TWe now come to perhaps the most central theorem in complex analysis (save possibly for the fundamental theorem of calculus), namely Cauchy’s theorem, which allows one to compute (or at least transform) a large number of contour integrals even without knowing any explicit antiderivative of . There are many forms and variants of Cauchy’s theorem. To give one such version, we need the basic topological notion of a homotopy:

Definition 1 (Homotopy)Let be an open subset of , and let , be two curves in .

- (i) If have the same initial point and final point , we say that and are
homotopic with fixed endpointsin if there exists a continuous map such that and for all , and such that and for all .- (ii) If are closed (but possibly with different initial points), we say that and are
homotopic as closed curvesin if there exists a continuous map such that and for all , and such that for all .- (iii) If and are curves with the same initial point and same final point, we say that and are
homotopic with fixed endpoints up to reparameterisationin if there is a reparameterisation of which is homotopic with fixed endpoints in to a reparameterisation of .- (iv) If and are closed curves, we say that and are
homotopic as closed curves up to reparameterisationin if there is a reparameterisation of which is homotopic as closed curves in to a reparameterisation of .In the first two cases, the map will be referred to as a

homotopyfrom to , and we will also say that can becontinously deformed to(either with fixed endpoints, or as closed curves).

Example 2If is a convex set, that is to say that whenever and , then any two curves from one point to another are homotopic, by using the homotopyFor a similar reason, in a convex open set , any two closed curves will be homotopic to each other as closed curves.

Exercise 3Let be an open subset of .

- (i) Prove that the property of being homotopic with fixed endpoints in is an equivalence relation.
- (ii) Prove that the property of being homotopic as closed curves in is an equivalence relation.
- (iii) If are closed curves with the same initial point, show that is homotopic to as closed curves if and only if is homotopic to with fixed endpoints for some closed curve with the same initial point as or .
- (iv) Define a
pointin to be a curve of the form for some and all . Let be a closed curve in . Show that is homotopic with fixed endpoints to a point in if and only if is homotopic as a closed curve to a point in . (In either case, we will callhomotopic to a point,null-homotopic, orcontractible to a pointin .)- (v) If are curves with the same initial point and the same terminal point, show that is homotopic to with fixed endpoints in if and only if is homotopic to a point in .
- (vi) If is connected, and are any two curves in , show that there exists a continuous map such that and for all . Thus the notion of homotopy becomes rather trivial if one does not fix the endpoints or require the curve to be closed.
- (vii) Show that if is a reparameterisation of , then and are homotopic with fixed endpoints in U.
- (viii) Prove that the property of being homotopic with fixed endpoints in up to reparameterisation is an equivalence relation.
- (ix) Prove that the property of being homotopic as closed curves in up to reparameterisation is an equivalence relation.

We can then phrase Cauchy’s theorem as an assertion that contour integration on holomorphic functions is a homotopy invariant. More precisely:

Theorem 4 (Cauchy’s theorem)Let be an open subset of , and let be holomorphic.

- (i) If and are rectifiable curves that are homotopic in with fixed endpoints up to reparameterisation, then
- (ii) If and are closed rectifiable curves that are homotopic in as closed curves up to reparameterisation, then

This version of Cauchy’s theorem is particularly useful for applications, as it explicitly brings into play the powerful technique of *contour shifting*, which allows one to compute a contour integral by replacing the contour with a homotopic contour on which the integral is easier to either compute or integrate. This formulation of Cauchy’s theorem also highlights the close relationship between contour integrals and the algebraic topology of the complex plane (and open subsets thereof). Setting to be a point, we obtain an important special case of Cauchy’s theorem (which is in fact equivalent to the full theorem):

Corollary 5 (Cauchy’s theorem, again)Let be an open subset of , and let be holomorphic. Then for any closed rectifiable curve in that is contractible in to a point, one has .

Exercise 6Show that Theorem 4 and Corollary 5 are logically equivalent.

An important feature to note about Cauchy’s theorem is the *global* nature of its hypothesis on . The conclusion of Cauchy’s theorem only involves the values of a function on the images of the two curves . However, in order for the hypotheses of Cauchy’s theorem to apply, the function must be holomorphic not only on the images on , but on an open set that is large enough (and sufficiently free of “holes”) to support a homotopy between the two curves. This point can be emphasised through the following fundamental near-counterexample to Cauchy’s theorem:

Example 7Let , and let be the holomorphic function . Let be the closed unit circle contour . Direct calculation shows thatAs a consequence of this and Cauchy’s theorem, we conclude that the contour is not contractible to a point in ; note that this does not contradict Example 2 because is not convex. Thus we see that the lack of holomorphicity (or

singularity) of at the origin can be “blamed” for the non-vanishing of the integral of on the closed contour , even though this contour does not come anywhere near the origin. Thus we see that the global behaviour of , not just the behaviour in the local neighbourhood of , has an impact on the contour integral.One can of course rewrite this example to involve non-closed contours instead of closed ones. For instance, if we let denote the half-circle contours and , then are both contours in from to , but one has

whereas

In order for this to be consistent with Cauchy’s theorem, we conclude that and are not homotopic in (even after reparameterisation).

In the specific case of functions of the form , or more generally for some point and some that is holomorphic in some neighbourhood of , we can quantify the precise failure of Cauchy’s theorem through the Cauchy integral formula, and through the concept of a winding number. These turn out to be extremely powerful tools for understanding both the nature of holomorphic functions and the topology of open subsets of the complex plane, as we shall see in this and later notes.

** — 1. Proof of Cauchy’s theorem — **

The underlying reason for the truth of Cauchy’s theorem can be explained in one sentence: complex differentiable functions behave locally like complex linear functions, which are conservative thanks to the fundamental theorem of calculus. More precisely, if is any complex linear function of , then has an antiderivative , and hence

for any rectifiable closed curve in the complex plane.

Perhaps the slickest way to make this intuition rigorous is through the following special case of Cauchy’s theorem.

Theorem 8 (Goursat’s theorem)Let be an open subset of , and be complex numbers such that the solid (and closed) triangle spanned by (or more precisely, theconvex hullof ) is contained in . (We allow the triangle to degenerate in that we allow the to be collinear, or even coincident.) Then for any holomorphic function , one haswhere is the closed polygonal path that traverses the vertices of the solid triangle in order.

*Proof:* Let us denote the triangular contour as . It is convenient (though odd-looking at first sight) to prove this theorem by contradiction. That is to say, suppose for contradiction that we had

for some . We now run the following “divide and conquer” strategy. We let , , be the midpoints of . Then from the basic properties of contour integration (see Exercise 16 of Notes 2) we can split the triangular integral as the sum of four integrals on smaller triangles, namely

(The reader is encouraged to draw a picture to visualise this decomposition.) By (2) and the triangle inequality (or, if one prefers, the pigeonhole principle), we must therefore have

where is one of the four triangular contours , , , or . Regardless of which of the four contours is, observe that the triangular region enclosed by is contained in that of . Furthermore, the diameter of is precisely half that of , where the diameter of a curve is defined by the formula

similarly, the perimeter of is precisely half that of . If we iterate the above process, we can find a nested sequence of triangular contours, each of which is contained in the previous one with half the diameter and perimeter, such that

for all . If we let be any point enclosed by , then from the decreasing diameters it is clear that the are a Cauchy sequence and thus converge to some limit , which is then contained in all of the closed triangles enclosed by any of the .

In particular, lies in and so is differentiable at . This implies, for any , that there exists a such that

whenever . We can rearrange this as

on . In particular, for large enough, this bound holds on the image on . In this case we can bound by , and hence by Exercise 16(v) of Notes 2,

From (1), the second integral vanishes. As each has half the diameter and perimeter of the previous, we thus have

But if one chooses small enough depending on and , we contradict (3).

Remark 9This is a rare example of an argument in which a hypothesis of differentiability, rather thancontinuousdifferentiability, is used, because one can localise any failure of the conclusion all the way down to a single point. Another instance of such an argument is the standard proof of Rolle’s theorem.

Exercise 10Find a proof of Goursat’s theorem that avoids explicit use of proof by contradiction. (Hint:use the fact that a solid triangle is compact, in the sense that every open cover has a finite subcover. For the purposes of this question, ignore the possibility that the proof of this latter fact might also use proof by contradiction.)

Goursat’s theorem only directly handles triangular contours, but as long as one works “locally”, or more precisely in a convex domain, we can quickly generalise:

Corollary 11 (Local Cauchy’s theorem for polygonal paths)Let be a convex open subset of , and let be a holomorphic function. Then for any closed polygonal path in , we have .

*Proof:* We induct on the number of vertices . The cases are trivial, and the case follows directly from Goursat’s theorem (using the convexity of to ensure that the interior of the polygon lies in ). If , we can split

The second integral on the right-hand side vanishes by Goursat’s theorem. The claim then follows from induction.

Exercise 12By using the (real-variable) fundamental theorem of calculus and Fubini’s theorem in place of Goursat’s theorem, give an alternate proof of Corollary 11 in the case that is a rectangle and the derivative of is continuous. (One can also use Stokes’ theorem in place of the fundamental theorem of calculus and Fubini’s theorem.)

We can amplify Corollary 11 using the fundamental theorem of calculus again:

Corollary 13 (Local Cauchy’s theorem)Let be a convex open subset of , and let be a holomorphic function. Then has an antiderivative . Also, for any closed rectifiable curve in , and whenever are two rectifiable curves in with the same initial point and same terminal point. In other words, is conservative on .

*Proof:* The first claim follows from Corollary 11 and the second fundamental theorem of calculus (Theorem 30 from Notes 2). The remaining claims then follow from the first fundamental theorem of calculus (Theorem 27 from Notes 2).

We can now prove Cauchy’s theorem in the form of Theorem 4.

*Proof:* We will just prove part (i), as part (ii) is similar (and in any event it follows from part (i)). Since reparameterisation does not affect the integral, we may assume without loss of generality that and are homotopic with fixed endpoints, and not merely homotopic with fixed endpoints up to reparameterisation.

Let be a homotopy from to . Note that for any and , lies in the open set . From compactness, there must exist a radius such that for all and . Next, as is continuous on a compact set, it is uniformly continuous. In particular, there exists such that

whenever and are such that and .

Now partition and as and in such a way that and for all and . For each such and , let denote the closed polygonal contour

(the reader is encouraged here to draw a picture of the situation; we are using polygonal contours here rather than the homotopy because we did not require any rectifiability properties on the homotopy). By construction, the diameter of this contour is at most , so the contour is contained entirely in the disk . This disk is convex and contained in . Applying Corollary 11 or Corollary 13, we conclude that

for all and . If we sum this over all and , and noting that the homotopy fixes the endpoints, we conclude after a lot of cancelling that

(again, the reader is encouraged to draw a picture to see this cancellation). However, from a further application of Corollary 11 we have

for , where is the restriction of to , and similarly for . Putting all this together we conclude that

as required.

One nice feature of Cauchy’s theorem is that it allows one to integrate holomorphic functions on curves that are not necessarily rectifiable. Indeed, if is a curve in , then for a sufficiently fine partition , the polygonal (and hence rectifiable) path will be contained in , and furthermore be homotopic to with fixed endpoints. One can then *define* when is holomorphic in and is non-rectifiable by declaring

where is *any* rectifiable curve that is homotopic (with fixed endpoints) to . This is a well defined definition thanks to the above discussion as well as Cauchy’s theorem; also observe that the exact open set in which the homotopy lives is not relevant, since given any two open sets containing the image of one can find a rectifiable curve which is homotopic to with fixed endpoints in , and hence in and separately. With this extended notion of the contour integral, one can then remove the hypothesis of rectifiability from many theorems involving integration of holomorphic functions. In particular, Cauchy’s theorem itself now holds for non-rectifiable curves. This reflects some duality in the integration concept ; if one assumes more regularity on the function , one can get away with worse regularity on the curve , and vice versa.

A special case of Cauchy’s theorem is worth recording explicitly. We say that an open set in the complex plane is simply connected if it is non-empty, connected, and if every closed curve in is contractible in to a point. For instance, from Example 2 we see that any convex non-empty open set is simply connected. From Theorem 4 we then have

Theorem 14 (Cauchy’s theorem, simply connected case)Let be a simply connected subset of , and let be holomorphic. Then for any closed curve in . In particular (by Exercise 31 of Notes 2), is conservative and has an antiderivative.

** — 2. Consequences of Cauchy’s theorem — **

Now that we have Cauchy’s theorem, we use it to quickly give a large number of striking consequences. We begin with a special case of the Cauchy integral formula.

Theorem 15 (Cauchy integral formula, special case)Let be an open subset of , let be holomorphic, and let be a point in . Let be such that the closed disk is contained in . Let be a closed curve in that is homotopic (as a closed curve, and up to reparameterisation) in to in . Then

Here we are already taking advantage of the ability to integrate holomorphic functions (such as , which is holomorphic on ) on curves that are not necessarily rectifiable.

Note the remarkable feature here that the value of at some point other than that on is completely determined by the value of on the curve , which is a strong manifestation of the “rigid” or “global” nature of holomorphic functions. Such a formula is certainly not available in the real case (Cauchy’s theorem is technically true on the real line, but there is no analogue of the circular contours available in that setting).

*Proof:* Observe that for any , the circles and are homotopic (as closed curves) in , and hence in . Since the function is holomorphic on , we conclude from Cauchy's theorem that

As is complex differentiable at , there exists a finite such that

for all in , and all sufficiently small . The length of this circle is of course . Applying Exercise 16(v) of Notes 2 we have

On the other hand, from explicit computation (cf. Example 7) we have

putting all this together, we see that

Sending to zero, we obtain the claim.

Note the same argument would give

if were homotopic to the curve : rather than , for some integer . In particular, if were homotopic to a point in , then the right-hand side would vanish.

Remark 16For various explicit examples of closed contours , it is also possible to prove the Cauchy integral formula by applying Cauchy’s theorem to various “keyhole contours”. We will not pursue this approach here, but see for instance Chapter 2 of Stein-Shakarchi.

Exercise 17 (Mean value property and Poisson kernel)Let be an open subset of , and let be a closed disk contained in .

- (i) If is holomorphic, show that
. Use this to give an alternate proof of Exercise 26 from Notes 1.

- (ii) If is harmonic, show that
. Use this to give an alternate proof of Theorem 25 from Notes 1.

- (iii) If is harmonic, show that
for any , where the

Poisson kernelis defined by the formula(

Hint:it simplifies the calculations somewhat if one reduces to the case , , and for some . Then compute the integral in two different ways, where is holomorphic with real part .)

The first important consequence of the Cauchy integral formula is the analyticity of holomorphic functions:

Corollary 18 (Holomorphic functions are analytic)Let be an open subset of , let be holomorphic, and let be a point in . Let be such that the closed disk is contained in . For each natural number , let denote the complex numberThen the power series has radius of convergence at least , and converges to inside the disk.

*Proof:* By continuity, there exists a finite such that for all on the circle , which of course has length . From Exercise 16(v) of Notes 2 we conclude that

From this and Proposition 7 of Notes 1, we see that the radius of convergence of is indeed at least .

Next, for any , the circle is homotopic (as a closed curve) in (and hence in to to small enough that lies in . Applying the Cauchy integral formula, we conclude that

On the other hand, from the geometric series formula (Exercise 12 of Notes 1) one has

for all , and thus

If we could interchange the sum and integral, we would conclude from (4) that

which would give the claim. To justify the interchange, we will use the Weierstrass -test (the dominated convergence theorem would also work here). We have the pointwise bound

by the geometric series formula and the hypothesis , the sum is finite, and so the -test applies and we are done.

Remark 19A function on an open set is said to be complex analytic on if, for every , there is a power series with a positive radius of convergence that converges to on some neighbourhood of . Combining the above corollary with Theorem 15 of Notes 1, we see that is holomorphic on if and only if is complex analytic on ; thus the terms “complex differentiable”, “holomorphic”, and “complex analytic” may be used interchangeably. This can be contrasted with real variable case: there is a completely parallel notion of a real analytic function (i.e., a function that, around every point in the domain, can be expanded as a convergent power series in some neighbourhood of that point), and real analytic functions are automatically smooth and differentiable, but the converse is quite false.

Recalling (see Remark 21 of Notes 1) that power series are infinitely differentiable (in both the real and complex senses) inside their disk of convergence, and working locally in various small disks in , we conclude

Corollary 20Let be an open subset of , and let be a holomorphic function. Then is also holomorphic, and is smooth (i.e. infinitely differentiable in the real sense).

In view of this corollary, we may now drop hypotheses of continuous first or second differentiability from several of the theorems in Notes 1, such as Exercise 26 from that set of notes.

Combining Corollary 20 with Proposition 28 of Notes 1 (with replaced by various rectangles in ), we obtain a form of elliptic regularity:

Corollary 21 (Elliptic regularity)Let be an open subset of , and let be a harmonic function. Then is smooth.

In fact one can even omit the hypothesis of continuous twice differentiability in the definition of harmonicity if one works with the notion of weak harmonicity, but this is a topic for a PDE or distribution theory course and will not be pursued further here.

Another immediate consequence of Corollary 18 is a version of the factor theorem:

Corollary 22 (Factor theorem for analytic functions)Let be an open subset of , and let be a point in . Let be a complex analytic function that vanishes at . Then there exists a unique complex analytic function such that for all .

*Proof:* For , we can simply define , and this is clearly the unique choice here. For equal to or near , we can expand as a Taylor series (noting that the constant term vanishes since ) and then set . One can check that these two definitions of agree on their common domain, so on gluing the two functions together one obtains a function that is complex differentiable (and hence analytic) on with the desired factorisation. Uniqueness at then follows from uniqueness at and continuity.

Yet another consequence is the important property of analytic continuation:

Corollary 23 (Analytic continuation)Let be a connected non-empty open subset of , and let , be complex analytic functions. If and agree on some non-empty open subset of , then they in fact agree on all of .

*Proof:* Let denote the set of all points in where and agree to all orders, that is to say that

for all . By hypothesis, is non-empty; by the continuity of the , is closed; and from analyticity and Taylor expansion (Exercise 17 of Notes 1) is open. As is connected, must therefore be all of , and the claim follows. (This is an example of the continuity method.)

There is also a variant of the above corollary:

Corollary 24 (Non-trivial analytic functions have isolated zeroes)Let be a connected non-empty open subset of , and let be a function which vanishes at some point but is not identically zero. Then there exists a disk in on which does not vanish except at ; in other words, all the zeroes of are isolated points.

*Proof:* If all the derivatives of at vanish, then by Taylor expansion vanishes in some open neighbourhood of , and then by Corollary 23 vanishes everywhere, a contradiction. Thus at least one of the is non-zero. If is the first natural number for which , then by iterating the factor theorem (Corollary 22) we see that for some analytic function which is non-vanishing at . By continuity, is also non-vanishing in some disk in , and the claim follows.

One particular consequence of the above corollary is that if two entire functions agree on the real line (or even on an infinite bounded subset of the complex plane), then they must agree everywhere, since otherwise would have a non-isolated zero, contradicting Corollary 24. This strengthens Corollary 23, and helps explain why real-variable identities such as automatically extend to their complex counterparts . Another consequence is that if an entire function is real-valued on the real axis, then one has the identity

for all complex , because this identity already holds on the real line, and both sides are complex analytic. Thus for instance

Next, if we combine Corollary 18 with Exercise 17 of Notes 1, as well as Cauchy’s theorem, we obtain

Theorem 25 (Higher order Cauchy integral formula, special case)Let be an open subset of , let be holomorphic, and let be a point in . Let be such that the closed disk is contained in . Let be a closed curve in that is homotopic (as a closed curve, up to reparameterisation) in to in . Then for any natural number , the derivative of at is given by the formula

Exercise 26Give an alternate proof of Theorem 25 by rigorously differentiating the Cauchy integral formula with respect to the parameter.

Combining Theorem 25 with Exercise 16(v) of Notes 2, we obtain a more quantitative form of Corollary 20, which asserts not only that the higher derivatives of a holomorphic function exist, but also places a bound on them:

Corollary 27 (Cauchy inequalities)Let be an open subset of , let be holomorphic, and let be a point in . Let be such that the closed disk is contained in . Suppose that there is an such that on the circle . Then for any natural number , we have

Note that the case of this corollary is compatible with the maximum principle (Exercise 26 of Notes 1).

The right-hand side of (5) has a denominator that improves when gets large. In particular we have the remarkable theorem of Liouville:

Theorem 28 (Liouville’s theorem)Let be an entire function that is bounded. Then is constant.

*Proof:* By hypothesis, there is a finite such that for all . Applying the Cauchy inequalities with and any disk , we conclude that

for any and . Sending to infinity, we conclude that vanishes identically. The claim then follows from the fundamental theorem of calculus.

This theorem displays a strong “rigidity” property for entire functions; if such a function is even vaguely close to being constant (by being bounded), then it almost magically “snaps into place” and actually is forced to be a constant! This is in stark contrast to the real case, in which there are functions such as that are differentiable (and even smooth and analytic) on the real line and bounded, but definitely not constant. Note that the complex analogue of the sine function is not a counterexample to Liouville’s theorem, since becomes quite unbounded away from the real axis (Exercise 16 of Notes 0). This also fits well with the intuition of harmonic functions (and hence also holomorphic functions) being “balanced” in that any convexity in one direction has to be balanced by concavity in the orthogonal direction, and vice versa (as discussed before Theorem 25 of Notes 1): any attempt to create an entire function that is bounded and oscillating in one direction will naturally force that function to become unbounded in the orthogonal direction.

Exercise 29Let be an entire function which is ofpolynomial growthin the sense that there exists a finite quantity and some exponent such that for all . Show that is, in fact, a polynomial.

Now we can prove the fundamental theorem of algebra discussed back in Notes 0.

Theorem 30 (Fundamental theorem of algebra)Letbe a polynomial of degree for some with non-zero. Then there exist complex numbers such that

*Proof:* This is trivial for , so suppose inductively that and the claim has already been proven for . Suppose first that the equation has no roots in the complex plane, then the function is entire. Also, this function goes to zero as , and so is bounded on the exterior of any sufficiently large disk; as it is also continuous, it is bounded on any disk and is thus bounded everywhere. By Liouville’s theorem, is constant, which implies that is constant, which is absurd (for instance, the derivative of is the non-zero function ). Hence has at least one root . By the factor theorem (which works in any field, including the complex numbers) we can then write for some polynomial , which by the long division algorithm (or by comparing coefficients) must take the form

for some complex numbers . The claim then follows from the induction hypothesis.

The following exercises show that can be alternatively defined as an algebraic closure of the reals (together with a designated square root of ), and that extending using a different irreducible polynomial than would still give a field isomorphic to the complex numbers, thus supporting the notion that the complex numbers are not an arbitrary extension of the reals, but rather a quite natural and canonical one.

Exercise 31Let be a field containing which is a finite extension of , in the sense that is a finite-dimensional vector space over . Show that is isomorphic (as a field) to either or . (Hint:if is some element of not in , show that for some irreducible polynomial with real coefficients but no real roots. Use this to set up an isomorphism between the field generated by and with . If there is an element of not in this field , show that there for some irreducible polynomial with coefficients in and no roots in , and contradict the fundamental theorem of algebra.)

Exercise 32A field is said to be algebraically closed if the conclusion of Theorem 30 with replaced by . Show that any algebraically closed field containing , contains a subfield that is isomorphic to (and which contains as a subfield, isomorphic to the copy of inside ). Thus, up to isomorphism, is the unique algebraic closure of , that is to say a minimal algebraically closed field containing .

Another nice consequence of the Cauchy integral formula is a converse to Cauchy’s theorem known as Morera’s theorem.

Theorem 33 (Morera’s theorem)Let be an open subset of , and let be a continuous function. Suppose that is conservative in the sense that for any closed polygonal path in . Then is holomorphic on .

*Proof:* By working locally with small balls in we may assume that is a ball (and in particular connected). By Exercise 31 of Notes 2, has an antiderivative . By definition, is complex differentiable at every point of (with derivative ), so by Corollary 20, is smooth, which implies in particular that is holomorphic on as claimed.

The power of Morera’s theorem comes from the fact that there are no differentiability requirements in the hypotheses on , and yet the conclusion is that is differentiable (and hence smooth, by Corollary 20); it can be viewed as another manifestation of “elliptic regularity”. Here is one basic application of Morera’s theorem:

Theorem 34 (Uniform limit of holomorphic functions is holomorphic)Let be an open subset of , and let be a sequence of holomorphic functions that converge uniformly on compact sets to a limit . Then is also holomorphic. Furthermore, for each natural number , the derivatives also converge uniformly on compact sets to for any natural number (In particular, converges pointwise to on .)

*Proof:* Again we may work locally and assume that is a ball (and in partiular is convex and simply connected). The are continuous, hence their locally uniform limit is also continuous. From Corollary 11 (or Corollary 14), we have on any closed polygonal path in , hence on taking locally uniform limits we also have for such paths. The holomorphicity of then follows from Morera’s theorem. The uniform convergence of to on compact sets follows from applying Theorem 25 to circular contours for and small enough that these contours lie in (note from compactness that one can take independent of ).

Actually, one can weaken the uniform nature of the convergence in Theorem 34 substantially; even the weak limit of holomorphic functions in the space of locally integrable functions on will remain harmonic. However, we will not need these weaker versions of this theorem here.

Exercise 35 (Riemann’s theorem on removable singularities)Let be an open subset of , let be a point in , and let be a holomorphic function on which is bounded near , in the sense that it is bounded on some punctured disk contained in . Show that has a removable singularity at , in the sense that is the restriction to of a holomorphic function on . (Hint:show that is conservative near , find an antiderivative, extend it to , and use Morera’s theorem to show that this extension is holomorphic. Alternatively, one can also proceed by some version of the Cauchy integral formula.)

Exercise 36 (Integrals of holomorphic functions)Let be an open subset of , and let be a continuous function such that, for each , the function is holomorphic on . Show that the function is also holomorphic on . (Hint:work locally and use Cauchy’s theorem, Morera’s theorem, and Fubini’s theorem.)

Exercise 37 (Schwarz reflection principle)Let be an open subset of that is symmetric around the real axis, that is to say whenever . Let be a continuous function on the set that is holomorphic in the open subset . Similarly, let be continuous on that is holomorphic in the open subset . Suppose further that and agree on . Show that and are both restrictions of a single holomorphic function .

The following two Venn diagrams (or more precisely, Euler diagrams) summarise the relationships between different types of regularity amongst continuous functions over both the reals and the complexes. The first diagram

describes the class of continuous functions on some interval in the real line; such functions are automatically conservative, but not necessarily differentiable, while differentiable functions are not necessarily smooth, and smooth functions are not necessarily analytic. On the other hand, when considering the class of continuous functions on an open subset of , the picture is different:

Now, very few continuous functions are conservative, and only slightly more functions are complex differentiable (and for simply connected domains , these two classes in fact coincide). Whereas in the real case, differentiable functions were considerably less regular than analytic functions, in the complex case the two classes in fact coincide.

** — 3. Winding number — **

One defect of the current formulation of the Cauchy integral formula (see Theorem 15 and the ensuing discussion) is that the curve involved has to be homotopic (as a closed curve, up to reparameterisation) to a circular arc , or at least to a curve of the form , for some integer . We now investigate what happens when this hypothesis is removed. A key notion is that of a *winding number*.

Definition 38 (Winding number)Let be a closed curve, and let be a complex number that is not in the image of . The winding number of around is defined by the integral

Here we again take advantage of the ability to integrate holomorphic functions on curves that are not necessarily rectifiable. Clearly the winding number is unchanged if we replace by any equivalent curve, and if one replaces the curve with its reversal , then the winding number is similarly negated. In some texts, the winding number is also referred to as the *index* or *degree*.

From the Cauchy integral formula we see that

when is homotopic in (as a closed curve, up to reparameterisation) to a circle , and more generally that

if is homotopic in (as a closed curve, up to reparameterisation) to a curve of the form , . Thus we see, intuitively at least, that measures the number of times winds counterclockwise about , which explains the term “winding number”.

We can now state a more general form of the Cauchy integral formula:

Theorem 39 (General Cauchy integral formula)Let be a simply connected subset of , let be a closed curve in , and let be holomorphic. Then for any that lies in but not in the image of , we have

*Proof:* By Corollary 22 (or Exercise 35), we have for some holomorphic function . Hence by Theorem 14 we have

The claim then follows from (6).

Exercise 40 (Higher order general Cauchy integral formula)With as in the above theorem, show thatfor every natural number . (

Hint:instead of approximating by , use a partial Taylor expansion of . Many of the terms that arise can be handled using the fundamental theorem of calculus. Alternatively, one can use differentiation under the integral sign and Lemma 44 below.)

To use Theorem 39, it becomes of interest to obtain more properties on the winding number. From Cauchy’s theorem we hav

Lemma 41 (Homotopy invariance)Let , and let be two closed curves in that are homotopic as closed curves up to reparameterisation in . Then .

The following specific corollary of this lemma will be useful for us.

Corollary 42 (Rouche’s theorem for winding number)Let be a closed curve, and let lie outside of the image of . Let be a closed curve such that

*Proof:* The map defined by is a homotopy from to ; by (7) and the triangle inequality, it avoids . The claim then follows from Lemma 41.

Corollary 42 can be used to compute the winding number near infinity as follows. Given a curve and a point , define the *distance*

and the *diameter*

Corollary 43 (Vanishing near infinity)Let be a closed curve. Then whenever is such that .

*Proof:* Apply Corollary 42 with equal to and equal to any point in the image of .

Corollary 42 also gives local constancy of the winding number:

Lemma 44 (Local constancy in )Let be a closed curve. Then is locally constant. That is to say, if does not lie in the image of , then there exists a disk outside of the image of such that for all .

*Proof:* From Corollary 42, we see that if is small enough and , then

where is the translation of by . But by a translation change of variables we see that

and the claim follows.

Exercise 45Give an alternate proof of Lemma 44 based on differentiation under the integral sign and using the fact that has an antiderivative away from .

As confirmation of the interpretation of as a winding number, we can now establish integrality:

Lemma 46 (Integrality)Let be a closed curve, and let lie outside of the image of . Then is an integer.

*Proof:* By Corollary 42 we may assume without loss of generality that is a closed polygonal path. By partitioning a polygon into triangles (and using Lemma 44 to move slightly out of the way of any new edges formed by this partition) it suffices to verify this for triangular . But this follows from the Cauchy integral formula (if is in the interior of the triangle) or Cauchy’s theorem (if is in the exterior).

Exercise 47Give another proof of Lemma 46 by restricting again to closed polygonal paths , and showing that the function is constant on by establishing that it is continuous and has vanishing derivative at all but finitely many points. (Note that exists for all but finitely many , so the integral here can be well defined.)

We now come to a fundamental and well known theorem about simple closed curves, namely the Jordan curve theorem.

Theorem 48 (Jordan curve theorem)Let be a non-trivial simple closed curve. Then there is anorientationsuch that the complex plane is partitioned into theboundary region, theexterior regionFurthermore the exterior region is connected and unbounded, and the interior region is connected, non-empty and bounded. Finally, if is any open set that contains and its interior, then is contractible to a point in .

This theorem is relatively easy to prove for “nice” curves, such as polygons, but is surprisingly delicate to prove in general. Some idea of the subtlety involved can be seen by considering pathological examples such as the lakes of Wada, which are three disjoint open connected subsets of which all happen to have exactly the same boundary! This does not contradict the Jordan curve theorem, because the boundary set in this example is not given by a simple closed curve. However it does indicate that one has to carefully use the hypothesis of being a simple closed curve in order to prove Theorem 48. Another indication of the difficulty of the theorem is its global nature; the claim does not hold if one replaces the complex plane by other surfaces such as the torus, the projective plane, or the Klein bottle, so the global topological structure of the complex plane must come into play at some point. For the sake of completeness, we give a proof of this theorem in an appendix to these notes.

If the quantity in the above theorem is equal to , we say that the simple closed curve has an *anticlockwise orientation*; if instead we say that has a *clockwise* orientation. Thus for instance, has an anticlockwise orientation, while its reversal has the clockwise orientation.

Exercise 49Let , be non-trivial simple closed curves.

- (i) If have disjoint image, show that either lies entirely in the interior of , or in the exterior.
- (ii) If avoids the exterior of , show that the interior of is contained in the interior of , and the exterior of contains the exterior of .
- (iii) If avoids the interior of , and avoids the interior of , and the two curves have disjoint images, show that the interior of is contained in the exterior of , and the exterior of contains the interior of .
(This is all visually “obvious” as soon as one draws a picture, but the challenge is to provide a rigorous proof. One should of course use the Jordan curve theorem extensively to do so. You will not need to use the final part of the Jordan curve theorem concerning contractibility.)

Exercise 50Let be a non-trivial simple closed curve. Show that the interior of is simply connected. (Hint:first show that any simple closed polygonal path in is contractible to a point in the interior; then extend this to closed polygonal paths that are not necessarily simple by an induction on the number of edges in the path; then handle general closed curves.)

Remark 51There is a refinement of the Jordan curve theorem known as the Jordan-Schoenflies theorem, that asserts that for non-trivial simple closed curve there is a homeomorphism that maps to the unit circle , the interior of to the unit disk , and the exterior to the exterior region . The proof of this improved version of the Jordan curve theorem will have to wait until we have the Riemann mapping theorem (as well as a refinement of this theorem due to Carathéodory). The Jordan-Schoenflies theorem may seem self-evident, but it is worth pointing out that the analogous result in three dimensions fails without additional regularity assumptions on the boundary surface, thanks to the counterexample of the Alexander horned sphere.

From the Jordan curve theorem we have yet another form of the Cauchy theorem and Cauchy integral formula:

Theorem 52 (Cauchy’s theorem and Cauchy integral formula for simple curves)Let be a simple closed curve, and let be an open set containing and its interior. Let be a holomorphic function.

- (i) (Cauchy’s theorem) One has .
- (ii) (Cauchy integral formula) If lies outside of the image of , then the expression vanishes if lies in the exterior of , equals if lies in the interior of and is oriented anti-clockwise, and equals if lies in the interior of and is oriented clockwise.

Exercise 53Let be a polynomial with complex coefficients and . For any , let denote the closed contour .

- (i) Show that if is sufficiently large, then .
- (ii) Show that if does not vanish on the closed disk , then .
- (iii) Use these facts to give an alternate proof of the fundamental theorem of algebra that does not invoke Liouville’s theorem.

In the case when the closed curve is a contour (which includes of course the case of closed polygonal paths), one can describe the interior and exterior regions, as well as the winding number, more explicitly.

Exercise 54 (Local structure of interior and exterior)Let be a simple closed contour formed by concatenating smooth curves together. Let be an interior point of one of these curves , thus for some $latex {a_i < t_i0}&fg=000000$ and . Recall from Exercise 24 of Notes 2 that for sufficiently small , the set can be expressed as a graph of the formfor some interval and some continuously differentiable function with . Show that if is oriented anticlockwise, and is sufficiently small then the interior of contains all points in of the form for some and , and the exterior of contains all points in of the form for some and and swapped.

Exercise 55 (Alexander numbering rule)Let be a simple closed contour oriented anticlockwise formed by concatenating smooth curves together. Let be a contour formed by concatenating smooth curves , with initial point and final point . Assume that there are only finitely many points where the images of and of intersect. Furthermore, assume at each of the points , , that one has a “smooth simple transverse intersection” in the sense that the following axioms are obeyed:

- (i) lies in the interior of one of the smooth curves that make up , thus for some .
- (ii) lies in the interior of one of the smooth curves that make up , thus for some .
- (iii) is only traversed once by , thus there do not exist in such that .
- (iv) The derivatives and are linearly independent over . In other words, we either have a
crossing from the rightin which for some and , or else we have acrossing from the leftin which for some and .Show that is equal to the total number of crossings from the left, minus the total number of crossings from the right.

Exercise 56Let be a non-empty connected open subset of . Show that is simply connected if and only if every holomorphic function on is conservative.

** — 4. Appendix: proof of the Jordan curve theorem (optional) — **

We now prove the Jordan curve theorem. We begin with a variant of Corollary 42 in which the curve is only required to have *image* close to the image of , rather than be close to in a *pointwise* (and uniform) sense. For any curve and any , let denote the -neighbourhood of .

Proposition 57Let be a non-trivial simple closed curve, and let . Suppose that is sufficiently small depending on and . Let be a closed curve (not necessarily simple) whose image lies in . Then is homotopic (as a closed curve, up to reparameterisation) to in , where is defined as the concatenation of copies of if is positive, the trivial curve at the initial point of if is zero, and the concatenation of copies of if is negative. In particular, from Lemma 41 one hasfor all .

*Proof:* After reparameterisation, we can take to have domain on the unit interval , and then by periodic extension we can view as a continuous -periodic function on .

As is compact, is uniformly continuous on , and hence also on . In particular, there exists such that

Fix this . Observe that the function is continuous and nowhere vanishing on the region . Thus, if is small enough depending on , we have the lower bound

whenever are such that . Using the -periodicity of , we conclude that if are such that

the there must be an integer such that

Note that this integer is uniquely determined by and .

Let be the domain of . By the uniform continuity of , we can find a partition of such that

for all and . Since the image of lies in , we can find, for each , a real number such that

Since is closed, we may arrange matters so that

From the triangle inequality and (12), (13) we have

Using (11), we conclude that for each , there is an integer such that

As is -periodic, we have the freedom to shift each of the by an arbitrary integer, and by doing this for in turn, we may assume without loss of generality that all the vanish, thus

for all . In particular, from (10) we have

whenever . Also, as is simple, we have from (14) that

for some integer . (Note that by enforcing (15), we no longer have the freedom to individually move or by an integer, so we cannot assume without loss of generality that vanishes.)

For , let denote the curve

from to ; similarly let denote the curve

from to . Observe from (16), (12), (13) that for each , the images of and both lie in , which will lie in if is small enough. We can thus form a homotopy from to by defining

for all and . Thus and are homotopic as closed curves in . But by Exercise 3, is homotopic up to reparameterisation as closed curves to in , and is similarly homotopic up to reparameterisation as closed curves to in , and the claim follows.

We can now prove Theorem 48. We first verify the claim in the easy (and visually intuitive) case that is a non-trivial simple closed polygonal curve. Removing the polygon from leaves an open set, which we may decompose into connected components as per Exercise 34 of Notes 2. On each of these components, the winding number is constant. Since each component has a non-empty boundary that is contained in , this constant value of must also be attained arbitrarily close to .

Now, a routine application of the Cauchy integral formula (see Exercise 59) shows that as crosses one of the edges of the polygon , the winding number is shifted by either or . Hence at each point on , the winding number will take two values in a sufficiently small neighbourhood of (excluding ). By a continuity argument, the integer is independent of . On the other hand, from Corollary 43 the winding number must be able to attain the value of zero. Thus we have for some . Dividing a small neighbourhood of (excluding itself) into the regions where the winding numbers are or , a further continuity argument shows that each of these regions lie in a single connected component. Thus there are only two connected components, one where the winding number is zero and one where the winding number is . From (43) the latter component is bounded, hence the former is unbounded, and the claim follows.

Now we handle the significantly more difficult case when is just a non-trivial simple closed curve. As one may expect, the strategy will be to approximate this curve by a polygonal path, but some care has to be taken when performing a limit, in order to prevent the interior region from collapsing into nothingness, or becoming disconnected, in the limit.

The first challenge is to ensure that there is at least one point outside of in which is non-zero. This is actually rather tricky; we will achieve this by a parity argument (loosely inspired by a nonstandard version of this argument from this paper of Kanovei and Reeken). Clearly, contains at least two points; by an appropriate rotation, translation, and dilation we may assume that contains the points and , with being both the initial point and the final point. Then we can decompose , where is a curve from to , and is a curve from to .

Observe from the simplicity of that whenever and are such that

Thus, by compactness, there exists such that one has the lower bound

separating from whenever , are such that (17) holds.

Next, for any natural number , we may approximate by a polygonal closed path with

for all . Although it is not particularly necessary, we can ensure that and . By perturbing the edges of the polygonal path slightly, we may assume that none of the vertices of lie on the real axis, and that none of the self-crossings of (if any exist) lie on the real axis; thus, whenever crosses the real axis, it does so at an interior point of an edge, with no other edge of passing through that point. Note that we do not assert that the curve is simple; with some more effort one could “prune” by deleting short loops to make it simple, but this turns out to be unnecessary for the parity argument we give below.

Let be the points on the real axis where crosses. By Exercise 59 below, the winding number changes by or as crosses each of the ; by Lemma 44, this winding number is constant otherwise, and by Corollary 43 it vanishes near infinity. Thus is even, and the winding number is odd between and for any odd .

Next, observe that each point belongs to exactly one of the polygonal paths or . Since each of these curves starts on one side of the real axis and ends up on the other, they must both cross the real axis an odd number of times. On the other hand, the crossing points can be grouped into pairs with odd. We conclude that there must exist an odd such that one of the lies in and the other lies in .

Fix such a . For sake of discussion let suppose that lies in and lies in . From (19) we have

and from (18) we have

for any . By the intermediate value theorem, we can thus (for large enough) find such that

and thus

or equivalently

We arrive at the same conclusion in the opposite case when lies in and lies in .

By Corollary 43 (and (19)), the are bounded in . By the Bolzano-Weierstrass theorem, we can thus extract a subsequence of the that converges to some limit . By continuity we then have

in particular does not lie in . By construction of , we know that is odd for all ; using Lemma 44 and Lemma 42 we conclude that is also odd. Thus we have found at least one point where the winding number is non-zero.

Now we can finish the proof of the Jordan curve theorem. Let be a non-trivial simple closed curve. By the preceding discussion, we can find a point outside of where the winding number is non-zero. Let be a sufficiently small parameter, and let be sufficiently small depending on . By compactness, one can cover the region by a finite number of (solid) squares of sidelength and sides parallel to the real and imaginary axes; by perturbation we may assume that no edge of one square is collinear to an edge of any other square. These squares all lie in , and in particular will not contain if is small enough; their union can easily be seen to be connected. The boundaries of these squares divide the complex plane into a finite number of polygonal regions (one of whom is unbounded). One of these regions, call it , contains the point . This region cannot contain any interior point of a square , since otherwise would be trapped inside a square of sidelength and hence not contain . In particular, avoids . The region cannot be unbounded, since one could then continuously move to infinity without ever meeting , contradicting Lemma 44, Corollary 43, and the non-vanishing nature of . The boundary of consists of one or more disjoint closed polygonal paths, whose edges consist of horizontal and vertical line segments. Actually, the boundary must consist of just one closed path, since otherwise the union of the squares would be disconnected, a contradiction. Let denote the path that bounds (traversed in either of the two possible directions). This path must be simple, because a crossing can only be formed by an edge of one square crossing an edge of another square at a point that is not on the corner of either of the two squares; as avoids both and , it can thus only occupy one quadrant of a neighbourhood of this crossing and so cannot bound all four edges of the crossing.

Applying the Jordan curve theorem to the polygonal path , we conclude that there is such that on , and for all outside of and . On the other hand, by Proposition 57 there is an integer such that is homotopic (as closed curves, up to reparameterisation) in to , so in particular

for all . Applying this to , we conclude that is either or . If we write (where , *a priori*, may depend on ), then , and we have

for . Thus takes only two values outside of . Sending , we conclude that is in fact independent of , and takes only the two values outside of .

We now define the interior and exterior regions by (9), (8), then we have partitioned into the interior, exterior, and . From Lemma 44 the interior and exterior are open, and from Lemma 43 the interior is bounded, and hence the exterior is unbounded. The point lies in the interior, so the interior is non-empty. The only remaining task to show is that the interior and exterior are connected. Suppose for instance that lie in the interior region. Then for small enough, lie outside of . From (20), (21) we conclude that lie in . As is connected, we can thus join to by a path in . As the region avoids , we see from Lemma 44 that the winding number stays constant on this path, and so the path remains in the interior region (9). This establishes the connectedness of the interior region; the connectedness of the exterior is proven similarly.

It remains to prove the contractibility of in any open set that contains and its interior. Once again, we begin with the simpler case when is a simple closed polygonal path. We induct on the number of edges in . The cases can be handled by direct calculation, so suppose that and the claim has been proven for all smaller values of . We may remove any edges of zero length from the polygon. If the interior of the polygon is convex, then the claim follows from Example 2, so we may assume that the interior is non-convex. This implies that one of the interior angles in the polygon exceeds (see Exercise 11 below), thus there are two adjacent edges whose interior angle exceeds . If one extends in the interior until it meets the polygon again, this wil divide the polygon into two subpolygons , each of which can be verified to have fewer than edges. By Exercise 49 (which does not use the contractibility part of the Jordan curve theorem), the interiors of and are contained in the interior of , and so by the induction hypothesis they are contractible to a point in . Using Exercise 3 we conclude that is contractible to a point in also.

Now suppose that is an arbitrary simple closed curve. Let be a small parameter. As before, we can find a simple polygonal path whose interior lies in the interior of , and such that is homotopic to in , and hence in if is small enough, for some . From the previous discussion we see that is contractible to a point in , and so is also. The claim then follows (after reversing the contour if necessary). This concludes the proof of the Jordan curve theorem.

Exercise 58Let be a simple closed polygonal path with all edges of positive length. Suppose that all interior angles of (that is, the angle that two adjacent edges make in the interior of the polygon) are less than or equal to . Show that the interior of is convex. (Hint:use a continuity argument to show that every line meets the interior of in at most one interval.)

Exercise 59Let be a non-trivial simple closed polygonal curve, and let be a point in the interior of an edge of (i.e., is not one of the two vertices of ). Let be two points sufficiently close to that lie on opposite sides of . Without using the Jordan curve theorem, show that . (Hint:replace by a “local” closed contour that is quite short, and a “global” closed contour which avoids the line segment connecting and . Then use the Cauchy integral formula.)

Exercise 60 (Jordan arc theorem)Let be a simplenon-closed curve. Show that the complement of in is connected. (Hint:first establish a variant of Proposition 57 for non-closed curves, in which is now set to zero. Then adapt the proof of the Jordan curve theorem.

Exercise 61Let be a bounded connected non-empty open subset of . Show that is simply connected if and only if the complement is connected. (Hint:suppose that there is a point in that is separated from infinity by . Show that there is some compact subset of that also separates from infinity. Then cover by small squares as in the proof of the Jordan curve theorem to locate a simple closed polygonal path in that separates from infinity.)

(The exercises below were added after the notes were first released; they will ultimately be moved to a more appropriate location, but are being placed here for now in order to not disrupt existing numbering.)

Exercise 62Let be a simply connected subset of , and let be a harmonic function. Show that has a harmonic conjugate , which is unique up to additive constants.

One can interpret Cauchy’s theorem through the lens of algebraic topology, and particularly through the machinery of homology and cohomology. We will not develop this perspective in depth in these notes, but the following exercise will give a brief glimpse of the connections to homology and cohomology.

Exercise 63Let be an open subset of . Define a -chain in to be a formal linear combinationof points (which we enclose in brackets to avoid confusion with the arithmetic operations on , in particular is

notidentified with ), where is a natural number and the are integers; these form an additive abelian group in the usual fashion. Similarly, define a-chainin to be a formal linear combinationof curves in , which (for very minor notational reasons) we will fix to have domain in the unit interval . Finally, define a

-chainin to be a formal linear combinationof -simplices , defined as continuous maps from the solid triangle .

Given a -chain in , we define its

boundaryto be the -chainand call a

-cycleif . Similarly, given a -chain , we define itsboundaryto be the -chainwhere is the curve on that maps to , and similarly for and . If is a -cycle, and is holomorphic, define the integral by

If lies outside of a -cycle , define the

winding number

- (i) Show that if is a -chain in , then .
- (ii) Show that if is a -chain in and is holomorphic, then .
- (iii) If is a -cycle in , and for all , show that . (
Hint:first perturb to be the union of line segments coming from a grid of some small sidelength . Observe that the winding number is constant whenever ranges in the interior of one of the squares in this grid. Then find another -cycle coming from summing boundaries of such squares such that for all in the interior of grid squares. Then show that and .)- (iv) If is a -cycle, and has an antiderivative, show that .
- (v) If is a -cycle, for all , is holomorphic, and is a point lying outside of any of the , show that

Exercise 64Let be a subset of the complex plane which is star-shaped, which means that there exists such that for any , the line segment is also contained in . Show that every star-shaped set is simply connected.

1

## 50 comments

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2 October, 2016 at 10:54 am

AnonymousIn the first line of remark 19, it seems clearer to add “of ” after “open set “.

[Added – T.]2 October, 2016 at 11:26 am

AnonymousIt seems that corollary 24, implies a stronger version of 23 in which on a subset of which has an accumulation point (because the difference has an accumulation point of zeros, hence by corollary 24 should vanish identically on ).

[Remark to this effect added, thanks – T.]2 October, 2016 at 12:50 pm

AnonymousIn theorem 34, it seems that the uniform convergence assumption can be relaxed (using the same proof!) to uniform convergence on compact subsets of (i.e. locally uniform convergence).

[Fair enough; I’ve reworded the theorem and proof accordingly -T.]3 October, 2016 at 12:06 am

AnonymousIn exercise 17, property (i) implies (via its application to an appropriate rotation of ) a stronger version of the maximum principle (that has no local maximum unless is constant). Similarly, property (ii) implies that has no local minimum or maximum unless it is constant.

[Extended the exercise to reflect a version of this – T.]3 October, 2016 at 1:59 am

AnonymousIn cases (iii), (iv) of definition 1, it seems clearer to add “of ” at the end.

[Corrected, thanks. -T.]3 October, 2016 at 12:29 pm

AnonymousDefinition 1(ii), last line should be {\gamma(s,0) =z_0, \gamma(s,1) = z_1}.

[Corrected, thanks -T.]3 October, 2016 at 10:18 pm

AnonymousIn theorem 15 cauchy integral formula( special case)

As {f} is complex differentiable at {z_0}, there exists a finite {M} such that

Shouldn’t it be( f(z)-f(zo) /(z-zo)) – f'(z) <=M

[Corrected, thanks – T.]4 October, 2016 at 5:10 am

Steven GubkinSince the integral of about the unit circle is (perhaps) the most important computation in complex analysis, I think it is worthwhile seeing the computation done in a few ways. I think, in addition to the direct computation, it is valuable to see the computation done using branches of the logarithm.

Namely, has no global antiderivative, but it does have an antiderivative on minus the positive imaginary axis, and on minus the negative imaginary axis. We can then break our computation of the integral about the circle into a sum of integrals about the top and bottom hemisphere. On each hemisphere we can use the fundamental theorem of calculus. Use the corresponding branch of the logarithm to see that each contributes a sum of to the integral. So the total integral is .

[I plan to return to this point in later notes when discussing branch cuts of multi-valued functions – T.]4 October, 2016 at 5:18 am

Steven GubkinAnother approach to the same integral is to realize that on the unit circle. So

Here I have used Stoke’s theorem on line 2, and on line 5 I used the fact that .

It is interesting to me that these two computations get the “” in two different ways: as arising from the circumference and area of the unit circle.

4 October, 2016 at 6:39 am

AnonymousIn corollary 22, it may be added that is uniquely determined (even under the requirement of being merely continuous at ).

[Fair enough – the corollary has been amended to reflect this – T.]4 October, 2016 at 10:47 pm

AnonymousThe links in exercise 17(i),(ii) are not working.

[Corrected, thanks – T.]5 October, 2016 at 1:12 pm

AnonymousIn the proof of theorem 15, in line 7 it should be “sufficiently small ” (instead of ).

[Corrected, thanks – T.]5 October, 2016 at 8:20 pm

Jhon ManugalFor a small enough square and that’s good enough. And we can do by adding over the sides of all the squares and the squares cancel out.

$$

I dunno that looks pretty good to me :-)

6 October, 2016 at 7:15 am

AnonymousThe remark which goes after cauchy’s integral theorem proof

“One nice feature of Cauchy’s theorem is that it allows one to integrate holomorphic functions on curves that are not necessarily rectifiable”

And then gamma is given to be a contour in U

Then cauchy integral for non rectifiable curve gamma is defined

I am not really sure, but I think gamma should be defined as just a curve in U as any contour is rectifiable

[Corrected, thanks – T.]7 October, 2016 at 3:07 am

AnonymousSomething seems to have gone wrong in the latex of exercise 54.

[Corrected, thanks – T.]7 October, 2016 at 6:20 am

Anonymousfor the definition of homotopy, since the interval for curves are [a,b], so the endpoint should be , that’s for (ii), for (i), the same.

[Corrected, thanks – T.]7 October, 2016 at 6:21 am

AnonymousIn the first diagram below exercise 37, the notion of “real analytic” is not defined.

[See Remark 19 – T.]8 October, 2016 at 2:07 am

AnonymousI meant that this notion is mentioned without definition (also in remark 19).

Perhaps a short definition (e.g. local representation by power series on real intervals) can make it clearer.

[Short definition added to Remark 19. -T.]7 October, 2016 at 6:50 pm

hFYI The article “Timothy Gowers – Elsevier — my part in its downfall” that linked to on the left column of your blog seems to have been removed by Gowers.

Does that mean Elsevier is not so bad now?

[The link seems to be working fine for me – T.]9 October, 2016 at 6:25 am

AnonymousIf is any curve with image in some open set , is there a complex measure (depended only on ) over such that for all functions which are analytic on ?

If such a measure exist, it can be used to extend the integral over to the larger class of -measurable functions.

[For rectifiable curves, such a measure can be constructed by the Riesz representation theorem. Perhaps there is a Hahn-Banach type construction to produce such a measure in the non-rectifiable case, but it might not be unique. -T]11 October, 2016 at 11:01 am

Math 246A, Notes 4: singularities of holomorphic functions | What's new[…] the previous set of notes we saw that functions that were holomorphic on an open set enjoyed a large number of useful […]

11 October, 2016 at 12:55 pm

AnonymousAs stated, 49 (iii) seems incorrect. Imagine $\gamma_2$ being a concentric circle with $\gamma_1$, having a larger radius.

[Corrected, thanks – T.]11 October, 2016 at 1:23 pm

Lior SilbermanAfter Theorem 34, I think it’s worth noting that even for notions of “convergence on average” such as or , the limit of a sequence of holomophic functions is holomorphic because one can average Cauchy’s integral formula over an annulus to obtain a kernel which is absolutely continuous with respect to Lebesgue measure.

[Added a remark to this effect – T.]18 October, 2016 at 9:32 pm

246A, Notes 5: conformal mapping | What's new[…] then on any ball in , is the real part of some holomorphic function thanks to Exercise 62 of Notes 3. By part (ii), is also holomorphic. Taking real parts we see that is harmonic on each ball in , […]

22 October, 2016 at 10:30 am

snarfblaatexercise 17(iii)–what do and refer to?

the short proof of cor.13–the references to the first and second fundamental theorems perhaps reversed?

exercise 12– is I think a closed rectangle, but the sequence of vertices in doesn’t loop back

[Corrected, thanks – T.]1 November, 2016 at 10:39 am

UCLA Math Grad LoungeExercise 3 iii) seems wrong to me. Taking U = C \ {0,1}, wouldn’t this exercise show that conjugation on the free group with 2 elements is trivial?

[Oops, there is a conjugation missing in this exercise – fixed now. -T.]18 February, 2017 at 5:55 am

AnonymousIn the second Euler diagram (the complex case), should “smooth” be in the same box as “analytic”?

25 February, 2017 at 10:07 am

Terence TaoNo; for instance, the function is smooth (in the real sense), but not (complex) analytic.

25 February, 2017 at 1:42 pm

AnonymousFair enough. I thought “smooth” means infinitely complex differentiable in this set of notes. It seems that one would never say “a smooth complex function” though.

21 February, 2017 at 7:48 am

Anonymous– According to the definition 1, is the figure considered as homotopic to a point?

– Is there a particular reason we need to consider “fixed-endpoints” homotopy instead of the general one that does not require the endpoints being fixed?

25 February, 2017 at 10:25 am

Terence TaoIt depends on what domain one is working in. For instance, in a simply connected domain, all closed curves are homotopic to a point (including a figure in the domain). But if the domain has “holes” in one or both of the regions enclosed by this figure , then this figure will not be homotopic to a point in that domain.

Homotopy without fixing the endpoints or requiring the curve to be closed is a rather trivial relation: see Exercise (vi). Related to this, Cauchy’s theorem (Theorem 4) is highly false if one allows the endpoints to vary.

5 March, 2017 at 8:09 am

AnonymousShould be a closed curve in instead of in Theorem 39?

[Corrected, thanks – T.]5 March, 2017 at 10:54 am

AnonymousIt shows “Formula does not parse” now.

[Corrected, thanks – T.]22 March, 2017 at 7:04 am

AnonymousTypo in Exercise 36:

the function

dz->dt

25 March, 2017 at 10:43 am

AnonymousIn the proof of Corollary 22, all the three are with different domains while the third one is defined by “gluing” the first two together. (Might be clearer if one using the notations and ).

Would you elaborate what “Uniqueness at {z_0} follows from continuity.” means? (This is a little confusing. I guess it means the uniqueness at for the in the statement of the Corollary, not the first in the proof. )

[Reworded – T.]25 March, 2017 at 11:01 am

AnonymousIn the proof of Corollary 23,

“by the continuity of the is closed”

might be clearer if a comma is added:

“by the continuity of the , is closed”

—

A remark on the “continuity method” could be added, again;-) http://www.tricki.org/article/Use_the_continuity_method

[Corrected, thanks – T.]25 March, 2017 at 11:17 am

AnonymousRegarding Corollary 24, can one define an analytic function on such that has zeros on a given infinite set of isolated points contained in .

[If the set of points is closed, then yes: see http://mathworld.wolfram.com/WeierstrassProductTheorem.html -T]5 June, 2017 at 10:04 am

Five-Value Theorem of Nevanlinna – Elmar Klausmeier's Weblog[…] Math 246A, Notes 3: Cauchy’s theorem and its consequences […]

26 September, 2017 at 4:10 am

AnonymousIn Exercise 36: Shouldn’t $f(t, z)$ be integrated over $dt$ instead of $dz$?

[Corrected, thanks – T.]29 October, 2017 at 7:19 am

AnonymousI have some dumb questions for Remark 16.

In Stein-Shakarchi Chapter 2, the keyhole contour is used for proving the Cauchy integral formula. But I’m quite uncomfortable with a hand-waving argument in their proof:

“use the continuity of F to see that in the limit, the integrals over the two sides of the corridor cancel out.”

How would you turn this into a rigorous estimate?

(In general, how often do you allow hand-waving arguments in your notes? )

29 October, 2017 at 7:38 am

AnonymousThe above is defined as

.

29 October, 2017 at 9:26 am

Terence TaoIn this particular case, this is not difficult; one simply parameterises both sides of the corridor to turn the integral into a Riemann (or Lebesgue) integral and then uses one of the standard convergence theorems (dominated convergence will certainly work, though this is overkill here since one in fact has uniform convergence).

8 November, 2017 at 4:53 pm

AnonymousAs I understand, this set of notes make the “minimum” assumption (being “rectifiable”) of a curve on which one wants to do complex integration. On the other hand, lots of the complex analysis textbooks (such as Stein-Shakarchi) make the stronger assumption that the curve is piecewise (infinitely) differentiable.

What could be a big picture in practice that (1) when one needs the full power of full generality; (2) when one would be OK with only the piecewise differentiable curves?

30 December, 2017 at 9:29 am

AnonymousIs the real line on the complex plane considered as a “closed” contour? How does one get a version of Cauchy integral formula for this one? For instance

30 December, 2017 at 4:04 pm

Terence TaoThe real line is not a closed contour in the complex plane, though if one can extend the -form holomorphically to the Riemann sphere then the extended real line becomes a closed contour that one can integrate the form on. In practice, one can often use the Cauchy integral formula to integrate on the real line by working with a large semicircular contour and sending the radius to infinity, but care must be taken in ensuring that the contribution of the semicircular arc becomes negligible in the limit. To this end, tools such as Jordan’s lemma are often employed.

31 December, 2017 at 4:34 pm

AnonymousWhat decay bound should obey so that Jordan’s lemma can be applied in the above example? In your lecture notes on Hilbert transform: http://www.math.ucla.edu/~tao/247a.1.06f/notes4.pdf, you suggest that would be enough. But it seems that one cannot get

here.

31 December, 2017 at 6:14 pm

Terence TaoFor arbitrary (which does not have a good analytic continuation to the upper half-plane), one does not expect to have this formula. Instead, one has the Plemelj formulae for general (and reasonably smooth and decaying) : https://en.wikipedia.org/wiki/Sokhotski%E2%80%93Plemelj_theorem

18 February, 2018 at 5:23 pm

JuhoLDear Terence.

Don’t you find it odd that the Goursat’s theorem takes such an “odd” proof? I mean that the most obvious guess might be to have a triangle (or rectangle) that is null-homotopic curve (t=1 is triangle and t=0 is a point inside the triangle) and show that the derivate of the integral in question with respect to homotopy parameter t is zero and then infer that the value of the integral is independent of t and it must be zero since the case t=0 is clearly zero.

But apparently this method is a dead-end since no-one has published it. Maybe that assumption of non-continuous derivative requires some deeper tricks.

19 February, 2018 at 3:20 pm

Terence TaoOne can certainly do this if one assumes enough regularity to be able to differentiate under the integral sign, and similar methods are used to establish conservation laws or monotonicity formulae in PDE or geometry. But as you say, it is the lack of continuity of derivative that forces one to move away from the most natural (but regularity-intensive) techniques and towards the proof by contradiction style of proof given here.

19 February, 2018 at 8:29 pm

JuhoLThanks for reply!

I had the following idea for the case of rectangle R. If we simply divide horizontal and vertical sides into n equal pieces we have a nice grid of similar subrectangles. And when we increase the number n, then it’s clear that every subrectangle R_i is inside a small neighbourhood of _some_ point z_i in R and we can estimate the integral over a subrectangle R_i with the error term E_i (z) of that neighbourhood of z_i (same estimate for every index i due to geometric construction of the grid). Since the integral over R is equal to the sum of integrals over all subrectangles we can use triangle inequality and have the same conclusion like with Goursat’s method.

This would be a student friendly direct geometric method that clearly indicates why we must assume derivative everywhere in R, and also it removes that annoying topological argument.