4.3 Contour Integration

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For some reason known only to historians, the term contour is used in Complex

Analysis to denote a curve, usually a simple closed curve, almost always

114 CHAPTER 4. INTEGRATION

a piecewise di_erentiable curve, in the plane. The term 'simple' means that it

does not cross itself, and we can always integrate over pieces that are smooth,

and add up the results (since integration is just adding up anyway!). So we

can include polygons as among the family of curves we can integrate over.

And integrating around such curves is called contour integration. If you had

to guess what it meant, you might come up with a lot of possibilities before

you hit on the actual meaning according to complex function theorists. More

bloody jargon, in short. Still, I suppose it is useful for frightening Law students

and other low forms of life who have never performed even the simplest

contour integrals. So that you won't be mistaken for such low life, we shall

now perform one. Watch closely.

Example 4.3.1 (Contour Integral)

Integrate 1=z around the unit circle, starting and _nishing at 1.

Solution

The fact that the function 1=z is not even de_ned at 0 and hence cannot be

di_erentiable there means that we cannot cheerfully claim that the answer is

zero, anyway, it isn't. First we do it the clunky way:

Put x + iy(t) = cos t + i sin t as a parametrisation of the unit circle, with

t 2 [0; 2_].

dx + i dy = 􀀀sin t + i cos t, and 1=z = _z=z_z and on the unit circle z_z = 1.

This gives:

ZS1

1=z dz

= Z 2_

0

(cos t 􀀀 i sin t)(􀀀sin t + i cos t) dt

= Z 2_

0

i(sin2 t + cos2 t) dt

= 2_i

Next we do it more neatly: z = eit parametrises the circle. dz = ieit dt

follows. So

ZS1

1=z dz

4.3. CONTOUR INTEGRATION 115

A

B

C

D

Figure 4.2: Any loop enclosing the (single) singularity has the same integral

= Z 2_

0

1

eit ieit dt

= i Z 2_

0

dt

= 2_i

2

This result measures some property of the singularity.

To see this, note that if I had gone around the origin in a di_erent loop but

in the same direction, once, I should have got exactly the same answer.

Exercise 4.3.1 Just to con_rm this, go around a square with vertices at

_1 _ i.

And to see this, look at _gure 4.2 which shows another loop going once around

the origin.

If we went around the circle, from A to D, but then went along the line DC,

then around the outer loop clockwise, then in to the circle by BA, we should

116 CHAPTER 4. INTEGRATION

get a value of RC 1=z dz = 0, since the function 1=z is analytic on the curve

and its interior (the region between the circle and the outer curve).

But the line CD and the line AB will cancel out if the two line segments

coincide, since they are traversed in opposite directions. Hence the integral

over the inner circle and the outer loop (traversed clockwise) sum to zero.

So the integral over the circle and over the outer loop traversed in the same

direction must be equal.

Thus I have shown:

Proposition 4.3.1 Let f be an analytic function with a singularity at a

point. Then the integral of f around any loop making one circuit of the

singularity is the same as the integral of f around any other loop making a

single circuit of the singularity in the same direction.

De_nition 4.3.1 We say that 1=z has a pole at the origin. More generally,

f has a pole at w if

lim

z!w jf(z)j = 1

So 1=(z 􀀀 1)(z 􀀀 i) has a pole at z = 1 and another at z = i. 1=z2 also has

a pole at 0.

Exercise 4.3.2 Find the integral for a loop around the singularity at 0 of

1=z2.

The above exercises should leave you prepared for the following:

Proposition 4.3.2 The function 1=(z􀀀z0) has a pole at z0 and the integral

of any single loop around z0 traversed anticlockwise is 2_i.

For any integer n 6= 1, and any simple closed loop c around z0 traversed

anticlockwise around z0,

Zc

1=(z 􀀀 z0)n dz = 0

Proof:

4.3. CONTOUR INTEGRATION 117

Let c be the loop z0 + eit, so dz = ieit. Then

Zc

1=(z 􀀀 z0)n dz = i Z 2_

0

eit

eint dt

= 2_i if n = 1

= 0 if n 6= 1

2

This allows us to use partial fractions to work out the integrals for loops

around a range of functions with singularities enclosed by the loops.

Example 4.3.2 Calculate the integral of z

z2􀀀1 around the circle centred on

the origin of radius 2, in the anticlockwise direction.

Solution

z

z2 􀀀 1

=

1

2 _ 1

z 􀀀 1

+

1

z + 1_

So

ZC

z

z2 􀀀 1

dz =

1

2 _ZC

1

z 􀀀 1

+

1

z + 1

dz_

=

1

2 ZC

1

z 􀀀 1

dz +

1

2 ZC

1

z 􀀀 1

dz

=

1

2

2_i +

1

2

2_i

= 2_i

Note that the loop contains both the singularities.

Exercise 4.3.3 I claim that the integral of an analytic function around a

loop containing k singularities is the same as the sum of the integrals of loops

around each one separately.

Produce an argument to show that my claim is correct, or produce a counterexample

to show I am blathering.

It is important to realise that all this works for functions which are analytic

except at a set of discrete singularities. It fails miserably when the function

is not analytic:

118 CHAPTER 4. INTEGRATION

Example 4.3.3 Integrate the function _z anticlockwise around the unit circle

and also around the square with vertices at _1 _ i, in the same sense.

Solution The circle _rst: z = cos t + i sin t ) dz = i(cos t + i sin t) so:

Z 2_

0

(cos t 􀀀 i sin t)i(cos t + i sin t) dt = 2_i

The right hand edge of the square: z = 1+ti ) dz = i dt:

Z 1

􀀀1

(1 􀀀 ti)i dt = 2i

The opposite edge: z = 􀀀1 􀀀 ti ) dz = 􀀀i dt so:

Z 1

􀀀1

(􀀀1 + ti) 􀀀 i dt = 2i

The bottom edge: z = t 􀀀 i ) dz = dt so:

Z 1

􀀀1

(t + i) dt = 2i

And the top edge: z = 􀀀t + i ) dz = 􀀀dt so:

Z 1

􀀀1 􀀀(􀀀t 􀀀 i)dt = 2i

So the result for the square is 8i and for the circle 2_i 2

It is immediate that the contour integral of a path in one direction is always

the negative of the integral in the opposite direction, and that this works

for functions which are not analytic as well as for those which are, since the

independence of parametrisation holds for all integrable functions, analytic

or not.

Exercise 4.3.4 Prove that reversing the direction of travel reverses the sign

of the answer for any integrable function.

It is also obvious that the integral along two paths which follow is the sum

of the integrals around each path separately, something we used in the last

example. This follows from the de_nition of the path integral- we are adding

up lots of little bits anyway.

4.4. SOME INEQUALITIES 119

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