Complex Numbers , Complex Functions and Contour Integrals

3. Contour Integrals

A complex function can be integrated along a path in the complex plane much in the same way a real
function of two real variables can be integrated along a path in R^2. In contrast to the real case, however, these
complex integrals (called contour integrals) are intimately linked with the poles of the integrand. Before stating
the precise relationship , let’s compute some contour integrals and make a few initial discoveries. We use the
fol lowing definition .

Definition. Suppose C is a path in the complex plane, and f(z) is a complex function. Given a parametric
representation z(t) of C, with t₀ ≤ t ≤ t₁, we define

Example. Let’s calculate   where C is the contour given below:


The curve C is given by the equation x = y² + 1. We must first find a parameterization of C. Suppose we
introduce the parameter t, and let y = t for −2 ≤ t ≤ 2. Since x = y² + 1, we then have x = t² + 1. We
therefore have a parameterization of the curve C, given by
z(t) = (t² + 1) + it

for −2 ≤ t ≤ 2.

To store this contour in MatLab we enter:

>> syms t real;
>> y = t;
>> x = t²+1;
>> z = simple(x + i*y)

z =

t²+1+i*t

Now that we have a parameterized path, MatLab can calculate the contour integral of any complex function
along this path (so long as that function doesn’t have a pole on the path). First we store the function and
integrand:

>> f = z²-2*z+1;
>> Integrand = f*diff(z,t);

We calculate the contour integral using the int command, and convert the result to a numerical answer
using the double command:

>> F = int(Integrand, 't, -2, 2);
>> F = double(F)
F =

0 +58.6667i

Example.
Thus, Of course, we could also calculate this integral explicitly:

As we will soon see, it is a theorem of complex analysis that the contour integral of an entire function depends
only on the endpoints of the path, and not on the path taken between them.

Example. To see this in action, let’s calculate   using the contour below, given by the
equation x = 5, shown below:

We repeat the same process used in the previous example:

>> syms t real;
>> y=t;
>> x=5;
>> z=simple(x+i*y);
>> f=z²-2*z+1;
>> Integrand=f*diff(z,t);
>> F=int(Integrand, 't, -2, 2);
>> F=double(F)
F =

0 +58.6667i

Example .
Thus, , and so it indeed appears that
Indeed, we can again manually compute

It turns out that a contour integral of a meromorphic function f around a closed loop C only depends on
the poles of f contained inside C. More precisely, we have the following theorem:

Theorem. Suppose f is a meromorphic function and C is a simple closed loop (e.g. a circle ). If p₁, . . . , p_k are
the poles of f lying in the interior of C, then

This remarkable theorem has several corollaries.
Corollary 1. If f is an entire function, then   for any closed loop C.
Proof. For a simple closed curve C, the statement follows immediately from the theorem. For an arbitrary
closed loop C, it is a consequence of a slightly more general version of the theorem.
Corollary 2. If f is an entire function, then   depends only on the endpoints of C.
Proof. Suppose C₁ and C₂ are two different paths with the same endpoints. Let C be the loop given by C₁−C₂.
Then

On the other hand, by Corollary 1 we have

since f doesn’t have any poles. The result follows.
 

Exercise 7. Let C be the contour given by x² + y² = 1, shown below. Observe that C can be parameterized
by x = cos(t), y = sin(t), for 0 ≤ t ≤ 2 .

For each of the functions f(z) given below, use MatLab to:

a) Compute   using MatLab;
b) Find the poles of f contained inside C;
c) Compute the residue of f at each of these poles; and
d) Compute and compare the result with your answer from part (a).

i) f(z) = z³ + 2z² + 1
ii) f(z) = z⁵ − z
iii)
iv)
v)