Having discussed differentiation of complex mappings in the preceding notes, we now turn to the integration of complex maps. We first briefly review the situation of integration of (suitably regular) real functions of one variable. Actually there are *three* closely related concepts of integration that arise in this setting:

- (i) The signed definite integral , which is usually interpreted as the Riemann integral (or equivalently, the Darboux integral), which can be defined as the limit (if it exists) of the Riemann sums
where is some partition of , is an element of the interval , and the limit is taken as the maximum mesh size goes to zero. It is convenient to adopt the convention that for ; alternatively one can interpret as the limit of the Riemann sums (1), where now the (reversed) partition goes leftwards from to , rather than rightwards from to .

- (ii) The
*unsigned definite integral*, usually interpreted as the Lebesgue integral. The precise definition of this integral is a little complicated (see e.g. this previous post), but roughly speaking the idea is to approximate by simple functions for some coefficients and sets , and then approximate the integral by the quantities , where is the Lebesgue measure of . In contrast to the signed definite integral, no orientation is imposed or used on the underlying domain of integration, which is viewed as an “undirected” set . - (iii) The
*indefinite integral*or antiderivative , defined as any function whose derivative exists and is equal to on . Famously, the antiderivative is only defined up to the addition of an arbitrary constant , thus for instance .

There are some other variants of the above integrals (e.g. the Henstock-Kurzweil integral, discussed for instance in this previous post), which can handle slightly different classes of functions and have slightly different properties than the standard integrals listed here, but we will not need to discuss such alternative integrals in this course (with the exception of some improper and principal value integrals, which we will encounter in later notes).

The above three notions of integration are closely related to each other. For instance, if is a Riemann integrable function, then the signed definite integral and unsigned definite integral coincide (when the former is oriented correctly), thus

and

If is continuous, then by the fundamental theorem of calculus, it possesses an antiderivative , which is well defined up to an additive constant , and

for any , thus for instance and .

All three of the above integration concepts have analogues in complex analysis. By far the most important notion will be the complex analogue of the signed definite integral, namely the contour integral , in which the directed line segment from one real number to another is now replaced by a type of curve in the complex plane known as a contour. The contour integral can be viewed as the special case of the more general line integral , that is of particular relevance in complex analysis. There are also analogues of the Lebesgue integral, namely the arclength measure integrals and the area integrals , but these play only an auxiliary role in the subject. Finally, we still have the notion of an antiderivative (also known as a *primitive*) of a complex function .

As it turns out, the fundamental theorem of calculus continues to hold in the complex plane: under suitable regularity assumptions on a complex function and a primitive of that function, one has

whenever is a contour from to that lies in the domain of . In particular, functions that possess a primitive must be conservative in the sense that for any closed contour. This property of being conservative is not typical, in that “most” functions will not be conservative. However, there is a remarkable and far-reaching theorem, the Cauchy integral theorem (also known as the Cauchy-Goursat theorem), which asserts that any holomorphic function is conservative, so long as the domain is simply connected (or if one restricts attention to contractible closed contours). We will explore this theorem and several of its consequences the next set of notes.

** — 1. Integration along a contour — **

The notion of a curve is a very intuitive one. However, the precise mathematical definition of what a curve actually is depends a little bit on what type of mathematics one wishes to do. If one is mostly interested in topology, then a good notion is that of a continuous (parameterised) curve. It one wants to do analysis in somewhat irregular domains, it is convenient to restrict the notion of curve somewhat, to the rectifiable curves. If one is doing analysis in “nice” domains (such as the complex plane , a half-plane, a punctured plane, a disk, or an annulus), then it is convenient to restrict the notion further, to the *piecewise smooth curves*, also known as contours. If one wished to get to the main theorems of complex analysis as quickly as possible, then one would restrict attention only to contours and skip much of this section; however we shall take a more leisurely approach here, discussing curves and rectifiable curves as well, as these concepts are also useful outside of complex analysis. In fact, we will structure our notes so that most of our theorems in fact apply to rectifiable curves (and several will in fact be applicable to arbitrary continuous curves).

We begin by defining the notion of a continuous curve.

Definition 1 (Continuous curves)Acontinuous parameterised curve, orcurvefor short, is a continuous map from a compact interval to the complex plane . We call the curvetrivialif , andnon-trivialotherwise. We refer to the complex numbers as theinitial pointandterminal pointof the curve respectively, and refer to these two points collectively as theendpointsof the curve. We say that the curve isclosedif . We say that the curve issimpleif one has for any distinct , with the possible exception of the endpoint cases or (thus we allow closed curves to be simple). We refer to the subset of the complex plane as theimageof the curve.

We caution that the term “closed” here does *not* refer to the topological notion of closure: for any curve (closed or otherwise), the image of the curve, being the continuous image of a compact set, is necessarily a compact subset of and is thus always topologically closed.

A basic example of a curve is the directed line segment from one complex point to another , defined by

for . (Thus, contrary to the informal English meaning of the terms, we consider line segments to be examples of curves, despite having zero curvature; in general, it is convenient in mathematics to admit such “degenerate” objects into one’s definitions, in order to obtain good closure properties for these objects, and to maximise the generality of the definition.) If , this is a simple curve, while for it is (a rather degenerate, but still non-trivial) closed curve. Another important example of a curve is the anti-clockwise circle of some radius around a complex centre , defined by

This is a simple closed non-trivial curve. If we extended the domain here from to (say) , the curve would remain closed, but would no longer be simple (every point in the image is now traversed twice by the curve).

Note that it is technically possible for two distinct curves to have the same image. For instance, the anti-clockwise circle of some radius around a complex centre defined by

traverses the same image as the previous curve (2), but is considered a distinct curve from . Nevertheless the two curves are closely related to each other, and we formalise this as follows. We say that one curve is a *continuous reparameterisation* of another , if there is a homeomorphism (that is to say, a continuous invertible map whose inverse is also continuous) which is endpoint preserving (i.e., and ) such that for all (that is to say, , or equivalently ), in which case we write . Thus for instance . The relation of being a continuous reparameterisation is an equivalence relation, so one can talk about the notion of a curve “up to continuous reparameterisation”, by which we mean an equivalence class of a curve under this relation. Thus for instance the image of a curve, as well as its initial point and end point, are well defined up to continuous reparameterisation, since if then and have the same image, the same initial point, and the same terminal point. It is common to depict an equivalence class of a curve graphically, by drawing its image together with an arrow depicting the direction of motion from the initial point to its endpoint. (In the case of a non-simple curve, one may need multiple arrows in order to clarify the direction of motion, and also the possible multiplicity of the curve.)

Exercise 2Let be a continuous invertible map.

- (i) Show that is continuous, so that is a homeomorphism. (
Hint:use the fact that a continuous image of a compact set is compact, and that a subset of an interval is topologically closed if and only if it is compact.)- (ii) If , show that and that is monotone increasing. (
Hint:use the intermediate value theorem.)- (iii) Conversely, if is a continuous monotone increasing map with and , show that is a homeomorphism.

It will be important for us that we do not allow reparameterisations to reverse the endpoints. For instance, if are distinct points in the complex plane, the directed line segment is *not* a reparameterisation of the directed line segment since they do not have the same initial point (or same terminal point); the map is a homeomorphism from to but it does not preserve the initial point or the terminal point. In general, given a curve , we define its *reversal* to be the curve , thus for instance is (up to reparameterisation) the reversal of , thus

Another basic operation on curves is that of concatenation. Suppose we have two curves and with the property that the terminal point of equals the initial point of . We can reparameterise by translation to , defined by . We then define the *concatenation* or *sum* by setting

for and

for (note that these two definitions agree on their common domain point by the hypothesis . It is easy to see that this concatenation is still a continuous curve. The reader at this point is encouraged to draw a picture to understand what the concatenation operation is doing; it is much simpler to grasp it visually than the above lengthy definition may suggest. If the terminal point of does not equal the initial point of , we leave the sum undefined. (One can define more general spaces than the space of curves in which such an addition can make sense, such as the space of -currents if one assumes some rectifiability on the curves, but we will not need such general spaces here.)

Concatenation is well behaved with respect to equivalence and reversal:

Exercise 3Let be continuous curves. Suppose that the terminal point of equals the initial point of , and the terminal point of equals the initial point of .

- (i) (Concatenation well defined up to equivalence) If and , show that and .
- (ii) (Concatenation associative) Show that . In particular, we certainly have
- (iii) (Concatenation and reversal) Show that .
- (iv) (Non-commutativity) Give an example in which and are both well-defined, but not equivalent to each other.
- (v) (Identity) If and denote the initial and terminal points of respectively, and is the trivial curve defined for any , show that and .
- (vi) (Non-invertibility) Give an example in which is not equivalent to a trivial curve. (It will however be homologous to a trivial curve, as we will discuss in later notes.)

Remark 4The above exercise allows one to view the space of curves up to equivalence as a category, with the points in the complex plane being the objects of the category, and each equivalence class of curves being a single morphism from the initial point to the terminal point (and with the equivalence class of trivial curves being the identity morphisms). This point of view can be useful in topology, particularly when relating to concepts such as the fundamental group (and fundamental groupoid), monodromy, and holonomy. However, we will not need to use any advanced category-theoretic concepts in this course.

Exercise 5Let and . For any integer , let denote the curve(thus for instance ).

- (i) Show that for any integer , we have .
- (ii) Show that for any non-negative integers , we have . What happens for other values of ?
- (ii) If are distinct integers, show that .

Given a sequence of complex numbers , we define the *polygonal path* traversing these numbers in order to be the curve

This is well-defined thanks to Exercise 3(ii) (actually all we really need in applications is being well-defined up to equivalence). Thus for instance would traverse a closed triangular path connecting , , and (this path may end up being non-simple if the points are collinear).

In order to do analysis, we need to restrict our attention to those curves which are rectifiable:

Definition 6Let be a curve. The arc length of the curve is defined to be the supremum of the quantitieswhere ranges over the natural numbers and ranges over the partitions of . We say that the curve is

rectifiableif its arc length is finite.

The concept is best understood visually: a curve is rectifiable if there is some finite bound on the length of polygonal paths one can form while traversing the curve in order. From Exercise 2 we see that equivalent curves have the same arclength, so the concepts of arclength and rectifiability are well defined for curves that are only given up to continuous reparameterisation.

Exercise 7Let be curves, with the terminal point of equal to the initial point of . Show thatIn particular, is rectifiable if and only if are both individually rectifiable.

It is not immediately obvious that any reasonable curve (e.g. the line segments or the circles ) are rectifiable. To verify this, we need two preliminary results.

Lemma 8 (Triangle inequality)Let be a continuous function. ThenHere we interpret as the Riemann integral (or equivalently, ).

*Proof:* We first attempt to prove this inequality by considering the real and imaginary parts separately. From the real-valued triangle inequality (and basic properties of the Riemann integral) we have

and similarly

but these two bounds only yield the weaker estimate

To eliminate this loss we can amplify the above argument by exploiting phase rotation. For any real , we can repeat the above arguments (using the complex linearity of the Riemann integral, which is easily verified) to give

But we have for any complex number , so taking the supremum of both sides in we obtain the claim.

Exercise 9Let be real numbers. Show that the interval is topologically connected, that is to say the only two subsets of that are both open and closed relative to are the empty set and all of . (Hint:if is a non-empty set that is both open and closed in and contains , consider the supremum of all such that .)

Next, we say that a non-trivial curve is *continuously differentiable* if the derivative

exists and is continuous for all (note that we are only taking right-derivatives at and left-derivatives at ).

Proposition 10 (Arclength formula)If is a continuously differentiable curve, then it is rectifiable, and

*Proof:* We first prove the upper bound

which in particular implies the rectifiability of since the right-hand side of (3) is finite. Let be any partition of . By the fundamental theorem of calculus (applied to the real and imaginary parts of ) we have

for any , and hence by Lemma 8 we have

Summing in we obtain

and taking suprema over all partitions we obtain (3).

Now we need to show the matching lower bound. Let be a small quantity, and for any , let denote the restriction of to . We will show the bound

for all ; specialising to and then sending will give the claim.

It remains to prove (4) for a given choice of . We will use a continuous version of induction known as the continuity method{continuity method}, which exploits Exercise 9.

Let denote the set of such that (4) holds for all . It is clear (using Exercise 7) that this set is topologically closed, and also contains the left endpoint of . If and , then from the differentiability of at , we have some interval such that

for all . Rearranging this using the triangle inequality, we have

Also, from the continuity of we have

for all , if is small enough. We conclude that

and hence

where is the restriction of to . Adding this to the case of (4) using (7) we conclude that (4) also holds for all . From this we see that is (relatively) open in ; from the connectedness of we conclude that , and we are done.

It is now easy to verify that the line segment is rectifiable with arclength , and that the circle is rectifiable with arclength , exactly as one would expect from elementary geometry. Finally, from Lemma 7, a polygonal path will be rectifiable with arclength , again exactly as one would expect.

Exercise 11Show that the curve defined by setting for and is continuous but not rectifiable. (Hint:it is not necessary to compute the arclength precisely; a lower bound that goes to infinity will suffice. Graph the curve to discover some convenient partitions with which to generate such lower bounds. Alternatively, one can apply the arclength formula to some subcurves of .)

Exercise 12(This exercise presumes familiarity with Lebesgue measure.) Show that the image of a rectifable curve is necessarily of measure zero in the complex plane. (In particular, space-filling curves such as the Peano curve or the Hilbert curve cannot be rectifiable.)

Remark 13As the above exercise suggests, many fractal curves will fail to be rectifiable; for instance the Koch snowflake is a famous example of an unrectifiable curve. (The situation is clarified once one develops the theory of Hausdorff dimension, as is done for instance in this previous post: any curve of Hausdorff dimension strictly greater than one will be unrectifiable.)

Much as continuous functions on an interval may be integrated by taking limits of Riemann sums, we may also integrate continuous functions on the image of a rectifiable curve :

Proposition 14 (Integration in rectifiable curves)Let be a rectifiable curve, and let be a continuous function on the image of . Then the “Riemann sums”

where ranges over the partitions of , and for each , is an element of , converge as the maximum mesh size goes to zero to some complex limit, which we will denote as . In other words, for every there exists a such that

whenever .

*Proof:* In real analysis courses, one often uses the order properties of the real line to replace the rather complicated looking Riemann sums with the simpler Darboux sums, en route to proving the real-variable analogue of the above proposition. However, in our complex setting the ordering of the real line is not available, so we will tackle the Riemann sums directly rather than try to compare them with Darboux sums.

It suffices to prove that the “Riemann sums” (5) are a Cauchy sequence, in the sense that the difference

between two sums of the form (5) is smaller than any specified if the maximum mesh sizes of the two partitions and are both small enough. From the triangle inequality, and from the fact that any two partitions have a common refinement, it suffices to prove this under the additional assumption that the second partition is a refinemnt of . This means that there is an increasing sequence of natural numbers such that for . In that case, the above difference may be rearranged as

where

By telescoping series, we may rearrange further as

As is continuous and is compact, is uniformly continuous. In particular, if the maximum mesh sizes are small enough, we have

for all and . From the triangle inequality we conclude that

and hence on summing in and using the triangle inequality, we can bound

Since is finite, and can be made arbitrarily small, we obtain the required Cauchy sequence property.

One cannot simply omit the rectifiability hypothesis from the above proposition:

Exercise 15Give an example of a curve such that the Riemann sumsfail to converge to a limit as the maximum mesh size goes to zero, so the integral does not exist in the sense of convergent Riemann sums even though the integrand is extremely smooth. (Of course, such a curve cannot be rectifiable, thanks to Proposition 14.)

Hint:the non-rectifiable curve in Exercise (11) is a good place to start, but it turns out that this curve does not oscillate wildly enough to make the Riemann sums here diverge, because of the decay of the function near the origin. Come up with a variant of this curve which oscillates more. Nevertheless, it is still possible to assign a meaning to integrals such as , as we shall see in the next set of notes once Cauchy’s theorem is established.

By abuse of notation, we will refer to the quantity as the *contour integral* of along , even though is not necessarily a contour (we will define this concept shortly). We have some easy properties of this integral:

Exercise 16Let be a rectifiable curve, and be a continuous function.

- (i) (Independence of parameterisation) If is another curve equivalent to , show that .
- (ii) (Reversal) Show that .
- (iii) (Concatenation) If for some curves , show that
- (iv) (Change of variables) If is continuously differentiable, show that
- (v) (Upper bound) If there is a pointwise bound of the form for all and some , show that
- (vi) (Linearity) If is a complex number and is a continuous function, show that
and

- (vii) (Integrating a constant) If has initial point and terminal point , show that
- (viii) (Uniform convergence) If , is a sequence of continuous functions converging uniformly to as , show that converges to as . (The requirement of uniform convergence can be relaxed substantially, thanks to tools such as the dominated convergence theorem, but this weaker convergence theorem will suffice for most of our applications.)
- (ix) (Change of variables, II) Let be an open neighbourhood of , let be a holomorphic function, and let be a continuous function. Show that
(For this question, you may assume without proof that holomorphic functions are continuously differentiable; this fact will be proven (without reliance on this part of the exercise) in the next set of notes.)

Exercise 17(This exercise assumes familiarity with the Riemann-Stieltjes integral.) Let be a rectifiable curve. Let denote the monotone non-decreasing functionfor , where is the restriction of to . For any continuous function , define the

arclength measure integralby the formulawhere the right-hand side is a Riemann-Stieltjes integral. Establish the triangle inequality

for any continuous . Also establish the identity

and obtain an alternate proof of Exercise 16(v).

The change of variables formula (iv) lets one compute many contour integrals using the familiar Riemann integral. For instance, if are real numbers and is continuous, then the contour integral along coincides with the Riemann integral,

and on reversal we also have

Similarly, if is continuous on the circle , we have

Remark 18We caution that if is real-valued, we cannot conclude that is also real valued, unless the contour lies in the real line. This is because the complex line element may introduce some non-trivial imaginary part, as is the case for instance in (6). For similar reasons, we haveand

in general. If one wishes to mix line integrals with real and imaginary parts, it is recommended to replace the contour integrals above with the line integrals

which are defined as in Proposition 14 but where the expression

appearing in (5) is replaced by

The contour integral corresponds to the special case (or more informally, ). Line integrals are in turn special cases of the more general concept of integration of differential forms, discussed for instance in this article of mine, and which are used extensively in differential geometry and geometric topology. However we will not use these more general line integrals or differential form integrals much in this course.

In later notes it will be convenient to restrict to a more regular class of curves than the rectifiable curves. We thus give the definitions here:

Definition 19 (Smooth curves and contours)Asmooth curveis a curve that is continuously differentiable, and such that for all . Acontouris a curve that is equivalent to the concatenation of finitely many smooth curves (that is to say, apiecewise smoothcurve).

Example 20The line segments and circles are smooth curves and hence contours. Polygonal paths are usually not smooth, but they are contours. Any sum of finitely many contours is again a contour, and the reversal of a contour is also a contour.

Note here that the term “smooth” differs somewhat here from the real-variable notion of smoothness, which is defined to be “infinitely differentiable”. Smooth curves are still only assumed to just be continuously differentiable; we do not assume that the second derivative of exists. (In particular, smooth curves may have infinite curvature at some points.) In practice, this distinction tends to be minor, though, as the smooth curves that one actually uses in complex analysis do tend to be infinitely differentiable. On the other hand, for most applications one does not need to control any derivative of a contour beyond the first.

The following examples and exercises may help explain why the non-vanishing condition imbues curves with a certain degree of “smoothness”.

Example 21 (Cuspidal curve)Consider the curve defined by . Clearly is continously differentiable (and even infinitely differentiable), but we do not view this curve as smooth, because vanishes at the origin. Indeed, the image of the curve is , which looks visibly non-smooth at the origin if one plots it, due to the presence of a cusp.

Example 22 (Absolute value function)Consider the curve defined by . This curve is certainly continuously differentiable, and in fact is four times continuously differentiable, but is not smooth because vanishes at the origin. The image of this curve is , which looks visibly non-smooth at the origin (in particular, there is no unique tangent line to this curve here).

Example 23 (Spiral)Consider the curve defined by for , and . One can check (exercise!) that is continuously differentiable, even at the origin ; but it is not a smooth curve because vanishes. The image of has some rather complicated behaviour at the origin, for instance it intersects itself multiple times (try to sketch it!).

Exercise 24 (Local behaviour of smooth curves)Let be a simple smooth curve, and let be an interior point of . Let be a phase of , thus for some . Show that for all sufficiently small , the portion of the image of near looks like a rotated graph, in the sense thatfor some interval containing the origin, and some continuously differentiable function with . Furthermore, show that as (or equivalently, that as ). (

Hint:you may find it easier to first work with the model case where and . The real-variable inverse function theorem will also be helpful.)

Exercise 25Show that a curve is a contour if and only if it is equivalent to the concatenation of finitely manysimplesmooth curves.

Exercise 26Show that the cuspidal curve and absolute value curves in Examples 21, 22 are contours, but the curve in Exercise 23 is not.

** — 2. The fundamental theorem of calculus — **

Now we establish the complex analogues of the fundamental theorem of calculus. As in the real-variable case, there are two useful formulations of this theorem. Here is the first:

Theorem 27 (First fundamental theorem of calculus)Let be an open subset of , let be a continuous function, and suppose that has an antiderivative , that is to say a holomorphic function with for all . Let be a rectifiable curve in with initial point and terminal point . Then

*Proof:* If were continuously differentiable, or at least *piecewise continuously differentiable* (the concatenation of finitely many continuously differentiable curves), we could establish this theorem by using Exercise 16 to rewrite everything in terms of real-variable Riemann integrals, at which point one can use the real-variable fundamental theorem of calculus (and the chain rule). But actually we can just give a direct proof that does not need any rectifiability hypothesis whatsoever.

We again use the continuity method. Let , and for each , let be the restriction of to . It will suffice to show that

for all , as the claim then follows by setting and sending to zero.

Let denote the set of all such that (7) holds for all . As before, is clearly closed and contains ; as is connected, the only remaining task is to show that is open in . Let be such that . As is differentiable at with derivative , and is continuous, there exists with such that

for any . On the other hand, if is small enough, we have that

for all , and hence by Exercise 16(v) we have

where is the restriction of to . Applying Exercise 16(vi), (vii) we thus have

Combining this with (8) and the triangle inequality, we conclude that

and on adding this to the case of (7) and again using the triangle inequality, we conclude that (7) holds for all . This ensures that is open, as desired.

One can use this theorem to quickly evaluate many integrals by using an antiderivative for the integrand as in the real-variable case. For instance, for any rectifiable curve with initial point and terminal point , we have

and so forth. If the curve avoids the origin, we also have

since is an antiderivative of on . If is a power series with radius of convergence , and is a curve in with initial point and terminal point , we similarly have

For the second fundamental theorem of calculus, we need a topological preliminary result.

Exercise 28Let be a non-empty open subset of . Show that the following statements are equivalent:

- (i) is topologically connected (that is to say, the only two subsets of that are both open and closed relative to are the empty set and itself).
- (ii) is path connected (that is to say, for any there exists a curve with image in whose initial point is and terminal point is )
- (iii) is polygonally path connected (the same as (ii), except that is now required to also be a polygonal path).
(

Hint:to show that (i) implies (iii), pick a base point in and consider the set of all in that can be reached from by a polygonal path.)

We remark that the relationship between path connectedness and connectedness is more delicate when one does not assume that the space is open; every path-connected space is still connected, but the converse need not be true.

Remark 29There is some debate as to whether to view the empty set as connected, disconnected, or neither. I view this as analogous to the debate as to whether the natural number should be viewed as prime, composite, or neither. In both cases I personally prefer the convention of “neither” (and like to use the term “unconnected” to describe the empty set, and “unit” to describe ), but to avoid any confusion I will restrict the discussion of connectedness to non-empty sets in this course (which is all we will need in applications).

In real analysis, the second fundamental theorem of calculus asserts that if a function is continuous, then the function is an antiderivative of . In the complex case, there is an analogous result, but one needs the additional requirement that the function is conservative:

Theorem 30 (Second fundamental theorem of calculus)Let be a non-empty open connected subset of the complex numbers. Let be a continuous function which isconservativein the sense that

whenever is a closed polygonal path in . Fix a base point , and define the function by the formula

for all , where is any polygonal path from to in (the existence of such a path follows from Exercise 28 and the hypothesis that is connected, and the independence of the choice of path for the purposes of defining follows from Exercise 16 and the conservative hypothesis (9)). Then is holomorphic on and is an antiderivative of , thus for all .

*Proof:* We mimic the proof of the real-variable second fundamental theorem of calculus. Let be any point in . As is open, it contains some disk centred at . In particular, if lies in this disk, then the line segment will have image in . If is any polygonal path from to , then we have

and

and hence by Exercise 16

(The reader is strongly advised to draw a picture depicting the situation here.) Let , then for sufficiently close to , we have for all in the image of . Thus by Exercise 16(v) we have

for sufficiently close to , which implies that

for any , and thus is an antiderivative of as required.

The notion of a non-empty open connected subset of the complex plane comes up so frequently in complex analysis that many texts assign a special term to this notion; for instance, Stein-Shakarchi refers to such sets as *regions*, and in other texts they may be called *domains*. We will stick to just “non-empty open connected subset of ” in this course.

The requirement that be conservative is necessary, as the following exercise shows. Actually it is also necessary in the real-variable case, but it is redundant in that case due to the topological triviality of closed polygonal paths in one dimension: see Exercise 33 below.

Exercise 31Let be a non-empty connected subset of , and let be continuous. Show that the following are equivalent:

- (i) possesses at least one antiderivative .
- (ii) is conservative in the sense that (9) holds for all closed polygonal paths in .
- (iii) is conservative in the sense that (9) holds for all
simpleclosed polygonal paths in .- (iv) is conservative in the sense that (9) holds for all closed contours in .
- (v) is conservative in the sense that (9) holds for all closed rectifiable curves in .
(Hint: to show that (iii) implies (ii), induct on the number of edges in the closed polygonal path, and find a way to decompose non-simple closed polygonal paths into paths with fewer edges. One should avoid non-rigorous “hand-waving” arguments, and make sure that one actually has covered all possible cases, e.g. paths that include some backtracking.) Furthermore, show that if has two antiderivatives , then there exists a constant such that .

Exercise 32Show that the function does not have an antiderivative on . (Hint:integrate on .) In later notes we will see that nevertheless does have antiderivatives on many subsets of , formed by various branches of the complex logarithm.

Exercise 33If is a continuous function on an interval, show that is conservative in the sense that (9) holds for any closed polygonal path in . What happens for closed rectifiable paths?

Exercise 34Let be an open subset of (not necessarily connected).

- (i) Show that there is a unique collection of non-empty subsets of that are open, connected, disjoint, and partition : . (The elements of are known as the connected components of .)
- (ii) Show that the number of connected components of is at most countable. (
Hint:show that each connected component contains at least one complex number with rational real and imaginary parts.)- (iii) If is a continuous conservative function on , show that has at least one antiderivative .
- (iv) If has more than one connected component, show that it is possible for a function to have two antiderivatives which do not differ by a constant (i.e. there is no complex number such that ).

Exercise 35 (Integration by parts)Let be holomorphic functions on an open set , and let be a rectifiable curve in with initial point and terminal point . Prove that

## 32 comments

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27 September, 2016 at 11:23 am

AnonymousIn lines 2-3 below (1), it seems that after “as the maximum mesh size” should appear “tends to zero”.

[Corrected, thanks – T.]27 September, 2016 at 12:29 pm

AnonymousBelow exercise 9, in the definition of , it should be .

[Corrected, thanks – T.]27 September, 2016 at 12:49 pm

AnonymousIn the proof of (4), the link to the “continuity method” is not working.

[Corrected, thanks – T.]27 September, 2016 at 1:11 pm

AnonymousIn the third displayed formula below (4), in the RHS should be .

[Corrected, thanks – T.]27 September, 2016 at 1:25 pm

AnonymousIn the last line of proposition 14, should be .

[Corrected, thanks – T.]27 September, 2016 at 1:34 pm

Friend of OryxFrom the proof of theorem 27: “But actually we can just give a direct proof that does not need any rectifiability hypothesis whatsoever.” Should this say “does not need any differentiability hypothesis”?

[Corrected, thanks – T.]27 September, 2016 at 1:51 pm

AnonymousIn the proof of proposition 14, the running index in the sum defining should be (instead of ). This typo appears also later in two formulas involving .

[Corrected, thanks – T.]27 September, 2016 at 2:28 pm

AnonymousHave you thought of putting the latex for the notes on GitHub? If you know git, it would make correcting the notes a bit easier.

27 September, 2016 at 3:33 pm

AnonymousIn theorem 30, in the line below the definition of , it should be . Also, in this line “in ” should be added after “polygonal path”.

[Corrected, thanks – T.]27 September, 2016 at 11:26 pm

AnonymousIn proposition 10 (arc length formula)

4th formula

I think it should be <= sign instead of =sign

[Corrected, thanks – T.]28 September, 2016 at 1:23 am

AnonymousIn the third displayed formula below (8), the second “dz” should be deleted.

[Corrected, thanks – T.]28 September, 2016 at 2:26 am

AnonymousIn the last displayed formula in the proof of theorem 27, the first “-” should be “+” (or should be interchanged).

[Corrected, thanks – T.]28 September, 2016 at 3:52 am

AnonymousIn the second example below theorem 27, the exponent “” should be ““.

[Corrected, thanks – T.]28 September, 2016 at 11:31 am

Jarred BarberIn Exercise 5, the domain of $\gamma_m$ should be $[0,1]$, since you already have the $2\pi$ factor in the exponent.

[Corrected, thanks – T.]28 September, 2016 at 4:33 pm

markmcaHi Terry, if your work could harness a volunteer grid of 400k+ PCs, we have petaflops of processing to spare. Would be our pleasure to see it doing some cutting-edge maths. Cheers, Mark M

28 September, 2016 at 8:50 pm

AnonymousFor the unsigned definite integral notation, with reals, is it defined only when is a real interval (i.e. ) and is meaningless otherwise?

29 September, 2016 at 11:52 am

Laurent C.In formula (6), is there a “r” missing just before the second integral ?

[Corrected, thanks – T.]30 September, 2016 at 6:23 am

AnonymousWhat is the essential difference between the signed and unsigned definite integrals? By convention, for . Can we also define similarly in the Lebesgue setting for ?

30 September, 2016 at 8:18 am

AnonymousSuch definition seems somewhat problematic since for the “interval” is (by definition) empty!

30 September, 2016 at 1:50 pm

AnonymousIn Proposition 10, you mean an sign instead of there?

[Corrected, thanks – T.]30 September, 2016 at 2:02 pm

AnonymousThe integral can be just defined as when $\gamma$ is piece-wise smooth. That is what Stein-Shakarchi does in the book. Proposition 14 considers the more general case that is only assumed to be continuous. Are there any advantages in applications for such generalization?

30 September, 2016 at 6:32 pm

AnonymousProposition 14 considers the case of rectifiable curves (not all continuous curves!) and exercise 15 shows that this integral definition can’t be extended for some unrectifiable curves (but perhaps it still may be extended for some other unrectifiable curves.)

30 September, 2016 at 7:35 pm

Topic boiIn the approach you are describing, I think we have to worry about proving that the value of the integral does not depend on how the contour is broken up into smooth pieces (a point that is rarely handled carefully). Is there a simple way to prove this? (I’m asking because I want to understand this point better myself.)

1 October, 2016 at 6:25 pm

CrustWhile that definition is less general in terms of $\gamma$ (which must be piecewise differentiable), it seems to me that it is much more general in terms of $f$ (interpreting the latter integral as a Lebesgue integral).

2 October, 2016 at 9:30 am

Terence TaoAs stated in the text, for most complex analysis applications, integration on piecewise smooth contours suffices. The added ability to integrate rectifiable curves has some use in geometric measure theory when one is working with fairly rough domains. In Notes 3 we extend integration of holomorphic functions further to arbitrary continuous curves, which is convenient for instance when developing a theory of winding numbers without any hypothesis of rectifiability.

1 October, 2016 at 10:58 am

Jhon ManugalThere is a typo in Proposition 14 ! The integral and the Riemann sum are within of one another whenever the mesh is smaller than .

[Corrected, thanks – T.]Consider the function: with and . Then clearly:

As the function is awfully chaotic. How can such the image curve be rectifiable? The more I think about it, I don’t know.

[I do not understand the question – what is the curve in question here, and what is the relevance of the ? -T.]2 October, 2016 at 12:44 am

AnonymousConcerning the statement just above exercise 15 (that one can’t simply omit the rectifiability hypothesis from proposition 14), is it still possible to slightly relax this hypothesis?

[Possibly, particularly if one assumes some strong regularity hypotheses on the integrand, but personally I think it is more productive instead to move away from the Riemann sum definition of integration if one wants to work with such rough curves. -T.]2 October, 2016 at 8:47 am

CrustIn Example 23, I think you want (instead of with ) in order to get .

[Corrected, thanks – T.]2 October, 2016 at 9:05 am

Math 246A, Notes 3: Cauchy’s theorem and its consequences | What's new[…] , 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, […]

6 October, 2016 at 8:43 am

Eulogio GarciaI will explain below relates to this post because it verifies that the integral calcules is not an absolute value ( mathematics need it). At any curve either derived or partial derivatives, the midpoint of the slope never coincides with the image of the function .

Function of all derivated is:

We note that this equation applies to any continuous function.

Therefore if we have:

$f(x + x_{n}}{2} \neq x + \frac{f(x_{n} – f(x)[(x_{n} – x) – x_{i}]}{x_{n} – x}$

7 October, 2016 at 2:53 am

Eulogio GarciaSorry for the tipo in the function F(x), where it is (x) is f(x).

So if the midpoint of the interval does not meet the following, it is not is correct the derivative.

9 October, 2016 at 2:58 pm

Fred Lunnonex. 15 —

for “possbile” read “possible”

ex. 31 —

for “Furthermoore” read “Furthermore”

[Corrected, thanks – T.]