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4. Complex integration and Cauchy’s theorem
§4.1 Introduction
Consider the real integral (^) ∫ b a
f (x) dx.
We often read this as ‘the integral of f from a to b’. That is, we think of starting at the point a and moving along the real axis to b, integrating f as we go. Now let z 0 , z 1 ∈ C. How might we define ∫ (^) z 1
z 0
f (z) dz?
We want to start at z 0 , move through the complex plane to z 1 , integrating f as we go. But in the complex plane there are lots of ways of moving from z 0 to z 1. Suppose γ is a path from z 0 to z 1 (we shall make precise what we mean by a path below, but intuitively just think of it as a continuous curve starting at z 0 and ending at z 1 ). Then, using similar ideas to those from Calculus and Vectors, we can define ∫
γ
f (z) dz.
A priori this looks like it will depend on the path γ. However, as we shall see, in complex analysis in many cases this quantity is independent of the path chosen.
§4.2 Paths and contours
First we need to make precise what we mean by a path.
Definition. A path is a continuous function γ : [a, b] → C where [a, b] is a real interval.
Remark. So, for each a ≤ t ≤ b, γ(t) is a point on the path. We say that the path γ starts at γ(a) and ends at γ(b).
Remark. Note that a path is a function. Sometimes, it is convenient to regard a path as a set of points in C, i.e. we identify the function γ with its image. If we think of the path γ as its image in C then we sometimes call the function γ(t) a parametrisation of the path γ. Note that the same path can have different parametrisations. For example
γ 1 (t) = t + it, γ 2 (t) = t^2 + it^2 , 0 ≤ t ≤ 1
are both parametrisations of the straight line that starts at 0 and ends at 1 + i. We shall see later (Proposition 4.3.1) that the distinction between paths (as functions) and paths (as subsets of C) is not particularly important.
As an example of a path, let z 0 , z 1 ∈ C. Define
γ(t) = (1 − t)z 0 + tz 1 , 0 ≤ t ≤ 1. (4.2.1)
Then γ(0) = z 0 , γ(1) = z 1 and the image of γ is the straight line joining z 0 to z 1. We sometimes denote this path by [z 0 , z 1 ]. See Figure 4.2.1.
z 0
z 1
Figure 4.2.1: The path γ(t) = (1 − t)z 0 + tz 1 , 0 ≤ t ≤ 1, describes the straight line joining z 0 to z − 1. We sometimes denote this path by [z 0 , z 1 ].
Definition. Let γ : [a, b] → C be a path. If γ(a) = γ(b) (i.e. if γ starts and ends at the same point) then we say that γ is a closed path or a closed loop.
Example. An important example of a closed path is given by
γ(t) = eit^ = cos t + i sin t, 0 ≤ t ≤ 2 π. (4.2.2)
This is the path that describes the circle in C with centre 0 and radius 1, starting and ending at the point 1. See Figure 4.2.2.
Definition. A path γ is said to be smooth if γ : [a, b] → C is differentiable and γ′^ is continuous. (By differentiable at a we mean that the one-sided derivative exists, similarly at b.)
All of the examples of paths above are smooth. We can use integrals to define the lengths of paths:
Definition. Let γ : [a, b] → C be a smooth path. Then the length of γ is defined to be
length(γ) =
∫ (^) b
a
|γ′(t)| dt.
Example. It is straightforward to check from (4.2.1) that
length([z 0 , z 1 ]) = |z 1 − z 0 |.
If γ(t) is the path given in (4.2.2)
length(γ) = 2π.
−R −ε^ ε^ R
Figure 4.2.3: The contour γ 1 + γ 2 + γ 3 + γ 4.
§4.3 Contour integration
Let f : D → C be a complex functions defined on a domain D. Let γ : [a, b] → D be a smooth path in D.
Definition. The integral of f along γ is defined to be ∫
γ
f (z) dz =
∫ (^) b
a
f (γ(t))γ′(t) dt.
We will often write
γ f^ for^
γ f^ (z)^ dz.
Remark. Strictly speaking we should write f (γ(t))γ′(t) = u(t)+iv(t) where u, v : [a, b] → R and define
γ f^ to be^
∫ (^) b a u(t)^ dt^ +^ i^
∫ (^) b a v(t)^ dt.
Example. Take f (z) = z^2 and γ(t) = t^2 + it, 0 ≤ t ≤ 1. Then f (γ(t)) = (t^2 + it)^2 = t^4 − t^2 + 2it^3 and γ′(t) = 2t + i. Hence ∫
γ
f (z) dz =
0
f (γ(t))γ′(t) dt =
0
(t^4 − t^2 + 2it^3 )(2t + i) dt
0
2 t^5 − 4 t^3 dt + i
0
5 t^4 − t^2 dt
[
t^6 − t^4
] 1
0
[
t^5 −
t^3
] 1
0
One can show that this definition is independent of the choice of parametrisation of the path.
Proposition 4.3. Let γ : [a, b] → C be a smooth path. Let φ : [c, d] → [a, b] be an increasing smooth bijection. Then γ ◦ φ : [c, d] → C is a path that has the same image as γ. Moreover, ∫
γ◦φ
f =
γ
f
for any continuous function f.
Proof. It is clear that both γ and γ ◦ φ have the same image. Thus γ and γ ◦ φ are different parametrisations of the same path. Note that ∫
γ◦φ
f =
∫ (^) d
c
f (γ(φ(t)))(γφ)′^ (t) dt
∫ (^) d
c
f (γ(φ(t)))γ′(φ(t))φ′(t) dt by the chain rule
∫ (^) b
a
f (γ(t))γ′(t) dt by the change of variables formula.
2
Remark. If φ in Proposition 4.3.1 is a decreasing smooth bijection then γφ has the same image as φ but the path traverses this in the opposite direction, i.e. γφ is a parametrisation of −γ. Following the above calculation we see that
γφ f^ =^ −^
γ f^ , corresponding to the fact stated below that
−γ f^ =^ −^
γ f^. Now suppose that γ = γ 1 + · · · + γn is a contour in D. We define ∫
γ
f =
γ 1
f + · · · +
γn
f.
The following basic properties of contour integration follow easily from this definition.
Proposition 4.3. Let f, g : D → C be continuous and let c ∈ C. Suppose that γ, γ 1 , γ 2 are contours in D. Then (i) ∫
γ 1 +γ 2
f =
γ 1
f +
γ 2
f ;
(ii) ∫
γ
(f + g) =
γ
f +
γ
g;
(iii) ∫
γ
cf = c
γ
f ;
(iv) ∫
−γ
f = −
γ
f.
Recall from real calculus (or, indeed, from A-level) that one way to calculate the integral of f is to find an anti-derivative, i.e. find a function F such that F ′^ = f. The Fundamental Theorem of Calculus then says that
∫ (^) b a f^ (x)^ dx^ =^ F^ (b)^ −^ F^ (a). We have an analogue of this in for the complex integral. We first need the following definition.
§4.4 The Estimation Lemma
There are two results about real integration that are obvious from considering the integral of f (x) over [a, b] as the area underneath the graph of f. Firstly
∣∣ ∣ ∣
∫ (^) b
a
f (x) dx
∫ (^) b
a
|f (x)| dx (4.4.1)
and secondly, if |f (x)| ≤ M then
∣ ∣∣ ∣
∫ (^) b
a
f (x) dx
∣ ≤^ M^ (b^ −^ a).^ (4.4.2)
Both of these results have analogies in the context of complex analysis. However, the proofs are surprisingly intricate. Here is the complex analogue of (4.4.1).
Lemma 4.4. Let u, v : [a, b] → R be continuous functions. Then
∣ ∣ ∣∣
∫ (^) b
a
u(t) + iv(t) dt
∫ (^) b
a
|u(t) + iv(t)| dt. (4.4.3)
Proof. Write (^) ∫ b a
u(t) + iv(t) dt = X + iY.
Then
X^2 + Y 2 = (X − iY )(X + iY )
=
∫ (^) b
a
(X − iY )(u(t) + iv(t)) dt
∫ (^) b
a
Xu(t) + Y v(t) dt + i
∫ (^) b
a
Xv(t) − Y u(t) dt.
However, X^2 + Y 2 is real, hence the imaginary part of the above expression must be zero, i.e. (^) ∫ b
a
Xv(t) − Y u(t) dt = 0
so that
X^2 + Y 2 =
∫ (^) b
a
Xu(t) + Y v(t) dt. (4.4.4)
Notice that the integrand in (4.4.4) is the real part of (X − iY )(u(t) + iv(t)). Recalling that for any complex number z we have that Re(z) ≤ |z|, we have that
Xu(t) + Y v(t) ≤ |(X − iY )(u(t) + iv(t))| = |X − iY ||u(t) + iv(t)| =
X^2 + Y 2 |u(t) + iv(t)|.
Hence
X^2 + Y 2 =
∫ (^) b
a
Xu(t) + Y v(t) dt
X^2 + Y 2
∫ (^) b
a
|u(t) + iv(t)| dt
and cancelling the term
X^2 + Y 2 gives ∣∣ ∣∣
∫ (^) b
a
u(t) + iv(t) dt
∣∣ = |X + iY | =
X^2 + Y 2 ≤
∫ (^) b
a
|u(t) + iv(t)| dt
as claimed. 2
We can now prove the following important result—the complex analogue of (4.4.2)— which we will use many times in the remainder of the course.
Lemma 4.4.2 (The Estimation Lemma) Let f : D → C be continuous and let γ be a contour in D. Suppose that |f (z)| ≤ M for all z on γ. Then (^) ∣ ∣∣ ∣
γ
f
∣ ≤^ M^ length(γ).
Remark. We shall use the Estimation Lemma in two different ways: (i) suppose f is a function which takes small (in modulus) values along a contour γ, then
γ f^ is small; (ii) if f is any continuous function and γ is a contour with small length, then
γ f^ is small.
Proof. Simply note that by Lemma 4.4.1 we have that ∣ ∣∣ ∣
γ
f
∫ (^) b
a
f (γ(t))γ′(t) dt
∫ (^) b
a
|f (γ(t))||γ′(t)| dt
≤ M
∫ (^) b
a
|γ′(t)| dt
= M length(γ). 2
Example. Let f (z) = 1/(z^2 + z + 1) and let γ(t) = 5eit^ be the circle of radius 5 centred at 0. We use the estimation lemma to bound
γ f^ (z)^ dz. First note that if z is a point on γ then |z| = 5. Hence |z^2 + z + 1| ≥ |z|^2 − |z + 1| ≥ 25 − 6 = 19.
Thus for z on γ we have that
|f (z)| ≤
Next we note that length(γ) = 2π × 5 = 10π. Thus, by the Estimation Lemma, ∣∣ ∣∣
γ
f (z) dz
∣∣ ≤ 10 π 19
Proposition 4.5. Let γ be a path in C \ { 0 }. Then there exists a parametrisation of γ, γ : [a, b] → C \ { 0 }, for which t 7 → arg γ(t) is a continuous function. Any other choice of parametrisation with a continuous choice of argument differs from this argument by a constant integer multiple of 2 π.
Example. For example, consider
γ(t) =
eit, 0 ≤ t ≤ π ei(t+2π), π < t ≤ 2 π.
Then γ describes the unit circle with centre 0 and radius 1. Here
arg γ(t) =
t, 0 ≤ t ≤ π t + 2π, π < t ≤ 2 π.
and this is not continuous. However, we can find a parametrisation of γ for which the argument is continuous, for example
γ(t) = eit, 0 ≤ t ≤ 2 π
and note that arg γ(t) = t, 0 ≤ t ≤ 2 π, is continuous.
Now consider the closed path γ. We can reinterpret the winding number w(γ, 0) of γ around 0 as the multiple of 2π giving the total change in argument along γ.
Proposition 4.5. Let γ be a closed path that does not pass through the origin. Then
w(γ, 0) =
2 πi
γ
z dz.
Example. Let γ(t) = e^4 πit, 0 ≤ t ≤ 1. Intuitively, this winds around the origin twice anticlockwise, and so should have winding number w(γ, 0) = 2. We can check this using Proposition 4.5.2 as follows:
1 2 πi
γ
z dz =
2 πi
0
e^4 πit^ 4 πie^4 πit^ dt
0
2 dt = 2.
Example. Let γ(t) = e−it, 0 ≤ t ≤ 2 π. In this case, γ winds around the origin once, but clockwise. Thus w(γ, 0) = −1. Again, we can check this using Proposition 4.5.2 as follows:
1 2 πi
γ
z dz =
2 πi
∫ (^2) π
0
e−it^ (−i)e−it^ dt
∫ (^2) π
0
2 π
dt = − 1.
Proof of Proposition 4.5.2. Intuitively this is clear: let γ : [a, b] → C \ { 0 } be a closed path that does not pass through the origin. Note that γ(a) = γ(b). Then (and we put quotes around the following to indicate that it does not work)
γ
z dz =
∫ (^) b
a
γ(t) γ′(t)
= [log(γ(t))]ba = (log |γ(b)| + i arg γ(b)) − (log |γ(a)| + i arg γ(a)) = i (arg γ(b) − arg γ(a)) = 2 πiw(γ, 0)”.
The reason that the above computation does not work is that 1/z does not have log(z) (or, indeed, the principal logarithm Log(z)) as an antiderivative on C \ { 0 }. This is because Log(z) is not continuous on C{ 0 }. However, Log(z) is continuous and is an anti-derivative for 1/z on the cut plane, where we remove the negative real axis from C. More generally, one can define a logarithm continuously on a cut plane where one removes any ray from C. For each α ∈ [−π, π) define the cut plane at angle α to be Cα = C \ {reiα^ | r > 0 },
i.e. the complex plane with the ray inclined at angle α from the positive x-axis removed. On Cα we can define arg z to be argα z = θ where
z = reiθ, r > 0 , α − 2(m + 1)π < θ ≤ α − 2 mπ
where we have the freedom to choose any m ∈ Z. (The case α = π, m = 0 corresponds to the usual principal value of the argument.) Let γ be a closed path that does not pass through the origin. In general, γ will not lie in one cut plane. Split γ up into pieces γ 1 ,... , γn defined on [t 0 , t 1 ],... , [tn− 1 , tn] so that each γr lies in a single cut plane, Cαr , say. Along each γr we will choose a value of the argument argαr which is continuous on Cαr and such that argαr γr (tr) = argαr+1 γr+1(tr), 0 ≤ r ≤ n − 1. Hence ∫
γr
z
dz = log γ(tr) − log(γ(tr− 1 ))
= log |γ(tr )| − log |γ(tr− 1 )| + i
argαr (γ(tr )) − argαr (γ(tr− 1 ))
Now ∫
γ
z
dz =
∑^ n
r=
γr
z
dz
∑^ n
r=
(log |γ(tr)| − log |γ(tr− 1 )|)
∑^ n
r=
argαr (γ(tr )) − argαr (γ(tr− 1 ))
The real parts cancel. The imaginary parts sum to
argαn (γ(tn)) − argα 0 (γ(t 0 )),
the total change in argument around γ, i.e. 2πw(γ, 0). 2
We can now state Cauchy’s Theorem.
Theorem 4.5.5 (Cauchy’s Theorem) Let f be holomorphic in a domain D and let γ be a closed contour in D that does not wind around any point outside D (i.e. w(γ, z) = 0 for all z 6 ∈ D). Then
γ f^ = 0.
Remark. The strength of Cauchy’s theorem is that we do not need to know if f has an anti-derivative on D. (If f did have an antiderivative on D then
γ f^ = 0 follows immediately from the Fundamental Theorem of Contour Integration; however, possessing an antiderivative is an extremely strong assumption on f. See Theorem 4.3.3 and the remarks following it.)
Remark. See Figure 4.5.2 for examples of the hypotheses of Cauchy’s Theorem.
D
D
D
(i) (ii) (iii)
γ
γ γ
Figure 4.5.2: In (i) and (ii), γ has winding number zero around every point outside D, so the hypotheses of Cauchy’s Theorem (Theorem 4.5.5 hold. In (iii) γ has winding number 1 around points inside the ‘hole’ in D, hence the hypothesis of Cauchy’s Theorem do not hold.
Proof. There are many proofs of Cauchy’s theorem; here we give one based on Green’s Theorem (see MATH10121 Calculus and Vectors). We assume that f has continuous partial derivatives. Green’s theorem states the following: suppose that γ is a piecewise smooth closed contour bounding a region Γ, g, h are C^1 functions defined on an open set containing Γ, then (^) ∫
γ
g(x, y) dx + h(x, y) dy =
Γ
∂h ∂x
∂g ∂y
dx dy. (4.5.2)
Let f be as in the hypotheses and write f (z) = f (x + iy) = u(x, y) + iv(x, y). Note that dz = dx + i dy. Then ∫
γ
f dz =
γ
(u + iv)(dx + i dy)
γ
u dx − v dy + i
γ
v dx + u dy
Γ
∂v ∂x
∂u ∂y
dx dy +
Γ
∂u ∂x
∂v ∂y
dx dy
= 0
as, by the Cauchy-Riemann equations, ∂u/∂x = ∂v/∂y and ∂u/∂y = −∂v/∂x hold every- where on Γ. 2
Remark. In many ways, this proof is cheating: Green’s Theorem is a deep theorem and not easy to prove. There are direct proofs of Cauchy’s theorem, but they are lengthy and difficult. (The idea is to build D up from smaller pieces, often starting with the case when D is a triangle; see Stewart and Tall, p.143.) Another reason for why the above proof is cheating is that Green’s theorem requires the partial derivatives in (4.5.2) to be continuous. In general, the statement of Cauchy’s theorem only requires the partial derivatives to exist in D (i.e. we do not need to assume that they are continuous). In fact, as we shall see, the existence of the partial derivatives on a domain forces them to be continuous (indeed, if the partial derivatives exist on a domain then the function is differentiable infinitely many times). However the proof of this fact uses Cauchy’s Theorem.
There are many variants of Cauchy’s theorem. Here we give just two simple modifica- tions. Our first variant deals with simply connected domains. Heuristically, a domain is simply connected if it does not have any holes in it. (For example, in Figure 4.5.2(i) the domain D is simply connected; however the domains D in Figures 4.5.2(ii) and (iii) are not simply connected as they have holes in them.) More precisely:
Definition. A domain D is simply connected if for all closed contours γ in D and for all z 6 ∈ D, we have w(γ, z) = 0.
Theorem 4.5.6 (Cauchy’s Theorem for simply connected domains) Suppose that D is a simply connected domain and f is a holomorphic function on D. Then for any closed contour γ we have that
γ f^ = 0. More generally, we can ask about integrating around sums of closed contours.
Theorem 4.5.7 (Generalised Cauchy’s Theorem) Let D be a domain and let f be holomorphic on D. Let γ 1 ,... , γn be closed contours in D. Suppose that w(γ 1 , z) + · · · + w(γn, z) = 0 for all z 6 ∈ D.
Then (^) ∫
γ 1
f + · · · +
γn
f = 0.
Remark. The hypotheses of the Generalised Cauchy Theorem (Theorem 4.5.7) give one way of extending Cauchy’s Theorem to non-simply connected domains. Consider the ex- ample in Figure 4.5.3. Here, if z is ‘outside’ D then clearly w(γ 1 , z) = w(γ 2 , z) = 0. If z is in the ‘hole’ in D then w(γ 1 , z) = 1, w(γ 2 , z) = −1 so that w(γ 1 , z) + w(γ 2 , z) = 0. Hence the hypotheses of the Generalised Cauchy Theorem hold.
Proof of Theorem 4.5.7. Suppose that γr starts and ends at zr ∈ D, 1 ≤ r ≤ n. Choose any z 0 ∈ D and contours σ 1 ,... , σn in D which join z 0 to z 1 ,... , zn, respectively. (See Figure 4.5.4.) Note that σ 1 + γ 1 − σ 1 is a closed contour that starts and ends at z 0 and, moreover, that for z 6 ∈ D we have w(σ 1 + γ 1 − σ 1 , z) = w(γ 1 , z). We see that
γ = σ 1 + γ 1 − σ 1 + · · · + σn + γn − σn
γ 3
γ 2 γ 1
σ 3
z 0 σ 1
σ 2
Figure 4.5.4: The path γ is formed by starting at z 0 , traversing σ 1 , then around γ 1 , then back along σ 1 , then along σ 2 , around γ 2 , back along σ 2 , along σ 3 , around γ 3 and back along σ 3 , ending at z 0.
Exercise 4. Find the values of (^) ∫
γ
x − y + ix^2 dz
where z = x + iy and γ is:
(i) the straight line joining 0 to 1 + i;
(ii) the imaginary axis from 0 to i; (iii) the line parallel to the real axis from i to 1 + i.
Exercise 4. Let
γ 1 (t) = 2 + 2eit, 0 ≤ t ≤ 2 π, γ 2 (t) = i + e−it, 0 ≤ t ≤ π/ 2.
Draw the paths γ 1 , γ 2. From the definition
γ f^ =^
∫ (^) b a f^ (γ(t))γ
′(t) dt, calculate
(i)
γ 1
dz z − 2
, (ii)
γ 2
dz (z − i)^3
Exercise 4. Evaluate
γ |z|
(^2) dz where γ is the circle |z − 1 | = 1 described anticlockwise.
Exercise 4. For each of the following functions find an anti-derivative and calculate the integral along any smooth path from 0 to i:
(i) f : C → C, f (z) = z^2 sin z; (ii) f : C → C, f (z) = zeiz^.
Exercise 4. Calculate
γ |z|
(^2) dz where
(i) γ denotes the contour that goes vertically from 0 to i then horizontally from i to 1 + i; (ii) γ denotes the contour that goes horizontally from 0 to 1 then vertically from 1 to 1 + i.
What does this tell you about possibility of the existence of an anti-derivative for f (z) = |z|^2?
Exercise 4. Calculate (by eye) the winding number around any point which is not on the path.
Figure 4.6.1: See Exercise 4.7.
Exercise 4. Prove Proposition 4.3.2(iv): Let D be a domain, γ a contour in D, and let f : D → C be continuous. Let −γ denote the reversed path of γ. Show that ∫
−γ
f = −
γ
f.
Exercise 4. Let
γ 1 (t) = −1 +
eit, 0 ≤ t ≤ 2 π,
γ 2 (t) = 1 +
eit, 0 ≤ t ≤ 2 π, γ(t) = 2 eit, 0 ≤ t ≤ 2 π.
