4.2 The Complex Integral

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Given f(z) = 2z let us try to integrate it along the straight line path from 0

to 1 + i.

As for second year integration along curves, I shall parametrise the curve:

x = t; y = t; t 2 [0; 1] takes us uniformly from 0 to 1+i. I put dz = dx+i dy;

then dx = dt = dy so we have

Z 1

0

2(t + it)(1 + i) dt

which is

(1 + i)2 Z 1

0

2tdt = (1+i)2 = 2i

Note that we could have got the same answer by writing

Z 1+i

0

2z dz = [z2]1+i

0 = (1+i)2 = 2i

This is using the fact that we know that 2z has an antiderivative, and we

put our faith in the Fundamental Theorem of Calculus. It seems to work in

this case.

More generally, suppose I gave you a curve in the complex plane, by giving

you a function c : [0; 1] 􀀀! C , and a complex function f. It makes sense to

do the usual business of confusing functions with values and write

c(t) = x(t) + i y(t)

I shall assume that both x and y are di_erentiable functions of t.

I can reasonably argue that now I have

dx = x_ dt; i dy = i y_ dt

and I can de_ne

Zc

f = Z t=1

t=0

f(x(t) + i y(t))(x_ + i y_) dt

108 CHAPTER 4. INTEGRATION

Example 4.2.1 Integrate the function f(z) = 2x+2iy around the unit circle,

starting and _nishing at 1. Compare with the value of integrating around the

same size circle shifted to have centre a + ib. What happens if the circle is

made bigger or smaller?

Solution

Put z = eit to get the unit circle; dz = ieit dt, so the integral is

Z 2_

0

2eitieit dt

= 2iZ 2_

0

e2itdt = 0

If the circle is of radius r and centre a + ib we have z = reit + a + ib, and

dz = ireit dt so we obtain

Z 2_

0

2(reit + a + ib)ireit dt

= 2ri Z 2_

0

e2itdt + r(a + ib) Z 2_

0

eitdt = 0 + 0 = 0

Exercise 4.2.1 I can integrate along curves which are straight lines and

indeed are along or parallel to the axes. If I integrate along the X-axis, with

the obvious parametrisation x(t) = t, I can calculate things such as

Z _=2

0

eit dt

Do it two ways: _rst by _nding an antiderivative to f and directly.

Explain carefully why you would expect these to agree.

You will recall something (I hope) of the integration of vector _elds over

curves from second year. You may remember that the value of the integral

of a vector _eld along a curve depends only on the set of points on the curve,

and not on the parametrisation of the curve. This is physically obvious:

The idea of integrating a vector _eld along a curve is that of driving along a

track and measuring the extent to which the gravity (Vector Field) helps you

4.2. THE COMPLEX INTEGRAL 109

when you are going down hill and costs you when you are going up hill. You

compute the projection of the force on the direction in which you are going

and multiply the value of the force by the distance you go in a very short

(in_nitesimal) time. Now travelling at di_erent speeds will make a di_erence

to the in_nitesimal distances, but they must all add up to the total distance

along the track. And the value of the assistance given by the force _eld

doesn't depend on the time.

The above argument is heuristic and would put some Pure Mathematicians

in a cold sweat until they noticed that a proof of the theorem can be made

which follows this heuristic argument quite closely.

So if we take

ZC

P dx + Q dy

and parametrise C by c : [0; 1] 􀀀! R2 with c given by t ; _ x(t)

y(t) _ we have

that the integral becomes

Z 1

0

P(x(t); y(t))

dx

dt

+ Q(x(t); y(t))

dy

dt

dt

And the answer will not be changed by altering the parametrisation, which

was only introduced to save us the hassle of chopping the curve up into little

bits and calculating the projection of the force on each little bit, and then

adding them all up; and then doing it again (and again!) for smaller, littler

little bits and taking the limit.

Now the integration of a complex function u + iv is in many ways similar to

this.

In integrating a complex valued function along a curve, we have

Zc

(u + i v)(dx + i dy)

= Zc

(u dx􀀀v dy) + i(v dx + u dy)

= Z 1

0

[u(x(t) + iy(t)) x_ 􀀀 v(x(t) + iy(t)) y_] dt

+ i Z 1

0

[v(x(t) + iy(t)) x_ + u(x(t) + iy(t)) y_] dt

110 CHAPTER 4. INTEGRATION

So the real part is the integral of the vector _eld _ u

􀀀v _ over the curve, and

the imaginary part is the integral of the vector _eld _ v

u _ over the curve.

Now since both of these are going to be independent of the parametrisation

of the curve for the same reasons as usual, it follows immediately that the

path integral in C is independent of the parametrisation.

You may also remember from second year that there are 'nice' vector _elds

which are derived from a potential _eld and have the much stronger property

that the integral along any curve of the _eld gives a result which depends

only on the end points of the curve, and is hence zero for closed curves.

And there are 'nasty' vector _elds where this ain't so. If you write down

a vector _eld 'at random', then it is 'nasty', for any sensible de_nition of

'at random'. It is cheering therefore to be able to tell you that the vector

_elds in the plane arising from analytic functions are all nice. This is the

Cauchy-Goursat Theorem:

Theorem 4.1 (Cauchy-Goursat)

If we integrate a function f which is analytic in a domain E _ C around a

piecewise smooth simple closed curve contained in E, the result is zero.

Idea of Proof: If f is analytic, then it satis_es the CR equations. Write

f(x + iy) = u+iv.

We want

ZC

[u dx 􀀀 v dy] + i [v dx + u dy] (4.1)

Now Let D be the region having the simple closed curve as its boundary:

@D = C. From Green's Theorem we have:

Z@D

F = ZD

dF

and if F = u dx􀀀v dy, which is the real part of the complex integral 4.1, we

have

ZC

[u dx 􀀀 v dy] = ZD􀀀@v=@x 􀀀 @u=@y = 0

by the CR equations.

4.2. THE COMPLEX INTEGRAL 111

Similarly, the imaginary part is also zero. 2

It follows immediately that if p : [0; 1] 􀀀! C is a piecewise smooth path from

0 to w in C , and if f is a complex function which is analytic on a ball big

enough to contain 0 and w, R1

0 f(p(t))(x_ +iy_) dt gives a result which depends

on w but not p. This is obvious, because if we could _nd a path with the

same end points but a di_erent value for the integral, we could go out along

one path, back along the other, and have a non-zero outcome, contradicting

the last theorem.

We use this to de_ne an inde_nite integral:

Theorem 4.2 (Antiderivatives)

For any f which is analytic on a domain E, de_ne

F : C 􀀀! C ; by F(w) = Z 1

0

f(c(t))(x_ + iy_) dt

where c is any smooth path which has c(0) = 0 and c(1) = w.

Then F is analytic and F0 = f

Proof

The proof is usually a _ddly argument from _rst principles. Since you will

have done similar things for the existence of the potential function for conservative

_elds, and this is pretty much the same idea, I shall skip it. 2

Corollary 4.2.1 If f is analytic and c : [0; 1] 􀀀! C is any smooth path in

C, then if F is the antiderivative provided by the above theorem,

Zc

f(z)dz = F(z(0)) 􀀀 F(z(1))

Proof:

This follows immediately from the construction of F. 2

It should be apparent that there is no need to start my construction of F

from the origin; anywhere else would do. The two antiderivatives would di_er

by a (complex) constant.

112 CHAPTER 4. INTEGRATION

P

Q

R

Figure 4.1: A path from i to 1 + i

This has given us a fairly satisfactory idea of what is involved in doing integration

for analytic complex functions. The key result is that integrating

an analytic function around a simple closed loop C gives zero. This has

implications for evaluating integrals around nasty curves:

Example 4.2.2 Evaluate the contour integral

ZC

1 + 2z2

z

where C is the curve starting at P = i and going to Q = 􀀀1 along the unit

circular arc centred at the origin in the anticlockwise direction, followed by a

straight line from Q to R at 1 + i.

Solution

The diagram _gure 4.1 shows the curve we have to integrate over. If we were

to join the endpoints by going to i from 1+i, the resulting closed curve would

not contain a singularity of the functions, which is analytic (being a ratio

of analytic functions), and the integral around the curve would therefore be

zero.

The integral therefore does not depend on the path, and the straight line path

4.3. CONTOUR INTEGRATION 113

from i to 1 + i given by c : [0; 1] 􀀀! C , t ; t + i gives the same answer as

the integral over the much more complicated curve asked for.

In fact we can rewrite the integral as

ZC

dz

z

+ ZC

2z dz (4.2)

and since the path does not do a circuit of a singularity, this is

[Logz]1

0 + [z2]1

0

= Log(1 + i) 􀀀 Log(i) + (1 + i)2

􀀀i2

= Log(1 􀀀 i) + 1 + 2i

= log(p2) 􀀀 i_=4 + 1 + 2i

= 1 + log(p2) + i(2 􀀀 _=4)

2

Exercise 4.2.2 Rework the last solution by substituting z = t + i in equation

4.2 and integrating along the path to con_rm that we agree on the answer.

Exercise 4.2.3 Suppose instead of going by the anticlockwise route along

the unit circle, the curve went the clockwise route and hence circumnavigated

the origin. How would you evaluate, quickly, the new path integral?

Things can be very di_erent when f stops being analytic, for example when

it has a singularity in the region enclosed by C. For a start, there is no

guarantee that integrating around such a loop will give zero, and it often

does not. For seconds, there is no guarantee that such a function will have

an antiderivative.

This is the start of some rather curious phenomena which will be investigated

in a separate section.

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