4.5 Some Solid and Useful Theorems

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Theorem 4.3 (The Cauchy Integral Formula)

If f is analytic in a region E _ C , and if C is any closed simple curve in E,

then for any w 2 E,

f(w) =

1

2_i ZC

f(z)

z 􀀀 w

dz

4.5. SOME SOLID AND USEFUL THEOREMS 121

Proof:

We certainly have that

ZC

f(w)

z 􀀀 w

dz = f(w) ZC

1

z 􀀀 w

dz = f(w) 2_i

Since the integral RC

f(w)

z􀀀w dz will remain constant no matter how small the

loop is, the limit as C shrinks to zero of the integral exists and is f(w)2_i.

But this is also the limit of

ZC

f(z)

z 􀀀 w

dz

which is also independent of the loop size.

Hence

ZC

f(z)

z 􀀀 w

dz = ZC

f(w)

z 􀀀 w

dz = f(w) ZC

1

z 􀀀 w

dz = f(w) 2_i

and the result is proved. 2

Theorem 4.4 (The Cauchy Integral Formula for Derivatives)

If f is analytic in a region E _ C , and if C is a simple closed curve in E

then for any z0 enclosed by C, the nth derivative of f exists and is given by:

fn(z0) =

n!

2_i ZC

f(z)

(z 􀀀 z0)n+1 dz

Proof:

We have for the original Cauchy formula:

f(w) =

1

2_i ZC

f(z)

z 􀀀 w

dz

for any w 2 C .

Parametrising the loop C by z(t), we can write this as

f(w) =

1

2_i Z 1

0

f(z(t))z_(t)

z 􀀀 w

dt

122 CHAPTER 4. INTEGRATION

We now treat the integral as a function of w and t and use Leibnitz Rule

which says we can di_erentiate through an integral sign to get

f0(w) =

1

2_i Z 1

0

@

@w _f(z(t) _ z(t)

z 􀀀w _ dt

This gives immediately:

f0(w) =

1

2_i Z 1

0

f(z(t))z_(t)

(z 􀀀 w)2

dt

We simply carry on doing this to get the required result. 2

An important corollary is:

Corollary 4.4.1 If f is analytic in a region E, then it has derivatives of all

orders in E and every derivative is also analytic in E. 2

This makes it clear that complex analytic functions are very special and

quite di_erent from continuously di_erentiable real functions. If you can

di_erentiate a complex function everywhere in a region, you can di_erentiate

the derivative in the region, and so on inde_nitely.

A second corollary follows also:

Corollary 4.4.2 If u : R2 􀀀! R is harmonic, then it has partial derivatives

of all orders, and all are harmonic functions. 2

There is a converse to the Cauchy-Goursat theorem:

Theorem 4.5 (Morera's Theorem)

If f : C 􀀀! C is continuous and satis_es the condition that for every closed

loop c

Zc

f(z) dz = 0

then f is analytic.

4.5. SOME SOLID AND USEFUL THEOREMS 123

Proof:

We can construct an antiderivative of f, F say, by the usual process of integrating

f from the origin (or some other convenient location) to the point w

to de_ne F(w). Then F has derivative f, which is by hypothesis continuous,

so F is analytic. Hence it has derivatives of all orders, each of which is also

analytic; f is the _rst of them. 2

We can also show the mean value theorem that says that for any circle centred

on a point w in the domain of an analytic function f, the mean value of all

the values of f on the circle is the value at the centre:

Theorem 4.6 (Gauss' Mean Value Theorem)

If f is analytic and w is any point, then for the circle w + Rei_ we have

f(w) =

1

2_ Z 2_

0

f(w+Rei_)d_

Proof:

By Cauchy's integral formula we have

f(w) =

1

2_i Zc

f(z)

z 􀀀 w

dz

where c can be taken to be w + Rei_ for _ 2 [0; 2_]. Substituting for z and

dz we get

f(w) =

1

2_i Z 2_

0

f(w + Rei_)iRei_

Rei_ d_

and some cancelling gives the result. 2

It is important to see that the integral R2_

0 f(w + Rei_)d_ is NOT a path

integral. If it were, it would be zero. We are not multiplying by a dz, which

being an in_nitesimal complex number has a direction associated with it, but

by a d_ which is a 'real in_nitesimal'.

You may have been told that in_nitesimals are wicked. This is obsolete.

Modern mathematicians just take them to be elements of a thing called the

'tangent bundle' and treat them pretty much the same way the great classical

124 CHAPTER 4. INTEGRATION

mathematicians did. Since I cannot explain the rationale properly in less than

a lecture course on manifolds, I shall rely on your vague intuitions.

The result of Gauss leads to another important property of analytic functions:

Theorem 4.7 (The Maximum Modulus Principle)

If f is analytic and non-constant in a connected region E, then jf(z)j attains

its maximum on the boundary of E.

Proof: Suppose that jf(z)j has a maximum at an interior point w. Then we

could _nd a circle C = w + Rei_ centred on w such that

(0) The disk with boundary C is in E,

(1) for every z 2 C, jf(z)j _ jf(w)j, and

(2) jf(w)j =___

1

2_ R2_

0 f(w + Rei_)d____

but we have

____

1

2_ Z 2_

0

f(w + Rei_)d_____

_

1

2_ Z 2_

0 jf(w + Rei_)jd_

But by (1) we have

1

2_ Z 2_

0 jf(w + Rei_)jd_ _

1

2_ Z 2_

0 jf(w)jd_ = jf(w)j

The two inequalities must mean that

Z 2_

0 jf(w)j 􀀀 jf(w) + Rei_)jd_ = 0

which can only happen if

jf(w)j = jf(w) + Rei_)j

for every point on the circle. But this must hold for every circle centred

on w of smaller radius than R, so jf(z)j must be constant in a disk shaped

neighbourhood of w.

Now we cover E with disks. each disk contained in E, with a disk centred at

every point of E. Since jf(z)j is constant in the _rst disk we can take any

4.5. SOME SOLID AND USEFUL THEOREMS 125

disk C0 of radius R0 intersecting the _rst disk, and observe that there is a

point w0 inside both disks and if we go through items (0), (1) and (2) above

replacing C by C0, R by R0 and w by w0 everything still holds. From which

we conclude that jf(z)j must also be constant (with the same value) on the

second disk.

This can be extended for all disks, and so jf(z)j is constant on E. This

contradicts the hypothesis. So jf(z)j cannot have an interior point of E as

its maximum. 2

Example 4.5.1 Find the maximum value of jz2 + 3z 􀀀1j on the unit disk

jzj _ 1.

Solution By the Maximum Modulus Principle, the value must be a maximum

on the boundary, jzj = 1. We can therefore put z = cos _ +i sin _, and try to

maximise

(cos 2_ + 3 cos _ 􀀀 1)2 + (sin 2_ + 3 sin _)2

since the maximum of a positive function occurs at the same place as the

maximum of its square. This simpli_es by elementary trigonometry to

11 􀀀 2 cos 2_

which has a maximum at _ = __=2. So z = _i is the location of the

maximum which has value p13. This may be con_rmed by plugging z = _i

into the original function and computing the modulus.

Theorem 4.8 (Cauchy's Inequalities)

For f analytic in a region containing the disk D of radius R centered on w,

and jf(z)j _ B for all z 2 D being a bound on jf(z)j on D, then the nth

derivative of f, fn has modulus bound:

jfn(w)j _

n!B

Rn

for all positive integers n.

Proof:

We have

fn(w) =

n!

2_i ZC

f(z)

(z 􀀀 w)n+1

dz

126 CHAPTER 4. INTEGRATION

Hence

jfn(w)j =____

n!

2_i Z 2_

0

f(w + Rei_(iRei_

Rn+1ei(n+1)_ d_____ and

jfn(w)j _

n!

2_Rn Z 2_

0 jf(w + Rei_

j d_

and since

Z 2_

0 jf(w + Rei_

j d_ _ 2_B

the result follows. 2

The extension of the Maximum Modulus Principle to the whole of C is obvious;

if f is entire (analytic on all of C ), then jf(z)j cannot have a maximum

at all, except in the rather uninteresting case where it is constant. Of course,

it might, in principle, be the case that although not achieving any maximum,

it 'saturates, that is, it gets closer and closer to some least upper bound. This

doesn't happen either:

Theorem 4.9 (Liouville's Theorem)

If f is an entire function which has jf(z)j bounded, then f is constant.

Proof: Take a circle of radius R around any point w 2 C .

By Cauchy's inequality for the _rst derivative we have

jf0(w)j _

B

R

where B is the bound for jf(z)j on all of C . Since this holds for all circles of

radius R, we see that

f0(w) = 0

This has to hold for all w 2 C . So f must be constant.

2

It is clear that these results for complex functions have implications for the

real and complex parts which are harmonic, and since any harmonic function

can be extended to a complex function by computing the conjugate harmonic

4.5. SOME SOLID AND USEFUL THEOREMS 127

function, we can deduce corresponding results for harmonic functions. For

example, we can deduce that the mean of the values on a circle is the value

of the function at the centre, and that the only bounded harmonic functions

de_ned on R2 are constant. When trying to solve Laplace's equation, every

little helps.

Exercise 4.5.1 Show that if u is a harmonic function of two variables, it

has the property that the mean value of u on a circle centred at w is u(w).

Exercise 4.5.2 Show that if u is a harmonic function of two variables and

E is a region in R2 , then the maximum value of ju(x; y)j is attained on the

boundary of E.

(It helps give some insight into the theorems for complex functions to see

what they say about the harmonic functions which are their components:

this makes particular sense with constraints on the modulus.)

Finally, the Fundamental Theorem of Algebra is going back to the roots of

Complex Analysis. It says that every polynomial of degree n has n roots,

generally complex, although some may be the same. So we count multiplicities.

Another way of putting this is that we can factorise any polynomial of

degree n into n linear factors (z 􀀀 r1)(z 􀀀 r2) _ _ _ (z 􀀀rn), where the roots rj

are generally complex. Now this is pretty much what the Complex Numbers

were invented for, in particular so that we could always factorise quadratics.

But there is more to the theorem than saying that if we take a real polynomial,

i.e. one with real coe_cients, then we can factorise it into complex

roots. What if we allow ourselves complex coe_cients? Well, it still works.

We can factorise all polynomials over C into linear factors

This is the Fundamental Theorem of Algebra (FTA):

Theorem 4.10 (Fundamental Theorem of Algebra)

A complex polynomial

P(z) = anzn +an􀀀1zn􀀀1 +_ _ _+a1z +a0

with n _ 1 can be factorised, uniquely up to order of terms as

an(z 􀀀 r1)(z 􀀀 r2) _ _ _ (z 􀀀rn)

128 CHAPTER 4. INTEGRATION

Proof:

We show _rst that P has at least one zero, that is there exists w 2 C such

that P(w) = 0.

If not then 1

P(z) is an entire function.

Now it is easy to see that

lim

jzj!1j

1

P(z) j = 0

since the anzn term of P dominates in C for the same reason that it does in

R. So we can _nd a disk of radius R centred on the origin, such that

jzj > R ) j

1

P(z)j _ 1

Now on the disk, j

1

P(z) j is a continuous function and the disk is compact so

there is some bound B which is attained by j

1

P(z) j on the disk. Actually on

its boundary.

It follows that j

1

P(z) j is bounded by the larger of B and 1 everywhere on C .

Hence, by Liouville's Theorem, 1

P(z) is constant, which is clearly not the case.

So P(z) has at least one zero, r1. Which means that (z 􀀀 r1) is a factor of

P(z), for we could certainly divide (z 􀀀r1) into P(z) by the usual rules, and

there could not be a non-zero remainder.

So P(z) = (z􀀀r1)Pn􀀀1(z) for a new polynomial of degree n􀀀1. This reduces

the degree of the polynomial by one. But the same argument as above applies

here also. So we can keep reducing the degree of the polynomial until it is

one, when the result is obvious.

2

It is also true that if the coe_cients of P are all real, then the roots must

come in conjugate pairs or be real. This certainly holds for P quadratic, for

if

z2 + az + b = (z 􀀀r1)(z 􀀀 r2)

we have immediately that

r1 + r2 = a; r1r2 = b

4.5. SOME SOLID AND USEFUL THEOREMS 129

and the _rst equation tells us that the imaginary parts of r1; r2 must have

equal and opposite values, and the second implies that the real parts have to

be the same.

It is easy to verify that if we multiply a quadratic with real coe_cients by a

linear term z 􀀀 r then we can get a cubic with real coe_cients only if r is

real.

Exercise 4.5.3 Complete the argument to show that if a polynomial in z

has real coe_cients, the roots must be real or come in conjugate pairs.

You are getting, in rather a lump, the results of about a century of exploration

of Complex Functions by some of the brightest guys in Europe. The impact of

it all, can be more than a bit mind numbing. Indeed if you don't feel smashed

by the weight of it all you have probably missed out on the meaning. This

is very dense, solid stu_ that needs a lot of thinking about to really absorb.

You are being told a lot of properties of these very special functions.

Exercise 4.5.4 Can you think of a well-known class of real functions which

have the property that they satisfy Liouville's Theorem?

You may be left wondering how they discovered all these results. Well, this

was before television, and mucking about with complex functions is rather

fun if you happen to be brilliant. There was certainly a lot to be found out.

And of all the ways of passing an idle hour known to man, just mucking

about with complex functions to see what happens has turned out to be one

of the most productive.

A very practical problem for people wanting to survive the exam is: how do

you get to know and feel comfortable with all these theorems?

The answer is, (1) you use them for solving problems, and (2) you work

through the proofs to see what the ideas are. Much of it is quite intuitive;

for instance the proof of the Cauchy Integral Formula depends strongly on

integrals around loops not changing as they shrink closer to a point inside the

loops. This in turn means that the functions have to be analytic except at the

point we are shrinking towards. This tells you what the assumptions in the

theorem are, which stops you doing something daft with the result. Settling

down somewhere quiet with a pen and lots of blank paper and making up

130 CHAPTER 4. INTEGRATION

your own problems, or working through a text book and doing the problems

there, is the best and surest way of feeling good about Mathematics. You

discover the reasons why Cauchy and Euler and Gauss did the original work:

there is a sense of triumph in getting something as fundamental as this sorted

out. It isn't easy, but when was anything worthwhile ever easy?

Our present culture is very di_erent from the one which produced the great

results of Mathematics and Science. It has taught you to regard anyone who

enjoys this sort of activity as qualifying for the title of King Nerd. A cynic

might say that the ideals of our culture are designed to reassure thickos, who

believe deeply that being really good at throwing balls in buckets makes you

a hero, while playing around with ideas makes you a nerd. This is because

there are a lot of thickos who are incapable of seeing the point of playing with

ideas, and you don't want a bunch of thickos going around feeling insecure

and inferior. Better by far if they focus their minds on watching other people

throw balls in buckets.

For the non-thickos:

Exercise 4.5.5 Show that if f is a non-constant complex function and jf(z)j >

1 for all z 2 E, and f is analytic in E, some region in C , then jf(z)j has its

minimum value on the boundary of E.

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