof, even difficult to see a path that it is indeed formalisable. When I do attempt to write a more formal proof (at the level of loose epsilon-delta arguments) based on your sketch, I find it extremely tedious and much less illuminating than a picture; for instance, it is very difficult to formalize and convince oneself it is worth to formalize the notions such as “perturbing… the curves”, “removing loops, self-intersections, or excursions…”. This seems to be in sharp contrast to other hard/soft analysis arguments.
Is there a way to make the “geometry/topology” part of the proof (or at least make it more look like formalisable) formal and yet without going to the “formidable” world of homology and not being tedious/dull? Or if one wants a cleaner and more formal/formalisable proof, one has to use the machinery of algebraic topology?
27 December, 2020 at 3:37 pm
Terence Tao
The formalism of algebraic topology is the best solution we currently have in the long-term, but for topological arguments involving bounded complexity objects such as polygonal paths one can also use the language of algorithms. All the operations discussed in Lemma 1 can be coded as a computer algorithm in some idealised computer language, and the main mathematical tasks are then to show that the algorithms terminate in finite time and produce objects with the stated properties, which can often be done with a suitable induction argument (for instance, some sort of induction on the “complexity” of a path as it is being simplified by removal of loops etc.). Of course, coding can also be tedious and/or dull if approached inefficiently, but this is again the gap between arguments that are formalisable in principle, and arguments that are fully formalised. And indeed topological arguments have proven somewhat challenging to formalise properly, though not impossibly so. See for instance this article by Hales on the (ultimately successful) months-long project to formalise the proof of the Jordan curve theorem: https://www.maths.ed.ac.uk/~v1ranick/papers/hales3.pdf
8 January, 2021 at 4:35 am
N is a number
Can the fact that any meromorphic function
which is meromorphic on
can be expressed as a ratio of two holomorphic functions on
be proved by chasing the Laurent expansion somehow? Or does it require some all new tool to prove this?
[The standard proof uses a version of the Weierstrass factorisation theorem. See Exercise 19 of 246B Notes 1. -T]