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

in the interior of this circle. In Corollary 18 of Notes 3, this was used to form a convergent Taylor series expansion

in the interior of this circle, where the coefficients could be reconstructed from the values of on the circle by the formula

Now suppose that is only known to be holomorphic outside of . Then the Cauchy integral formula no longer directly applies, because the interiors of contours such as are no longer contained in the region where is holomorphic. To deal with this issue, we use the following convenient decomposition.

Lemma 1 (Cauchy integral formula decomposition in annular regions)Let be a holomorphic function. Let , be simple closed anticlockwise contours 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 decompositionon , 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

*Proof:* We begin with uniqueness. Suppose we have two decompositions

on , where and holomorphic, and both going to zero at infinity. Then the holomorphic functions and agree on the common domain and are hence restrictions of a single entire function . But goes to zero at infinity and is hence bounded; applying Liouville’s theorem (Theorem 28 of Notes 3) we see that vanishes entirely. This gives on and on .

Now for existence. Suppose that we can establish the identity (1) for in . Then we can define on by

and on by

noting from (1) that this consistently defines on

From Exercise 36 of Notes 3 we see that is holomorphic. Similarly if we define on by

and on by

One can then verify that obey all the required properties.

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

and

and so to prove (1), it suffices to show that

By the factor theorem (Corollary 22 of Notes 3) it thus suffices to show 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

taking limits as we obtain (2) as claimed.

Exercise 2Let be a simple closed anticlockwise contour, and let be simple closed anticlockwise contours in the interior of whose images are disjoint, and such that the interiors are also disjoint. Let be an open set containing and the regionLet be holomorphic. Show that for any , one has

(

Hint:induct on using Lemma 1.)

Exercise 3 (PainlevĆ©’s theorem on removable singularities)Let be 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 open set that contains an annulus of the form

for some and . From Lemma 1, we can split , where is holomorphic in , and is holomorphic in the exterior region , with going to zero as . From Corollary 18 of Notes 3, one has a Taylor expansion

for some coefficients that is absolutely convergent in the disk . One cannot directly apply this Taylor expansion to . However, observe that the function is holomorphic in the punctured disk , and goes to zero as one approaches zero. By Riemann’s theorem (Exercise 35 from Notes 3), this function may be extended to to a holomorphic function that vanishes at the origin. Applying Corollary 18 of Notes 3 again, we conclude that there is a Taylor expansion

for some coefficients that is absolutely convergent in the punctured disk . Changing variables, we conclude that

for all in (3), with the doubly infinite series on the right-hand side being absolutely convergent. This series is known as the Laurent series in the annulus (3). The coefficients may be explicitly computed in terms of :

Exercise 4 (Fourier inversion formula)Let be 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 formulafor all integers , whenever is a simple closed curve in the annulus with . Also establish the bounds

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:

On the other hand, in the exterior region , the Taylor expansion is no longer convergent. Instead, if one writes and uses the geometric series formula, one instead has the Laurent expansion

in this region.

Exercise 6Find the Laurent expansions for the function in the regions , . (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 7The function has a Laurent expansionand 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)Let be an open subset of . A function defined on outside of a singular set is said to be meromorphic on if

- (i) is closed and discrete (i.e., all points in are isolated); and
- (ii) 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)Let denote 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)If is a meromorphic function, and , define theorderof at as follows:

- (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 .
Establish the following facts:

- (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.
In the language of abstract algebra, the above facts are asserting that is a valuation on the field .

The behaviour of a holomorphic function near an isolated singularity depends on the type of singularity.

Theorem 11Let be 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 12Let be 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 13Let be holomorphic on a disk except for a singularity at . Let be an integer, and suppose that there exist , such that one has the upper boundfor 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 Lemma 4 and a limiting argument to evaluate the Laurent coefficients for .) In particular, if one hasfor 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 ascoordinate charts) from each to an open subset of . Furthermore, the atlas is said to besmoothif for any , thetransition 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 besmoothif, for any and , the maps are smooth; if is invertible and and are both smooth, we say that is a diffeomorphism, and that and arediffeomorphic. Two smooth atlases on are said to beequivalentif 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 asmooth structureon . Asmooth -dimensional real manifoldis 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 besmooth, 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 15A 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 16Verify that the unit circle is 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)Let be a Hausdorff topological space. Aholomorphic atlason is an open cover of together with a family of homeomorphisms (known ascoordinate charts) from each to an open subset of , such that, for any , thetransition 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 beholomorphicif, for any and , the maps are holomorphic; if is invertible and and are both holomorphic, we say that is acomplex diffeomorphism, and that and arecomplex diffeomorphic. Two holomorphic atlases on are said to beequivalentif 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 structureon . ARiemann surfaceis 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 MĆ¶bius transformations, as is for instance exemplified by this excellent video. (We may cover MĆ¶bius 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 again adopt the convention .

We can then identify meromorphic functions with holomorphic functions on the Riemann sphere:

Exercise 18Let be open connected, let be a discrete subset of , and let be a function. Show that the following are equivalent:

- (i) is meromorphic on .
- (ii) is the restriction of a holomorphic function to the Riemann sphere, which is not identically equal to .
Furthermore, if (ii) holds, show that 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 19Let be a holomorphic map from the Riemann sphere to itself. Show that either is identically equal to , 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)Let be a polynomial of one complex variable, which by the fundamental theorem of algebra we may write asfor 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

which is convergent in some punctured disk . The coefficient plays a privileged role and is known as the residue of at ; we denote it by . Clearly this is quantity is local in the sense that it only depends on the behaviour of in a neighbourhood of ; in particular, it does not depend on the domain so long as remains inside of that domain. By convention, we also set if is holomorphic at (i.e., if ).

We then have

Theorem 21 (Residue theorem)Let be 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 . Thenwhere only finitely many of the terms on the right-hand side are non-zero.

*Proof:* The image of is contained in some large ball; restricting and to this ball, we may assume without loss of generality that is both discrete and compact, and thus finite (by the Bolzano-Weierstrass theorem).

Next, we reduce to the case where all the residues vanish. We introduce the rational function defined by

From Laurent expansion around each singularity we see that for all , thus . Also, from the definition of winding number (see Definition 38 of Notes 3) we have

Setting , it thus suffices to show that

As is simply connected, is homotopic in (as closed contours) to a point. Let denote the homotopy. We would like to mimic the proof of Cauchy’s theorem (Theorem 4 of Notes 3) to conclude (9). The difficulty is that the homotopy may pass through points in . However, note from the vanishing of the residue that one has a Laurent expansion of the form

for some coefficients , in some punctured disk , with both series being absolutely convergent in this punctured disk. From term by term differentiation (see Theorem 15 of Notes 1) we see that has an antiderivative in this punctured disk, namely

(note how crucial it is that the term is absent in order to form this antiderivative). The absolute convergence of the series on the right-hand side in can be seen from the comparison test. From the fundamental theorem of calculus, we thus conclude that is conservative on . Also, for any that is *not* in , we see from Cauchy’s theorem that is conservative on for some radius . Putting this together using a compactness argument, we conclude that there exists a radius , such that for all in the image of the homotopy , the function is conservative in .

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 contours)Let be a simple closed anticlockwise contour 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 haveIf 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 curve is 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)Let be a simple closed anticlockwise contour in . Suppose that is holomorphic on an open set containing the image andexteriorof , outside of a finite that does not intersect the image of . Suppose also that converges to a finite limit in the limit . Show thatIf 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 25Let be holomorphic on an open set outside of a singular set , and let be an isolated point of .

- (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

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

proofof the residue theorem!)

The residue theorem can be applied in countless ways; we give only a small sample of them below.

Exercise 26Use the residue theorem to give an alternate proof of the fundamental theorem of algebra, by considering the integral for a polynomial of degree and some large radius .

Exercise 27Let be a Dirichlet polynomial of the formfor 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. Let be 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 haveWe 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 decompositionwhere 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

for a holomorphic function , at least away from the zeroes of . Inspired by this formal calculation, we refer to the function as the log-derivative of . Observe the product rule and quotient rule, when applied to complex differentiable functions that are non-zero at some point , gives the formulae

which are of course consistent with the formal calculation (10), given how we expect the logarithm to act on products and quotients. Thus, for instance, if , are polynomials that are factored as

and

for some non-zero complex numbers , distinct complex numbers , and positive integers , then the log-derivative of the rational function is given by

In particular, the log-derivative of is meromorphic with poles at , with a residue of at each zero of , and a residue of at each pole of .

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 29Let be 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

Since is holomorphic at , the claim (ii) follows. The claim (v) is proven similarly using a factorisation , and using (12) in place of (11). The claims (iii), (iv) then follow from (i), (ii) respectively after removing the singularity.

Remark 30Note 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)Let be a simple closed anticlockwise contour. 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 havewhere is the closed contour .

*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

and the second identity also follows.

We isolate the special case of the argument principle when there are no poles for special mention:

Corollary 32 (Special case of argument principle)Let be a simple closed anticlockwise contour, 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 contour is also invariant with respect to continuous perturbations, so long as zeroes never cross the contour itself. More precisely:

Corollary 33 (Stability of number of zeroes)Let be an open set. Let , be simple closed anticlockwise contours 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

But the curves and are homotopic as closed curves in , using the homotopy defined by

(note that this avoids the origin by hypothesis). The claim then follows from Lemma 41 of Notes 3.

Informally, the above corollary asserts that zeroes of holomorphic functions cannot be created or destroyed, as long as they are confined within a closed contour.

Example 34Let be 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 35When 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 zeroesminusthe 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)Let be a simple closed contour, 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)Let be 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 38Use Rouche’s theorem to obtain another proof of the fundamental theorem of algebra, by showing that a polynomial with and has exactly zeroes (counting multiplicity) in the complex plane. (Hint:compare with inside some large circle .)

Exercise 39 (Inverse function theorem)Let be 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 thatfor 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 40Let be 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)Let be 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
injectiveholomorphic 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. Let be 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

which, by Euler’s formula, can be written as

where

denotes all the possible arguments of in polar form. These arguments are a coset of the group , and so the complex logarithm is a coset of the group . For instance, if , then

and

The complex exponential never vanishes, so by our definitions we see that is the empty set. As such, we will usually omit the origin from the domain when discussing the complex exponential.

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

for any . If is complex differentiable at some point , then by differentiating (14) at using the chain rule, we see that

and hence by (14) again we have

(which is of course consistent with the real-variable formula ). If now is a closed contour in , and is differentiable on the entire image of , then the fundamental theorem of calculus then tells us that

On the other hand, is equal to . We thus conclude that for any branch of the complex logarithm, the set on which is complex differentiable cannot contain any closed curve that winds non-trivially around the origin. Thus for instance one cannot find a branch of that is holomorphic on all of , or even on a neighbourhood of the unit circle (or any other curve going around the origin).

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

and hence by the quotient rule

As is connected, must therefore be constant; by construction we have , and thus

for all . In other words, is a branch of the complex logarithm.

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

where is the *standard branch* of the argument, defined as the unique argument in in the interval . This branch of the logarithm is continuous on , and hence (by the exercise below) is holomorphic on this region, and is thus an antiderivative of here. Similarly if one replaces the negative real axis by other rays emenating from the origin (or indeed from arbitrary simple curves from zero to infinity, see Exercise 44 below.)

Exercise 43Let be 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 44Let be 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 . From the fundamental theorem of calculus, one has

for any curve that avoids the negative real axis. Of course, the contour does not avoid this negative axis, but it can be approximated by (non-closed) contours that do. More precisely, one has

where is the map . As each avoids the negative real axis, we thus have

We observe that has a jump discontinuity of on the negative real axis, and specifically

and

which gives an alternate derivation of the identity (15). More generally, the identity

for any closed curve avoiding the origin can be interpreted using the standard branch of the logarithm as a version of the Alexander numbering rule (Exercise 55 of Notes 3): each crossing of across the branch cut triggers a jump up or down in the count towards the winding number, depending on whether the crossing was in the anticlockwise or clockwise direction.

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 function by the formula

It is easy to see that is indeed holomorphic with for all . Thus for instance we have the standard branch of the root function, which is holomorphic away from the negative real axis. More generally, one can define a “standard branch of ” for any complex by the formula , for instance the standard branch of can be computed to be .

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 45Compute the improper integralby applying the residue theorem to the function for some branch of with branch cut on the

positivereal axis, and using a “keyhole” contour that is a perturbation ofthe 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 46Let be 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 .
- (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)Let be a continuous map between two connected topological spaces . We say that is acovering mapif, 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 acovering spaceof .

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 48The exponential map is a covering map, because for any element of written in polar form as , one can pick (say) the neighbourhoodof , 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

nota 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 isnota 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)Let be 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 aliftof 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 , where is the restriction of to . Clearly is closed in and contains ; using the covering map property it is not difficult to show that is also open 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 . 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)Let be 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 aliftof 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 51It is also possible to establish the above corollary using the monodromy theorem and analytic continuation.

Exercise 53Let be 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 54Show that there exists a holomorphic function such that for all . (Hint:use the Schwartz 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)Let be 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)Let be 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

defined by

Clearly is holomorphic on the right-half plane and avoids . We also observe that obeys the periodicity property

As the right-half plane is simply connected, we may (as before) express for some holomorphic function , and then write for some holomorphic function that avoids all numbers of the form for natural numbers and integers . Using Bloch’s theorem as before, we see that for any disk in the right-half plane , we have for some absolute constant . We cannot set to infinity any more, but we can make as large as the real part of , giving the bound

In particular, on integrating along a line segment from to , and using the boundedness of on the compact set , we obtain a bound of the form

for some , and all and . Taking cosines using the formula , we obtain a polynomial type bound

On the other hand, from (16) one has

and hence

for all in the right half-plane. The set is discrete, the function is continuous, and the right half-plane is connected, so this function must in fact be constant. That is to say, there exists an integer such that

for all in the upper half plane. Equivalently, the function is periodic with period . From (17) and the triangle inequality we conclude that

We now upgrade this bound on (18) by exploiting the quantisation of pole orders (Exercise 13). As the function is periodic with period on the right half-plane, we may write

for some function , which is holomorphic thanks to the chain rule. From (18) we have

when . Applying Exercise 13 (with, say, and ), we conclude that has a removable singularity and is thus in particular bounded on (say) the disk . From (19) we conclude that is bounded on the region ; taking exponentials, we conclude that is also bounded on this region. Since , we conclude that is bounded on , and thus by Riemann’s theorem (Exercise 35 from Notes 3) has a removable singularity at the origin. But by taking Laurent series, this implies that has a pole of order at most at the origin, contradicting the hypothesis that the singularity of at the origin was essential.

Exercise 57 (Montel’s theorem)Let be 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 ameromorphic normal familyto 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)Let be 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)Let be 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 .

## 66 comments

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11 October, 2016 at 12:52 pm

AnonymousIs 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 TaoCertainly; 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 SilbermanIn 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 …

onthe Riemann sphere, but I think you instead mean that can be extended to a function $f\colon U\to S\cup\{\infty\}$tothe Riemann sphere.[Corrected, thanks – T.]11 October, 2016 at 2:52 pm

AnonymousIt 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

AnonymousSince 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

AnonymousIt 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. ScorseseYou 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 TaoWell, 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

AnonymousIn 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

AnonymousCorrection: 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

AnonymousIn 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

AnonymousThere is a typo in the term.

[Corrected, thanks – T.]14 October, 2016 at 1:32 am

AnonymousEvery 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

savefor 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

AnonymousAh, 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

AnonymousLet 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 TaoNot 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

AnonymousIs 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

AnonymousEvery algebraic function has only finitely many branches on the Riemann sphere. Is the converse also true?

16 October, 2016 at 8:34 am

Terence TaoI *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

AnonymousA 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 TaoAh, 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 GubkinAnother 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

AnonymousFew minor typos

1. In the formula preceding (2) in the proof of Lemma 1, at the end should be removed.

2. Last paragraph in the proof of Lemma 1, should be .

3. Same sentence, “vanishing winding number in the .. exterior of should be instead. Actually, I am not sure where the vanishing of winding numbers is used here.

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, should be .

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

AnonymousIn definition 14, a formula does not parse.

[Corrected, thanks – T.]16 October, 2016 at 10:10 pm

AnonymousI 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

AnonymousAs 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).

This simple necessary and sufficient condition (that the sum of the orders of the zeros of 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 .

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

AnonymousIs it possible to extend exercise 2 to the case of infinitely many contours?

17 October, 2016 at 1:05 pm

Terence TaoYes, 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

AnonymousIf 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

AnonymousThird 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

AnonymousIn theorem 31, the summands in the LHS of (13) should have running index .

[Corrected, thanks – T.]18 October, 2016 at 3:08 am

AnonymousIn 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 ElencwajgDear 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 AsimovA 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

AnonymousNo! (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!)

one of simplest counter-examples seems to be the error function which (according to the above discussion) is an entire function with nowhere vanishing derivative, that attains every complex number infinitely many times.

27 October, 2016 at 12:56 am

AnonymousNo! (e.g. the exponential function).

27 October, 2016 at 1:02 am

AnonymousMy error! (the exponential function omits ).

30 October, 2016 at 7:54 pm

AnonymousIt 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

AnonymousDear 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

AnonymousI 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

AnonymousTypo, Exercise 20: “polynomial of one complex polynomial”

3 November, 2016 at 2:34 am

AnonymousIn the beginning of line 7.

3 November, 2016 at 11:23 am

AnonymousIn the third line there is a latex typo.

[Corrected, thanks – T.]3 November, 2016 at 3:10 pm

AnonymousThe typo is still there.

[Hopefully fixed now – T.]10 November, 2016 at 7:42 am

AnonymousIn 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

JDIn 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

AnonymousSmall 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

AnonymousI 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

AnonymousOne can also see in Ahlfors that

15 March, 2017 at 9:34 pm

Terence TaoOne can interpret integrals such as using integration of differential forms (or line integrals) rather than contour integrals.

20 March, 2017 at 4:09 pm

AnonymousIn 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

AnonymousIn 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

AnonymousIn 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 proof of Lemma 1 again, why do we need to put the extra term in the definition of and ? How is Exercise 36 of Notes 3 relevant.

[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

AnonymousIn 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

AnonymousThank you for the comments!

For 1, , and is either on or in establishing (2), so … ?

For 3, I understand that

since for small enough , .

How should one show that

where ?

31 March, 2017 at 9:23 am

jura05Dear 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

jura05Sorry, 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 HruskaHi, 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.]