complex analysis - houston

8/10/2019 Complex Analysis - Houston

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Complex Analysis 2002-2003

cK. Houston 2003

1 Complex Functions

In this section we will define what we mean by a complex func-tion. We will then generalise the definitions of the exponential,sine and cosine functions using complex power series. To dealwith complex power series we define the notions of conver-gent and absolutely convergent, and see how to use the ratiotest from real analysis to determine convergence and radiusof convergence for these complex series.

We start by defining domains in the complex plane. Thisrequires the prelimary definition.

Definition 1.1The-neighbourhood of a complex numberzis the set of com-plex numbers{w C :|z w|< } where is positive number.Thus the -neigbourhood of a pointz is just the set of pointslying within the circle of radius centred at z. Note that itdoesnt contain the circle.

Definition 1.2A domain is a non-empty subset D of C such that for everypoint inD there exists a-neighbourhood contained inD.

Examples 1.3The following are domains.

(i) D = C. (Take c C. Then, any > 0 will do for an -neighbourhood ofc.)

(ii) D = C\{0}. (Take c D and let = 12

(|c|). This gives a-neighbourhood ofc inD.)

(iii) D ={z :|z a| < R} for some R > 0. (Take c C and let= 1

2(R |c a|). This gives a-neighbourhood ofc inD.)

Example 1.4The set of real numbers R is not a domain. Consider anyreal number, then any -neighbourhood must contain somecomplex numbers, i.e. the-neighbourhood does not lie in thereal numbers.

We can now define the basic object of study.

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Definition 1.5LetD be a domain in C. Acomplex function, denotedf :DC, is a map which assigns to each z inD an element ofC, this

value is denotedf(z).

Common Error 1.6Note thatf is the functionand f(z)is the valueof the functionatz. It is wrong to sayf(z)is a function, but sometimes peopledo.

Examples 1.7(i) Letf(z) =z2 for all z C.

(ii) Let f(z) = |z| for all z C. Note that here we have acomplex function for which every value is real.

(iii) Letf(z) = 3z4 (5 2i)z2 +z 7 for all z C. All complexpolynomials give complex functions.

(iv) Letf(z) = 1/z for all z C\{0}. This function cannot beextended to all ofC.

Remark 1.8Functions such as sin x for x real are not complex functionssince the real line in C is not a domain. Later we see how toextend the concept of the sine so that it is complex functionon the whole of the complex plane.

Obviously, if f and g are complex functions, then f + g,fg, and f g are functions given by (f+g)(z) = f(z) +g(z),(f g)(z) = f(z)g(z), and (f g)(z) = f(z)g(z), respectively.We can also define (f /g)(z) = f(z)/g(z) provided thatg(z)= 0on D. Thus we can build up lots of new functions by theseelementary operations.

The aim of complex analysis

We wish to study complex functions. Can we define differenti-ation? Can we integrate? Which theorems from Real Analysiscan be extended to complex analysis? For example, is there aversion of the mean value theorem? Complex analysis is es-sentially the attempt to answer these questions. The theorywill be built upon real analysis but in many ways it is easierthan real analysis. For example if a complex function is dif-ferentiable (defined later), then its derivative is also differen-tiable. This is not true for real functions. (Do you know an ex-ample of a differentiable real function with non-differentiablederivative?)

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Real and imaginary parts of functions

We will often use z to denote a complex number and we will

have z = x+ iy where x and y are both real. The value f(z)is a complex number and so has a real and imaginary part.We often use u to denote the real part and v to denote theimaginary part. Note thatu and v are functions ofz.

We often write f(x+ iy) = u(x, y) +iv(x, y). Note that u isa function of two real variables, x and y. I.e. u : R2 R.Similarly forv.

Examples 1.9(i) Letf(z) =z2. Then,f(x + iy) = (x + iy)2 =x2 y2 + 2ixy. So,

u(x, y) =x2 y2 and v(x, y) =xy.(ii) Let f(z) =|z|. Then, f(x+ iy) = x2 +y2. So, u(x, y) =

x2 +y2 and v(x, y) = 0.

Exercises 1.10Find u and v for the following:

(i) f(z) = 1/z forz C\{0}.(ii) f(z) =z3.

Visualising complex functions

In Real Analysis we could draw the graph of a function. Wehave an axis for the variable and an axis for the value, and sowe can draw the graph of the function on a piece of paper.

For complex functions we have a complex variable (thatstwo real variables) and the value (another two real variables),so if we want to draw a graph we will need 2 + 2 = 4 realvariables, i.e. we will have to work in 4-dimensional space.Now obviously this is a bit tricky because we are used to 3space dimensions and find visualising 4 dimensional spacevery hard.

Thus, it is very difficult to visualise complex functions. How-ever, there are some methods available:

(i) We can draw two complex planes, one for the domain andone for the range.

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(ii) The two-variable functions u and v can be visualised sep-arately. The graph of a function of two variables is a sur-face in three space.

u(x, y) = cos x+ sin y and v(x, y) =x2 y2

(iii) Make one of the variables time and view the graph assomething that evolves over time. This is not very helpful.

Defining ez, cos z and sin z

First we will try and define some elementary complex func-tions to play with. How shall we define functions such as ez,cos zandsin z? We require that their definition should coincidewith the real version when z is a real number, and we wouldlike them to have properties similar to the real versions of thefunctions, e.g. sin2 z+ cos2 z= 1 would be nice. However, sineand cosine are defined using trigonometry and so are hard togeneralise: for example, what does it mean for a triangle tohave an hypotenuse of length 2 + 3i? The exponential is de-fined using differential calculus and we have not yet defineddifferentiation of complex functions.

However, we know from Real Analysis that the functionscan be described using a power series, e.g.,

sin x= x x3

3! +

x5

5! =

n=0

(1)n x2n+1

(2n+ 1)!.

Thus, for z C, we shall define the exponential, sine andcosine ofzas follows:

ez :=n=0

zn

n!,

sin z :=n=0

(1)n z2n+1

(2n+ 1)!,

cos z :=

n=0(1)n z

2n

(2n)!.

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Thus,

e

3+2i

=

n=0 (3 + 2i)n

n! = 1 + (3 + 2i) +

(3 + 2i)2

2! +

(3 + 2i)3

3! +. . .

These definitions obviously satisfy the requirement that theycoincide with the definitions we know and love for real z, buthow can we be sure that the series converges? I.e. when weput in az, such as 3 + 2i, into the definition, does a complexnumber comes out?

To answer this we will have to study complex series andas the theory of real series was built on the theory of realsequenceswe had better start with complex sequences.

Complex Sequences

The definition of convergence of a complex sequence is thesame as that for convergence of a real sequence.

Definition 1.11A complex sequencecnconvergesto c C, if given any >0,then there exists Nsuch that|cn c|< for all nN.

We writecnc orlimncn= c.Example 1.12

The sequencecn= 4 3i7 n

converges to zero.Consider

|cn 0|=|cn|=4 3i7

n = 4 3i7n =

25

49

n=

5

7

n.

So

|cn 0|< (5/7)n < n log(5/7)

log

log(5/7).

So, given any we can choose N to be any natural numbergreater than log / log(5/7). Thus the sequence converges tozero.

Remark 1.13Notice thatan =|cn c| is a real sequence, and thatcn c ifand only if the real sequence|cn c| 0. Hence, we are sayingsomething about a complex sequence using real analysis.

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Paradigm 1.14The remark above gives a good example of the paradigm1 wewill be using. We can apply results from real analysis to pro-

duce results in complex analysis. In this case we take themodulus, but we can also take real and imaginary parts.

This is a key observation. Note it well!

Lets apply the paradigm. The next proposition shows thata sequence converges if and only its real and imaginary partsdo.

Proposition 1.15Letcn = an+ ibn where an and bn are real sequences, and c =a+ib. Then

cn

c

an

a and bn

b.

Proof. [] Ifcnc, then|cn c| 0. But0 |an a|=|Re(cn) Re(c)|=|Re(cn c)| |cn c|.

So by the squeeze rule|ana| 0, i.e. an a. Similarly,bnb.

[] Suppose an a and bn b, then|ana| 0, and|bn b| 0. We have

0 |cn c|=|(an a) +i(bn b)| |an a| + |bn b|.

The last inequality follows from the triangle inequality appliedtoz= anaand w = i(bnb). Because |ana| 0 and |bnb| 0we deduce|cn c| 0, i.e. cnc. HTTLAM 1.16Try not to use the definition of convergence to prove that asequence converges.

Example 1.17

n2 +in3

n3

+ 1

= n2

n3

+ 1

+ i n3

n3

+ 10 + i.1 =i.

Exercises 1.18(i) Which of the following sequences converge(s)?

(n+ 1)5

n5i and

5 12i

6

n.

(ii) Show that the limit of a complex sequence is unique.

1Paradigm: a conceptual model underlying the theories and practice of a scientific subject.(Oxford English Dictionary).

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Complex Series

Now that we have defined convergence of complex sequences

we can define convergence of complexseries.

Definition 1.19A complex series

k=0wk convergesif and only if the sequence

sn formed by its partial sumssn=n

k=0wk converges.

That is, the following sequences converges

s0 = w0

s1 = w0+w1

s2 = w0+w1+w2

s3 = w0+w1+w2+w3...

Lets apply the paradigm and give a result on complex seriesusing real series.

Proposition 1.20Letwk =xk+iyk wherexk and yk are real for allk. Then,

k=0wk converges

k=0xk and

k=0yk converge.

In this casek=0

wk =k=0

xk+ik=0

yk.

Proof. Let an =n

k=0xk, bn =n

k=0yk, and sn =n

k=0wk, andapply Proposition 1.15. The second part of the statementcomes from equating real and imaginary parts.

Example 1.21The series

n=0

(1)nin!

converges. Let xk = 0 and yk = (1)k

k! .

Then xk = 0, obviously, and (1)kk! =e1.Thus

n=0

(1)nin!

converges toi/e.

In real analysis we have some great ways to tell if a seriesis convergent, for example, the ratio test and the integral test.Can we use the real analysis tests in complex analysis? Thenext theorem says we can, but first let us make a definition.

Definition 1.22We say

k=0wk is absolutely convergent if the real series

k=0 |wk| converges.

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This definition is really the same as in Real Analysis, it hasmerely been extended to complex numbers in a natural way.

Now for a very important theorem which says that if a series

is absolutely convergent, then it is convergent.

Theorem 1.23If

k=0 |wk| converges, then

k=0wk converges.

This is a fantastic tool. Remember it. The assumption sayssomething about arealseries (we know lots about these!) andgives a conclusion about acomplexseries. Thus, we can applythe ratio test or comparison test to the real series and saysomething about the complex series. Great!

Proof. Let wk = xk +iyk, with xk and yk real. Then k=0

|wk

|convergent implies thatk=0 |xk| is convergent (because 0|xk| =|Re(wk)| |wk| and we can apply the comparison test).So the real series

k=0xk converges absolutely and we know

from Real Analysis I that this implies that

k=0xk converges.Similarly, the series

k=0yk converges.

Then,

k=0wk =

k=0xk+i

k=0yk, by Proposition 1.20.

HTTLAM 1.24When asked to show a series converges, show it absolutelyconverges.

Remark 1.25Note that the converse to Theorem 1.23 is not true. We al-

ready know this from Real Analysis. For example,

k=0(1)k

k

converges but

k=0

(1)kk = k=0 1k diverges.We now prove an infinite version of the triangle inequality.

Lemma 1.26Suppose that

k=0wk converges absolutely. Then

k=0 wk

k=0 |wk|.Proof. Forn1,

k=0

wk

=k=0

wknk=0

wk+nk=0

wk

k=0

wknk=0

wk

+nk=0

wk

k=0wk

n

k=0wk

+n

k=0|wk| .

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As n then obviously,|k=0wknk=0wk| 0, hence theresult.

Definition 1.27Acomplex power seriesis a sum of the form

mk=0ckz

k, whereck C andm is possibly infinite.Such a power series is a function ofz. Much of the theory ofdifferentiable complex functions is concerned with power se-ries, because as we shall see later, any differentiable complexfunction can be represented as a power series.

Radius of Convergence

Just as with real power series we can have complex powerseries that do not converge on the whole of the complex plane.

Example 1.28Consider the series

0 z

n, where z C. We know for z = 1this series does not converge because then we have

0 1

n =0 1 = 1 + 1 + 1 +. . . .We also know it converges for z = 0, because

0 0

n =0 0 = 0 + 0 + 0 + = 0. Hopefully, you remember from

Real Analysis I that for realzthe power series converges onlyfor1< z 1 and for|z| = 1 we dont know what willhappen. So

zn converges absolutely, and hence converges,

for|z| R.

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Proof. The proof is similar to that for real power series usedin Real Analysis I. Stewart and Tall also have a good proof, seep56-57.

HTTLAM 1.30Given a power series, immediately ask What is its radius ofconvergence?

Exercise 1.31Show that

0 z

n/nhas radius of convergence 1.

In the last exercise note that forz=1 the series converges,but for z = 1 the series diverges, (both these fact should bewell known from Real Analysis). This tells us that for|z| = 1we can get some values ofzfor which the series converges andsome for which the series diverges.

Sine, cosine, and exponential are defined for all complexnumbers

Let us now return to showing that the sine, cosine and expo-nential functions are defined on the whole ofC.

Example 1.32(Ill do this example in great detail. The next example will bemore like the solution I would expect from you.)

The seriesez = n=0 znn! converges for all z C.For anyz C letan =

znn!

. We wantn=0an to converge, sowe use the ratio test on this real series. We havean+1an

= zn+1(n+ 1)! znn!

=

zn+1zn n!(n+ 1)!

= |z|n+ 1

0 as n

.

The last part is true because for fixed zthe real number|z| isof course a finite constant.

So by the ratio test

n=0an =

n=0

znn!

converges. Thus byTheorem 1.23 the series

n=0

zn

n!converges for all z C.

The following is an example with some of the small detailmissing. This is how I would expect the solution to be given ifI had set this as an exercise.

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Example 1.33The seriessin z=

n=0(1)n z

2n+1

(2n+ 1)!converges for all z C.

Letan= (1)n z2n+1(2n+ 1)! . Thenan+1an = z2(n+1)+1(2(n+ 1) + 1)!

z2n+1(2n+ 1)!

=

z2n+3(2n+ 3)!

z2n+1

(2n+ 1)!

=

z2

(2n+ 3)(2n+ 2)

= |z|

2

(2n+ 3)(2n+ 2) 0asn .

So by the ratio test the complex series converges absolutely,and hence converges.

Exercise 1.34Prove thatcos zconverges for all z.

Properties of the exponential

We have defined the exponential function and shown that isdefined on all of C, lets now look at its properties. Most ofthese you may already from Numbers and Proofs, but theproofs may not have been rigorous.

Theorem 1.35(i) ez =ez, for all z C.

(ii) eiz = cos z+i sin z, for all z C.(iii) ez+w =ezew, for all z , w C.(iv) ez

= 0, for all z

C.

(v) ez = 1/ez, for all z C.(vi) enz = (ez)n, for allz C and n Z.

(vii)|ez|= eRe(z), for all z C.(viii)|eiy|= 1, for all y R.Proof. (i) We have

ez =

n=0(z)n

n! =

n=0(zn)

n! =

n=0zn

n! =ez.

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(ii) Exercise. (Just putizinto the power series and separatethe real and imaginary parts.)

(iii) This will be delayed until we deal with differentiability.

(iv) Note thatez and ez both exist. We have

ezez = ezz by (iii),

= e0

= 1, by calculation.

Thus,ez cannot be zero.(v) This is obvious from the proof of (iv).(vi) Follows from repeated application of (iii).(vii) We have

|ez

|2

= ez

ez

by definition,= ezez by (i),

= ez+z by (iii),

= e2Re(z)

=

eRe(z)2

by (vi).

As both|ez|andeRe(z) are real and positive we deduce that (vii)is true.

(viii) From (vii) we get|eiy|= eRe(iy) =e0 = 1. Corollary 1.36

(i) e2i = 1.

(ii) (De Moivres Theorem) (cos + i sin )n = cos n+i sin n forall R.

The proofs are left as simple exercises. Part (i) is one of thebest theorems in mathematics. It relates so many differentimportant numbers, e,

1, , and of course 1 and 2, in asimple expression.

Warning! 1.37

We have not shown that ezw = (ez)w for all z, w C. This isbecause we have not yet defined ab for all complex a and b.Consider z = 2i and w = i. Then (ez)w = (e2i)i = 1i. Whatcould 1i be? 2

Exercise 1.38Prove that

sin z=eiz eiz

2i and cos z=

eiz +eiz

2 .

2The astute reader may say define it to bee2.

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Proof. [] Writez= x+iy, so we getex+iy =ex(cos y+i sin y) =w.Now let us take the modulus of both sides:

|ex+iy| = |w||ex||eiy| = |w|

|ex| = |w|ex = |w|

logeex = loge |w| (using the real log function)x = ln |w|.

Now suppose thatw = r(cos + i sin )for some realr and , i.e.r =|w| and = arg(w). Then ez = w implies thatr = ln |w| andy= + 2k for somek Z. Sozhas the form in the statement.

[] Ifz= ln |w| +i(arg(w) + 2k) for some k Z thenez = eln |w|+i(arg(w)+2k) =eln |w|ei arg(w)e2ik

= |w|ei arg(w) e2ik =|w|ei arg(w) =w.

Example 1.41Solve ez = 1 +i

3.

Solution: Let w = 1 + i

3. The modulus of w is |w| =

12 +

32

=

4 = 2. By drawing a picture (or through careful

use of calculator) we can see thatarg(w) = 3+ 2n,n Z. So

z= ln 2 +i

3+ 2n

, n Z.

HTTLAM 1.42Notice how well working out the modulus and argument servesus. Conclusion: calculate modulus and argument.

Common Error 1.43Dont forget the2k with the argument.

Exercise 1.44Solve e2iz =i. (Its noti(/2 + 2k).)

So does the theorem allow us to define the log of a complexnumber? Yes, if we define the log to be the complex numberwith < arg(w) . (The point is that if we have aw thenthe proposition gives us lots ofzs to choose from. Ifz is suchthatez =w, then z+ 2i will work just as well (ez+2i =eze2i =ez.1 = ez = w). Thus, there is some ambiguity and we makea choice.) However, if we are trying to solve an equation andtake the log of both sides using this definition, then we may be

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losing solutions. So in fact the best definition is to make thefunction multi-valued. This is something we will not go into ingreat depth just now.

Another worked example

Example 1.45Solve the equation sin z= 2.

Solution: We can rewrite this as eiz eiz

2i . Letw = eiz. Then

the equation becomes 1

2i

w 1

w

= 2. So,

w2

1 = 4iw

w2 4iw 1 = 0w =

4i (4i)2 + 42

= 4i 12

2= 2i 3= (2

3)i.

Now, eiz =w = (2 3)i, so

iz = ln |w| +i arg(w)= ln |(2 3)i| +i

2+ 2n

= ln |2

3| +i

2

+ 2n

z = 1

i

ln |2

3| +i

2

+ 2n

= i

ln |2

3| +i

2+ 2n

=

2

+ 2n

i ln |2

3|

= 2+ 2n i ln(2

3).

(The last equality is true because ln(2+

3)> 0 and ln(23)>0.)

HTTLAM 1.46Note that in the above example we replaced eiz with anothercomplex numberw, because we could then get a polynomialequation.

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Exercise 1.47Show that

sin z= 0 z=k, k Z,cos z= 0 z=1

2(2k+ 1), k Z.

These results will be used later.

Summary

Paradigm: Complex analysis is developed by reducing toreal analysis, often through taking the modulus.

We define exponential, sine and cosine by power series.

Ifn=0 |wk| converges, then n=0wk converges. Apply the ratio test, comparison test, etc, to the modulus

of terms of a complex series to determine convergence.

For power series use the ratio test to find radius of con-vergence.

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2 Complex Riemann Integration

In a later section we define contour integration, that is inte-

gration over a complex variable. This notion is fundamental incomplex analysis. But let us first generalise integration anddifferentiation to complex-valued functions of arealvariable.

A complex-valued function of a real variable is a map f :S C, where S R. E.g. If f(t) = (2 + 3i)t3, t R, thenf(1) = 2 + 3i C.

Such a function is different to a complex function. A complex-valued function of a real variable takes a real number and pro-duces a complex number. A complex function takes a complexnumber from a domain and produces a complex number.

Differentiation of complex valued real functions

Suppose thatf : R C is given byf(t) = (1 + 3i)t2. If we definef(t) using the standard definition:

f(t) = lim0

f(t+) f(t)

, ( R),then we get f(t) = 2(1 + 3i)t. That is, in this case, the rulef(t) =nctn1 forf(t) =ctn, holds even though c is complex.

Basically, all similar rules work in this way, any constantscan be real or complex. So, for instance,

d

dtect = cect,

d

dtsin(ct) = c cos(ct),

d

dtcos(ct) = c sin(ct).

Exercise 2.1Let(x) = 3x3 + 2ix i+ tan((4 + 2i)x). Then,

(x) =

Remark 2.2This is not the same as differentiation with respect to a com-

plex variable.3 That will come later.

Complex Riemann Integrals

Now we shall integrate complex-valued functions with respectto one real variable. We shall do this with a bit more care thandifferentiation.

3They do behave in much the same way though.

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So a lot of the time we can use standard integrals to calculatecomplex-valued integrals with respect to a real variable t.

Exercise 2.6Calculate

20 t

2 it3 cos(2t) dt.

Triangle inequality for C-RI

We need the following result later. Its format should be fa-miliar from real analysis, the only difference here is that thefunctions can be complex-valued.

Lemma 2.7Ifg : [a, b] C is C-RI and|g| is R-RI, then b

a

g(t) dt b

a

|g(t)| dt.

Proof. If LHS = 0, then the statement is trivial. Hence, as-sumeLH S= 0. Let =| b

ag|/ b

ag. (Hence||= 1).

So, ba

g(t) dt

= ba

g(t) dt

= b

a

Re (g(t))dt+i b

a

Im(g(t)) dt

=

ba

Re (g(t))dt, because LHS is real,

ba

|g(t)| dt, asRe(z) |z|,

=

ba

|| |g(t)| dt

=

ba

|g(t)| dt, as||= 1.

Summary

We can integrate and differentiate complex-valued func-tions of real variables in the same way as real-valuedfunctions of real variables.

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3 Contours

In the next section define integration along a contour in the

complex plane.4 This is a fairly abstract process, the mean-ing of which usually takes a little time to understand. Fortu-nately, it is easy to doas it has similar properties to Riemannintegration of one real variable, and you have years of experi-ence of that.

In case you think that it is too abstract and not relevant toreal problems, then consider the integral 2

0

ecos cos(n sin n) d.

This is a seriously nasty integral! Imagine trying to solve itvia the methods we know. Using contour integration we shallshow that it is very simple to calculate.

First though we will define contours.

Contours

Definition 3.1Acontour (also called apath) is a continuous map : [a, b]C which is piecewise smooth, i.e. there exista = a0< a1 < a2 < < an= b such that

(i) |[aj1, aj]is differentiable, for all j ,(ii) is continuous on[aj1, aj], for all j .

(The left and right derivatives ofataj may differ.)We say is closed if(a) =(b).

Warning! 3.2A contour is not a complex function. It is a complex-valuedfunction of a real variable. Its image is usually some curve inthe plane.

Examples 3.3(i) Straight line from to : This is : [0, 1] C given by

(t) =+t( ).(ii) Circle of radius r based at the origin: : [0, 2] C given

by(t) =reit.

(iii) Circle of radius r based at z0: : [0, 2] C given by(t) =z0+re

it.

4Those of you who have done MATH2360 or MATH2420 will see that this is just a lineintegral.

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(iv) Circular arc of radius r based atw: (t) =w+ reit, 1t2. (So 0t2 gives the circle above).

(v) Let : [1, /2] C be given by(t) =

t+ 1, for 1t0,eit for0t/2.

Draw the image:

We draw an arrow to show the direction we go in.

(vi) Let: [0, 4] C be given by(t) =eit. Then the image ofis the unit circle centred at zero. The contour goes roundthe circle twice. This subtlety will be important later.

HTTLAM 3.4Given a contour, try to draw its image.

Common Error 3.5

There is often confusion between a contour and its image. Acontour is not a set of points in the complex plane, it is a map.Consider the contours (ii) and (vi) above, taking r = 1 in

(ii). They have the same image, the unit circle. However, thecontours are different, one maps from [0, 2], the other[0, 4].

We do use the notation z later, by which we mean z([a, b]). Strictly speaking, writingzis incorrect because is not a set.

Since contours are complex-valued functions of a real vari-ables, we can differentiate them, etc, with ease.

Summary

A contour is a continuous map : [a, b] C which ispiecewise smooth. It is a complex-valued function of areal variable.

Straight line from to : : [0, 1] C given by (t) =+t( ).

Circle of radius r based at z0: : [0, 2] C given by(t) =z0+re

it.

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4 Contour Integration

We now come to probably the most important definition in

complex analysis: contour integral. It is central to the mod-ule. If you dont understand this section, then the rest of thecourse will be a complete mystery to you.

Definition 4.1Letf :D C be a continuous complex function and: [a, b]C be a contour. Then, the integral off along is

f=

f(z) dz:=

ba

f((t))(t) dt.

Note that f((t)) and

(t) are complex-valued functions of areal variable, and hence so is their product. Thus we canintegrate this product.

Example 4.2Let(t) =t+it2 for0t2 and f(z) =z. Then, (t) = 1 + 2it,and

f =

20

(t+it2)(1 + 2it) dt

= 2

0

t+ 2it2 +it2 + 2i2t3 dt

=

20

t+ 3it2 2t3 dt

=

1

2t2 +

3it3

3 2t

4

4

20

=

1

2t2 +it3 t

4

2

20

= 1

222 +i23 2

4

2=

6 + 8i.

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Example 4.6Letbe as in Example 3.3(v). Find

z2 dz.

z2 dz = /21

(t)2(t) dt

=

01

(t+ 1)2.1 dt+

/20

eit2

ieit dt

=

01

(t+ 1)2 dt+

/20

ie3it dt

=

1

3(t+ 1)3

01

+

i

3ie3it/20

= 13 0+ i3i[i 1]= i

3.

Remarks 4.7(i) Suppose f : D C is a complex function such thatf(x)

is real forx real, for example, sin x. If we take : [a, b]Cgiven by(t) =t foratb, then

f(z) dz=

b

a

f(t)(t) dt=

b

a

f(t) dt.

Thus, by taking a contour along the real line, we can seethat contour integration includes the theory of real inte-gration as a special case.

(ii) From a purely formal viewpoint, we can justify the defini-tion of contour integral by saying that we are replacing zby(t) so we need to replace dz by(t)dt, (which can bethought of as (dz/dt)dt).

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Fundamental Example

Take the function defined byf(z) = (z

w)n where n

Z, (so

forn < 0 the map is not defined atw). Let be a circle withcentrew and radiusr >0, i.e. (t) =w+reit,0t2. Then,

(z w)n dz = 20

((t) w)n (t) dt

=

20

w+reit wn .ireit dt

=

20

rneint .ireit dt

= irn+1 2

0

ei(n+1)t dt

=

rn+1

n+ 1

ei(n+1)t

20

= 0, ifn=1,i20

1.dt = 2i, ifn =1.

Thus

1

z wdz= 2i and

(z w)n dz= 0forn=1.Note this well, this innocuous looking calculation will be

used to devastating effect later!

Summary The integral off along is

f=

f(z) dz=

ba

f((t))(t) dt.

1

z wdz= 2i, (t) =w+reit, 0t2.

(z w)n dz= 0forn=1, (t) =w+reit,0t2.

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5 Properties of Contour Integration

There are a number of well known properties of ordinary inte-

gration: ba

f(x) +g(x) dx =

ba

f(x) dx+

ba

g(x) dx, , R, ba

f(x) dx =

ca

f(x) dx+

bc

f(x) dx, a < c < b, ba

f(x) dx = ab

f(x) dx,

b

a

f(x) dx = 1(b)

1

(a)

f((y))(y) dy (change of variables).

All of these have analogues in contour integration. We shallnow describe them.

In the following the functions will be continuous on somedomain D and the contours will be maps into D.

Linearity.

Iff , g are continuous on D, a contour in D,, C, then

f+ g=

f+

g.

So this has the same format as in integration over a real vari-able.

The proof follows directly from the linearity property of Rie-mann integration. The details are as follows:

f+ g =

ba

(f+ g)((t))(t) dt

= b

a

(f((t)) +g((t))) (t) dt

=

ba

f((t))(t) dt+

ba

g((t))(t) dt

=

f+

g.

Integration over joins

What if produce a contour by doing one after another?

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Definition 5.1If1 : [a, b] C and 2 : [b, c] C are two contours such that1(b) = 2(b), then their join (or sum) is the contour 1+ 2 :

[a, c] C given by

(1+2) (t) =

1(t) atb,2(t) btc.

The join is continuous by the glue rule.

Example 5.2Consider Example 3.3(v). The contour : [1, /2] C is givenby

(t) = t+ 1, for 1t0,eit for0

t

/2.

This can be produced from 1(t) = t+ 1 for1 t 0 and2(t) =e

it for0t/2.Proposition 5.3For any continuousf

1+2

f=

1

f+

2

f.

Again, this follows from applying the definition and using a

C-RI property: ba = ca + bc .Example 5.4Find

1

(z w)dz, where is the boundary of the square withcorners w l il, starting atw + l li and going anticlockwise.

So =1+2+3+4. By Proposition 5.3 we have

dz

z w =4i=1

i

dz

z w .

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To compute1

let1(t) =w+l+it,ltl. We have,

1 dzz w = l

l

t(t)

1(t) wdt

=

ll

i

l+itdt

=

ll

l itl2 +t2

i dt

=

ll

t

l2 +t2dt +i

ll

l

l2 +t2dt

= 0 +i tan1t

ll

l

= i

2.

Similarly2

=3

=4

=

2i (check!). So,

dz

z w = 2i.

Notice that this coincides with the value of the integral whenis a circle, see the Fundamental Example.

Reverse contour

Definition 5.5If : [a, b] C is a contour, then its reverse is the contour()(t) =(a+b t), foratb.

The point is that we do backwards. Instead of starting at(a) we end there, etc.:

()(a) =(a+b a) =(b),()(b) =(a+b b) =(a).

Proposition 5.6For any continuousf, we have

f=

f.

Proof. Again we apply definitions and use RI properties, butin this case we also need a simple change variables: s= a+bt,

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sodt =ds. Note that()(t) = (1)(a+b t). We have,

f = b

a f(()(t))()

(t) dt

=

ba

f((a+b t))(1)(a+b t) dt

= ab

f((s))(s) (ds)

=

ab

f((s))(s) ds

= ba

f((s))(s) ds

=

f.

Example 5.7See Exercise 4.4(ii) and (iii). Then 3=2 as follows. We havea= 0 and b = 1, so

(2)(t) =2(0 + 1 t) =2(1 t) = (1 t) +i(1 t) =3(t).This explains why

3f3(z) dz=

2f2(z) dz.

Reparametrisation

Now we shall do the analogue of a change of variables.

Definition 5.8Let : [a, b] C be a contour and let : [c, d] [a, b] be afunction such that

(i) is continuously differentiable ,

(ii) (s)> 0 for all s[c, d],(iii) (c) =a, and (d) =b.

Then the contour: [c, d] C defined by(s) =((s))is calledareparametrisationof.

Example 5.9Let(t) = eit for0t2. Let(s) = 2s for0s1. Then(s) =e2is for0s1.

We think of reparametrised contours as equivalent since wehave the following.

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Summary

Iff, g are continuous onD,a contour inD,,

C, then

f+ g=

f+

g.

Iffis continuous on D,a contour in D, then1+2

f=

1

f+

2

f.

If: [a, b] C is a contour, then its reverse is the contour()(t) =(a+b t), foratb.

Forfcontinuous, we have

f=

f.

Ifis a reparametrisation of, thene

f=

f,

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6 The Estimation Lemma

Recall for a real continuous function that ba

f(x) dx sup

x[a,b]

{|f(x)|} (b a).

We would like a complex version of this, that is, is there somebound on|

f(z) dz|? It is obvious that the first part in the

product above can be generalised, but what does ba cor-respond to? Those of you familiar with measure theory willknow it is the length of the interval from a to b, and so thegeneralisation to complex analysis is the length of the contour.

Definition 6.1Let : [a, b] C be a contour. The length of, denoted L(),is defined to be

L() =

ba

|(t)| dt.

This really does measure the length of the curve. Intuitivelyspeaking, (t) is the velocity of a contour, so |(t)| is thespeed, and the integral of speed over time gives the lengthof a path.

Example 6.2Let(t) = + t()with0t1 be the straight line contourfrom to.

We have

L() =

10

|(t)| dt= 10

| | dt=| | 10

dt=| |[t]10= | |.

So, the length of the contour from to is||, which isreassuring.

Exercise 6.3Show that the length of the contour given by traversing onceround the the circle of radius r based at the origin is 2.

The next theorem is similar to one from real analysis:

Lemma 6.4 (Estimation Lemma)Letf :D C be a continuous complex function and: [a, b]D be a contour. Suppose that|f(z)| M for all z. Then

f(z) dz

M L().

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Proof. We have,

f(z) dz = b

a

f((t))(t) dt

ba

|f((t))(t)| dtby Lemma 2.7

ba

M|(t)|dt= ML().

Example 6.5Show that

ez

z+ 1dz RR 1

where describes the semi-circle from iR toiR in the lefthalf-plane{z : Re(z)0}and R >1.

Solution: Ifz=x+iy lies on the image of, thenx0 andso|ez| = eRe(z) = ex e0 = 1. Also, whenz lies on ,|z| = R soR=|z+ 1 1| |z+ 1| + 1. Thus|z+ 1| R 1. Hence,

ez

z+ 1

1

R 1 , z.

We know thatL() = R as is semi-circle of radius R, so bythe Estimation Lemma (6.4)

ez

z+ 1dz

1R 1L() = RR 1 .Remarks 6.6

(i) The constant M always exists: On the image of thefunction|f(z)| will always be bounded because the mapt |f((t))| is a continuous real function on [a, b], andhence is bounded. So M = supz{|f(z)|} will do as abound. Anything bigger than this is also useful.

(ii) We can show some integral is zero by finding an M thattends to zero or some contour, which can be made smaller,so thatL()0.

Termwise integration of series

The lemma will be useful in a number of contexts. To beginwith, we use it to prove that we can integrate certain seriesterm-by-term. (We know that an infinite series can be differ-entiated term-by-term.)

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Corollary 6.7 (Term-by-term integration of series)Letbe a contour in a domainD. Letf :D C andfk :D Cbe continuous complex functions, k

N. Suppose that

(i)

k=0fk(z) converges to f(z) for all z;(ii) there exist real constants Mk such that|fk(z)| Mk for all

z;(iii)

k=0Mk converges.

Then,k=0

fk(z) dz=

k=0

fk(z) dz=

f(z) dz.

Proof. The second equality is obvious. We show the first. LetM= k=0Mk. We have

f(z) dznk=0

fk(z) dz

=

f(z)

nk=0

fk(z)

dz

sup

z

f(z) nk=0

fk(z)

L()

= supz

k=n+1

fk(z)

L()

supz

k=n+1

|fk(z)| L()

k=n+1

Mk

L()

=

M

nk=0

Mk

L()

0 L(), as n ,= 0.

Thus, the first equality is true.

Remark 6.8I could have defined uniform convergence and so on for com-plex series in order to state the theorem. Rather than wastetime doing so I just used a version of the Weierstrass M-test inthe assumptions above. Hopefully, you remember from RealAnalysis II that this implies uniform convergence.

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Example 6.9For any contour the integral

ez dz can be calculated by

term-by-term integration of the series forez.

Letfk(z) =zk

k!, then ez =

k=0

fk(z), so condition (i) is fulfilled.

We have

|fk(z)|=zkk! =|z|kk! .

But|z| =|(t)| and (t) is a continuous map from [a, b] to C soits image must lie within an origin-centred circle of radius R,for some large enough R. Thus|z| R for all z.

Hence, letMk =Rk

k!, then |fk(z)| Mk for allz. Therefore,

condition (ii) holds.Also,

0

Mk =0

Rk

k! =eR,

so

0 Mk converges. Condition (iii) holds.Thus,

ez dz=

k=0

zk

k! =

k=0

zk

k!dz.

This of course begs the question, how do we find zk

k!dz? Ob-

viously, we can apply the definition of integration, but thereare better ways, such as the Fundamental Theorem of Calcu-lus, as we shall see in the next two sections.

Summary

The length of, denoted L(), is defined to be

L() =

b

a

|(t)| dt.

Suppose that|f(z)| M for allz. Then

f(z) dz

M L(). We can integrate term-by-term complex series that satisfy

a WeierstrassM-test type condition.

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7 Complex Differentiation

If we were inventing the theory of differentiation of complex

functions for the the first time, then we might be tempted todefine complex differentiable to mean that the real and imag-inary parts of the function are differentiable with respect tox and y. This would give us a theory, but not a great one; itwould be the same as theory of differentiable maps from R2 toR2.

Instead we develop a theory that really uses the complexnumbers. This gives a much richer theory. It looks a lotbetter, but more importantly, it allows us to solve some hardproblems concerning real functions in a simple manner.

Continuity

Lets first give a definition of continuity.

Definition 7.1Suppose thatf :D C is a complex function on a domain D.We sayf is continuous at ciflimzcf(z) =f(c)

This is not remarkably different to real analysis.

Remark 7.2This is equivalent to either of the following.

(i) For all >0, there exists >0 such that|f(z) f(c)| < ,whenever|z c|< .

(ii) For any sequencezn, we haveznc implies thatf(zn)f(c).

The proofs of these facts are the same as in real analysis.

Definition 7.3We say f is continuous on D if f is continuous at c for allc

D.

We need some examples and the next result helps supplysome.

Proposition 7.4Suppose f(x+iy) = u(x, y) + iv(x, y). Then, f is continuous ifand only ifu and v are continuous.

Proof. Use Proposition 1.15.

Thus, f(z) = zn is continuous. Sums and products of contin-uous functions are continuous, and so on.

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Differentiation

Now we come to the crucial definition, that of differentiation.

It doesnt look different to the real situation, but the conse-quences are far more profound.

Definition 7.5Let f : D C be a complex function. Then, f is complexdifferentiable at c if

limh0

f(z+h) f(z)h

, h C

exists. We write f(c) for this limit.We say f is complex differentiable on D if it is complex

differentiable atc for all cD.The definition looks the same as in the real case, but the factthath can go to zero from anydirection in the complex planemakes a huge difference, it imposes considerable restrictions.

Remark 7.6Complex differentiable functions are also called holomoprhicoranalytic.

Example 7.7The function f : C C given byf(z) = z2 is differentiable onC

. Letc C. Then

limh0

(c+h)2 c2h

= limh0

c2 + 2ch+h2 c2h

= limh0

2c+h

= 2c.

That is, f(c) = 2c, as expected.

More generally, we have that, if f(z) = bzn for some complexconstantb, then f(z) = nbzn1. The proof of this is the same

as the real situation, at least symbolically. Of course, math-ematicians do not want to do anything as clumsy or ugly asusing first principles. We would use a theorem such as thefollowing.

Proposition 7.8Letf and g be complex functions on the domain D.

(i) Iff and g are differentiable atcD, then so are(a) f+ g, and (f+ g)(c) =f(c) +g(c);

(b) f g, and (f g)(c) =f(c)g(c) +f(c)g(c);

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(c) f /g, and (f /g)(c) = g(c)f(c) f(c)g(c)

g(c)2 provided g(c)=

0.

(ii) (Chain Rule) Suppose f :D C and g: E C are complexfunctions withf(D)E. Iffis differentiable atc and g isdifferentiable atf(c), then g f is differentiable atc with(g f)(c) =g (f(c))f(c).

Proof. These are proved in the same way as the real case.

Warning! 7.9Unlike continuity, complex differentiablity isnt the same asbeing differentiable with respect to two real variables. Thereis a connection as we see shortly. In fact it is this difference

that makes complex analysis so different to real analaysis.

Exercise 7.10Show that iff :D C is differentiable atc, thenf is continu-ous atc.

We dont yet know that we can differentiate series term-by-term and so cant immediately prove thatez, cos z and sin zare differentiable, or that their derivatives are what we ex-pect them to be. It is possible to work find their derivativesfrom first principles, (you can try this as an exercise if you

are keen), but will delay a proof till later and just state thefollowing.

Theorem 7.11The elementary functions have the expected derivatives:

(i) d

dzez =ez, for all z C.

(ii) d

dzsin z= cos z for all z C.

(iii) d

dzcos z=

sin z for all z

C.

Exercise 7.12Where are the following funtions not differentiable?

z2, 1

z,

1

z2,

sin z

z(z2 + 1), tan z.

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Summary

The complex function f is continuous atc if limzcf(z) =

f(c).

The complex functionfis complex differentiable atc if

limh0

f(z+h) f(z)h

, h C.

Differentiablity = continuity. The elementary functions have the expected derivatives.

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8 Cauchy-Riemann Equations

If f is a differentiable complex function, then the real andimaginary parts are differentiable as real functions. (But notvice versa, see earlier warning.)

Theorem 8.1Supposef(x+iy) =u(x, y) +iv(x, y), wherez= x+iy, and thatfis differentiable atc = a+ib. Then the partial derivatives ux,uy,vx and vy all exist at(a, b) and

f(c) =ux(a, b) +ivx(a, b) =vy(a, b) iuy(a, b).

Proof. We know f(c+h) f(c)

h f(c) as h0. Leth = k +il,

we getu(a+k, b+l) u(a, b) +iv(a+k, b+l) iv(a, b)

k+il f(c) ask+il0.

So if we leth0 through realvalues, i.e. l= 0, thenu(a+k, b) u(a, b) +iv(a+k, b) iv(a, b)

k f(c) as k0.

Therefore, u(a+k, b) u(a, b)

k Re(f(c)) as k 0. So ux(a, b)

exists and equalsRe(f(c)). Likewise,vx(a, b) =I m(f(c)).

The second equation follows from letting h

0 through

imaginary values, i.e. k= 0. And now for a result that is fundamental in all complex

analysis courses.

Corollary 8.2 (Cauchy-Riemann equations)Iff(z) =u(x, y) +iv(x, y) is differentiable, then

ux= vy and vx=uy.Proof. Equate real and imaginary parts in the theorem above.

The two equations are called theCauchy-Riemann equationsafter two of the founders of complex analysis.

Note that the corollary says that iff is differentiable, thenthe equations hold, but says nothing of the converse, which isnot true anyway!

Examples 8.3(i) Letf(z) =z2 = (x + iy)2. Then,u(x, y) =x2 y2 andv(x, y) =

2xy. We see that

ux= vy = 2x and uy =vx=2y,so the C-R equations hold.

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(ii) Letf(z) =|z|2 =x2 +y2. Thenu = x2 +y2 and v = 0. Thus,ux = 2x, vy = 0,

uy = 2y, vx = 0.

The only place where the C-R equations are satisfied isx= y = 0.

So f is differentiable nowhere, except possibly at 0. Is itdifferentiable at0? Well,

f(0 +h) f(0)h

=|h|2 0

h =

hh

h =h0 ash0,

so f(0) = 0.

Exercise 8.4Show that the functionf(z) =|z| is not differentiable anywherein C.

Remark 8.5The preceding exercise shows that complex differentiability isimposing a stronger condition real differentiability, becausefor real points of C the function f(x) =|x| is differentiable,provided x= 0. (You know this last fact well, I hope!).

The condition is stronger because we require f(c) to existash0 from all directions, not just real ones.

Remark 8.6The converse to Corollary 8.2 is false. For example, let

f(z) =

0, ifx = 0ory = 0,1, otherwise.

Then, the equations are satisfied, (check!), but f is not con-tinuous, so cant be differentiable.

However, ifux, uy, vx, vy exist nearc and are continuous atc, and satisfy the C-R equations, then f is differentiable atc.This fact wont be used later, but is useful to know.

Harmonic functions

Definition 8.7Letw(x, y) be aC2-function of two real variables5. Then w is aharmonic function if it satisfies Laplaces equation:

wxx+wyy = 0.

Example 8.8Letw(x, y) =x2y2. Thenwxx = 2and wyy =2, sowxx+wyy = 0.

5ACk-function is a function that is differentiablek times

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Laplaces equation is important in potential theory and manyother areas, and so, since they are solutions of it, harmonicfunctions are of particular interest. Complex analysis provides

many examples of harmonic functions, as the next theoremshows.

Theorem 8.9If f : D C is complex differentiable, and u and v are C2-functions, thenu and v are both harmonic:

uxx+uyy = 0and vxx+vyy = 0.

Proof. We use the C-R equations:

uxx+uyy = (ux)x+ (uy)y = (vy)x+ (vx)y =vxy vyx = 0.Similarly forv.

Conversely, given an harmonic function u there is locally(i.e. in some -neighbourhood) a harmonic function v suchthatf(x+iy) = u(x, y) +i(v, y) is complex differentiable. Suchav is called a harmonic conjugate ofu.

Example 8.10Letu(x, y) = 2xy, thenu is harmonic. We can constructv usingthe C-R equations (theyre very useful, arent they!) Because

2y= ux= vy = v= y2

+g(x) whereg is a function ofx,and

2x= uy =vx = v=x2 +h(y) whereh is a function ofy,we can deduce thatv = y2 x2 +C whereCis a constant.

Summary

(Cauchy-Riemann equations) If f(z) = u(x, y) +iv(x, y) isdifferentiable, then

ux= vy and vx=uy.

The function w: R2 R is a harmonic function if it satis-fies Laplaces equation:

wxx+wyy = 0.

For every harmonic u there is a harmonic function v(calleda harmonic conjugate ofu) such thatf(x+iy) = u(x, y) +i(v, y) is complex differentiable.

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9 Fundamental Theorem of Calculus for Com-

plex Functions

One of the best theorems in Real Calculus is the FundamentalTheorem of Calculus. We now see this in a contour integrationsetting.

Theorem 9.1 (Fundamental Theorem of Calculus)Letf :D C be a continuous complex function and: [a, b]D be a contour. Suppose there exists a complex differentiableF :D C such thatF =f. Then,

f(z) dz= F((b)) F((a)).

Proof. Let a = a0 < a1


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How common are antiderivatives for continuous complexfunctions? Do they always exist? For real continuous func-tions we know that the Riemann integral can be found and

this will be an antiderivative. Unfortunately, the analogue isnot true for continuous complex functions, not even differen-tiable ones. Consider this corollary of the FTC and the follow-ing example.

Corollary 9.3With the assumptions of the above theorem suppose that isany closed contour. Then

f(z) dz= 0.

Proof. The definition of a closed contour is that (a) = (b).So,

f(z) dz= F((b)) F((b)) = 0.

Example 9.4Letf(z) = 1/z. Then, fis differentiable on D = C\{0}. Letbethe unit circle round the origin, traversed once anti-clockwise.Then, by the fundamental example we know

f(z) dz=

1z

dz= 2i.

Therefore, if there existed anF :D Csuch thatF =f, thenthe corollary would be contradicted. Hence, antiderivatives donot always exist.

Property of exponential

We can now prove a result we would expect to be true by anal-ogy with real analysis: if a function has zero derivative, then

it is constant.First we need a definition.

Definition 9.5A domainD is called connectedif for each pair, D, thereexists a contour: [a, b]D such that(a) = and (b) =.So a domain is connected if we can draw a curve from anypoint to any other.

Examples 9.6(i) The domains D = C and D = C\{0}are both connected.

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(ii) The domain D = C\{z : z is real} is not connected. Thereis no way to construct a contour starting below and fin-ishing above the real line, without crossing that line. But

the real line is not in D.

Theorem 9.7Suppose that D is is connected domain, and f : D C isanalytic, such thatf(z) = 0 for all zD. Then,fis constant.Proof. Take any and in D. Then, as D is connected, thereexists a path : [a, b]D, such that(a) = and (b) =. Bythe FTC

f() f() =f((b)) f((a)) =

f =

0 = 0.

Thus f() = f(). Since these were general points of D, wededuce thatfis constant.

This theorem allows us to prove the property ofez in Theo-rem 1.35 we did not prove earlier.

Corollary 9.8For all complex numberszand w we have ez+w =ezew.

Proof. Letf(z) = ezez, where is any complex number. Bythe product rule we getf(z) =ezez +ezez = 0. Thus, bythe theorem, the function is constant. So, if we let = w +z,then we get

f(w) = f(0)

ewe(w+z)w = e0ew+z+0

ewez = ew+z.

* Existence of antiderivatives

The next proposition gives some conditions equivalent to theexistence of antiderivatives.

Proposition 9.9Let f : D C be a continuous complex function on a con-nected domain D. The following statements are equivalent.

(i) There exists an F :D C such thatF =f.(ii)

f= 0 for every closed contour in D.

(iii)

fonly depends on the end points offor any contour

in D.

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Proof. We shall prove (i)(ii)(iii)(i).(i)(ii): This is just Corollary 9.3.(ii)

(iii): Suppose that and are two points inD, then we

want to prove that the integral off over any two contours 1and 2 both starting at and finishing atare equal.

Consider the contour 12. This is a closed contour be-cause it starts and finishes at. (Draw a picture!) Hence, (ii)implies that

12

f= 0. We have

0 =

12

f=

1

f2

f.

Hence (iii) is true.(iii)

(i): We shall define a function F and show that it is

differentiable. Fixz0D. Since D is connected, for any pointzD, there exists a contour fromz0 toz. We define

F(z) =

f(w) dw.

[Note that this really is a function ofz because has zas itsendpoint.]

Now we show that F(z) = f(z). Let z1 D be any point.Then by definitionFis differentiable atz1 with derivativef(z1)if

limh0

F(z1+h)

F(z1)

h =f(z1).So, as D is a domain, there exists an 1-neighbourhood atz1 for some > 0. If|h| < 1, then there exists a contour : [0, 1]D which gives the straight line from z1 toz1+h, i.e.(t) =z1+ht.

Using our definition ofF, we get

F(z1+h) =

+

f=

f+

f.

Therefore,

F(z1+h) F(z1)h

= 1

h

f+

f

f

= 1

h

f(w) dw.

We have,

f(z1) dw= f(z1)

dw= f(z1) [w]z0+hz0

=f(z1)h.

Hence,

F(z1+h) F(z1)h

f(z1) =

f(w) f(z1)h

dw.

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We want the LHS to tend to zero as h 0, so we use theEstimation Lemma on the RHS. Since f is continuous, givenany > 0, there exists a > 0 such that

|w

z1

|< implies

that|f(w) f(z1)| < . We can assume that < 1. So, when|z|< and w, we havef(w) f(z1)h

< |h| .The length of is |h|, so we deduce from the Estimation Lemmathat

f(w) f(z1)h

dw

|h| |h|.

Thus, F(z1+h) F(z1)h f(z1) for all|h|< .Since was arbitrary we deduce that the LHS is zero. Thisimplies thatF(z1) =f(z1). So,F is what we were seeking.

Summary

Fundamental Theorem of Calculus: Suppose f : D Cis a continuous complex function and there exists F suchthatF =f. Then,

f(z) dz= F((b)) F((a)).

Not all functions have an antiderivative. A domain D is called connected if for each pair, D,

there exists a contour: [a, b]D such that(a) = and(b) =.

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10 Differentiability of Power Series

We have successfully defined functions such as exp, sin and

cos by power series. Now we would like to show that they aredifferentiable. We do this by investigating the differentiabilityof power series with positive radius of convergence. We willshow later that if a function is differentiable at a point z0,then near that point it can be given as a power series, so thisinvestigation wil be useful for more than just the elementaryfunctions.

We prove:

A power series is differentiable in the obvious way: term-by-term.

The derivative of a power series has the same radius ofconvergence.

Power series are infinitely differentiable. The coefficients of the power series can be given in terms

of the derivatives.

The coefficients of the power series are unique.Suppose that

n=0anz

n is a series. Then, the obvious can-didate for the derivative is the one produced by term-by-term

differentiation6

, i.e. n=0nanzn1. But, initially, we do do noteven know that this sequence converges.

So, we begin by proving that a power series and its obviousderivative have same radius of convergence, and then showthat this obvious derivative really is the derivative of the series.

Lemma 10.1The series

n=0anz

n and

n=0nanzn1 have the same radius of

convergence.

Proof. The following proof is taken from Priestley p20.Ifn=0nanzn1 had radius of convergence R= 0, then it isnot difficult to show that n=0anzn has the same radius of

convergence.Therefore, we prove the converse: If

n=0anz

n had radius ofconvergenceR=, then so does n=0nanzn1.

Fixz with 0


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The series

n(|z|/)n is easily shown to converge, by the ratiotest. (Exercise!).

Just as in the real case, if the series bn converges, thenbn 0 as n and so there exists a constantM such that|bn| M for all n. Thus, there exists a constantM such thatfor all n

n

|z|

M,

thus,

|nanzn1| M|z| |ann|,

and so by the comparison test,

nanz

n1 is absolutely conver-gent. Hence, the result.

Now, lets show that this really is the derivative of the series.

Theorem 10.2Suppose thatf(z) =

n=0anz

n has radius of convergence R >0. Then,f(z) =

n=0nanz

n1, for all|z|< R.Proof. Again we follow Priestleys proof. We know thatg(z) =

nanzn1 is well-defined for|z| < R. We shall show thatf(z)

exists and is equal to g(z).[If we are thinking like mathematicians, then we know a

good way of doing this is to show that

|f(z)

g(z)

| = 0. In

other words we wantlimh0 f(z+h) f(z)h g(z) = 0.

Certainly, this is true iff(z+h) f(z)h g(z) 0 ash0.

This is our method of attack.]

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For anyzand h such that|z|< R and|z+h|< R, we have

f(z+h) f(z)h g(z) =

n=0

an(z+h)n

zn

h

n=0

nanzn1

=

n=1

an

(z+h)n zn

h nzn1

= |h|

n=1

an

nk=2

n

k

znkhk2

use binom thm |h|

n=1

1

2n(n 1)|an|

n2m=2

n 2

m

|z|n2m|h|m

|z|n=1

12n(n 1)|an|(|z| + |h|)n2.

Fixzand choose with|z|< < R. By Lemma 10.1 n=1n(n1)|an|n2 converges toK say. For|h|< |z|,f(z+h) f(z)h g(z)

12K|h|.So ash0 we get

f(z+h)f(z)

h g(z)

0 as required.

The proofs are quite technical, probably the most technicalin the course, and I wouldnt expect you to reproduce themin exams, but I would like you to understand them. Thesetwo results will certainly be very important to the course, sounderstand what they mean:

The derivative of a series can be found by term-by-termdifferentation and the resulting series has the same radius ofconvergence.

Examples 10.3(i) We can show that

0 z

n = 11z for|z|< 1. So, differentiat-ing both sides gives 0 nzn1 = 1(1z)2 for|z|


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We have seen that, iff is defined on the disc of convergence,then so is f, and f is a series. We can differentiate f term-by-term to getf and so on. This suggests the following:

Corollary 10.5Iff(z) =

n=0anz

n has radius of convergence R >0, then f isinfinitely differentiable.

Proof. By induction using the theorem.

Note that this is to be expected as a similar statement is truefor real power series. This result is one of the reasons thatpower series are so great. Later, we will see that any com-plex function differentiable at a point can be given by a powerseries.

Corollary 10.6Iff(z) =

n=0anz

n has radius of convergence R >0, then ak =

f(k)(0)

k! for allk.

Proof. By induction we can prove

f(k)(z) =n=0

n(n 1) . . . (n k+ 1)anznk

and this of course holds for|z|< R.Then,f(k)(z) = (k(k 1) . . . (k k+ 1)ak)

+ ((k+ 1)((k+ 1) 1) . . . ((k+ 1) k+ 1)ak+1z) +. . .If we putz= 0 into this, then we getf(k)(0) =k!ak.

Again, you should already know this is true for real series.

Lemma 10.7 (Uniqueness Lemma)Suppose for some R >0,

n=0anz

n =n=0bnz

n, for all|z|< R.Then,an= bn for all n.

Proof. Let f(z) =

n=0anzn =

n=0bnz

n. Then, by Corollary10.6,

an=f(n)(0)

n! and bn=

f(n)(0)

n! .

Hence,an= bn.

Note that the Uniqueness Lemma is not trivial. A priori wedont know that we can equate coefficients for infinite series.

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Remark 10.8Just like in Real Analysis it is possible to prove that the prod-uct of two power series S1 and S2 is again a power series with

radius of convergence at least the minimum of those forS1andS2. Its coefficients can be given in terms of those ofS1 and S2.

However, we shall delay proving this as later work will givea particularly simple proof of this fact. (See Theorem 15.6.)

Power series about points other than zero

So far all our power series have been centred at the origin.This is rather limiting. Just as in real analysis giving an ex-pansion about a different point is very useful. Lets do thatnow.

Suppose we are interested in the pointz0 C. If we leth bea complex variable, then in a neighbourhood ofz0 we can getz0+ h to give any point. For example, ifh is zero we getz0. If|h|< Rfor someR, then we have a disc around z0.

Suppose we have f(z0+ h) =

0 anhn, with the series con-

vergent for|h|< R. I.e. a power series in habout0, but whichdefines the function faroundthe pointz0.

Letz= z0+ h, wellz0 is a constant and h is a single variablesozis a variable. By substitution,f(z) =

0 an(zz0)n, which

converges for|h|=|z z0|< R.Thus we can deal with power series centred at a particular

pointz0 and we will have a disc of convergence centred atz0.

The main point is that we can prove results about powerseries centred at zero and just translate to another point. Sofor example, if we have the power series an(z z0)n, then wecan leth= zz0 to get the series

anh

n and we can apply ratiotest, etc., to that. Then we translate back to say somethingabout the series

an(z z0)n.

For example, we can differentiate term-by-term:

Example 10.9If f(z) =

an(zz0)n converges for|zz0| < R, then f(z) =

nan(z z0)n1 converges for|z z0|< R.

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Proof: Leth = z z0. Theng(h) =

anhn converges for all

|h|=|z z0|< R asf(z) =g(z z0). We have

f(z) = ddz

g(zz0) =g (zz0) ddz

(zz0) =g (zz0) = nan(zz0)n.Summary

Letf(z) =

n=0anzn have radius of convergence R >0.

fcan be differentiated term-by-term to get the derivative. f has the same radius of convergence as f. fis infinitely differentiable.

The coefficients an are unique and an= f(n)(0)n!

for alln.

We can translate these results to series of the followingform: f(z) =

n=0an(z z0)n, for|z z0|< R.

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11 Win a Million Dollars!

The Riemann Zeta function

We define the Riemann Zeta function as follows:

(s) =n=1

1

ns.

So, for example if s = 1, then we get (1) =

n=11n

which weknow does not converge, so (1) is not defined. If s = 2 weget(2) =

n=1

1n2

which does converge, and so on for all realnumberss.

If we let s be complex, then what happens? We first we

need to define n

s

when n is a real number and s is a complexnumber, the only number we have done this for ise. We knowthatn = elog n and so we could define ns by

ns =es log n.

(Since obviously we wantns = (elog n)s =es log n.)Riemann was not the first to study this type of function but

did leave us an interesting hypothesis.

The Riemann Hypothesis

We now come to the million dollar question. Where do theroots of the function lie? I.e. the s C such that(s) = 0.

What Riemann found is that all roots (apart from some triv-ial real ones) seem to have Re(s) = 1

2, and so he conjectured

that all non-trivial roots have the property of being on thisline. This is called the Riemann Hypothesis and it is regardedas being the most important unsolved problem in pure math-ematics.

Anyone who proves that the hypothesis is true can claim amillion dollars from the Clay Institute. You can find the detailson the Web at http://www.claymath.org/ under the heading

Millenium Prize Problems.There are six other problems from various areas of mathe-

matics for which a million dollar prize is offered. One of them(the Poincare Conjecture) is explained in Homotopy and Sur-faces in Year 3.

For futher reading see Prime Time by Erica Klarreich inNew Scientist, Vol 168 issue 2264, 11/11/2000, p32, (EdwardBoyle Library Floor 11), for the connection with number theoryand the distribution of primes.

See also the classic text by E.C. Titchmarsh, The theory ofthe Riemann-Zeta function.

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12 Winding numbers

Cauchys theorem is one of the most remarkable theorems in

mathematics. To state it we need the notions of winding num-ber and interior point. Both these notions are intuitively sim-ple. We start with the winding number.

Definition 12.1Let be a closed contour and w a point not on . Then thewinding numberofaboutw, writtenn(, w), is the net num-ber of times thatwinds aboutw, with anticlockwise countedpositively.

Example 12.2

The winding numbers for points in the regions enclosed by thecontour are shown below.

Exercise 12.3Find the winding numbers for points in the various regions.

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method. In the examples above note that when passing fromone region to another via an edge (rather than via a crossingof two line) the winding number for points in the regions only

changes by 1.Now, consider the complicated contour image drawn below

with a line passing through it.

Start at the left side of the line. Obviously the winding num-ber there is 0. As we go from left to right on the line we willcross the contour. We apply the following rules:

If we cross the contour so that it is travelling up, then thewinding numberdecreasesby 1.

If we cross the contour so that it is travelling down, thenthe winding numberincreasesby 1.

In the above diagram, the contour is travelling up when wefirst meet it, so the point in the region we pass into havewinding number1. At the next crossing the contour is goingdown, and so the winding number of points in the next regionis 0. We can carry this out for all regions on the line. To getother regions we can use different lines.

Exercise 12.6

Find the winding numbers for the points in the diagram.

Justification of the method

We can justify the method by considering two points w and zthat lie on opposite sides of a contour line that goes up. Wecan assume that the contour starts and finishes at a pointnearw and z. (If it didnt we can do some reparametrisationand join work.)

Now, take a loop C round w. Then, n(+ C, w) = n(, z), asthe diagram shows. Thus,

n(+C, w) = n(, z)

n(, w) +n(C, w) =

n(, w) + 1 =

n(, w) = n(, z) 1.This shows that if we pass fromztow, then the winding num-ber decreases by 1. Similarly, one can show an increase by 1for a contour locally heading down.

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Warning! 12.7We have been using the imageof the contour to calculate thewinding number of a point. Recall that the contour and its

image are different. Effectively, we have assumed that thecontour is traversed only once. If we go round the contourtwice, then the numbers calculated by eye have to be doubled.More generally, if we go round k times, then we multiply theby eye calculations byk.

Interior points

The method of counting the number of times a contour wrapsround a point helps define the notion of an interior point.

Definition 12.8Letbe a closed contour. The interiorofis the set of pointsw C\for which n(, w)= 0.

We denote the set of interior points ofby Int().

Example 12.9The interior points of the contour in the next diagram areshaded.

Example 12.10

Let(t) =w+Re2it, 0t1. Then, Int() ={z :|z w|< R}.

Summary

The winding number is the number of times a contourwraps round a point in an anticlockwise direction.

Letbe a closed contour, w C\. Then

n(, w) = 1

2i

dz

z w .

It is easy to calculate a winding number by eye. An interior point is any point with non-zero winding num-

ber.

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13 Cauchys (Fantastic) Theorem

We now come to the fundamental theorem in complex analy-

sis. There is no analogue in real analysis. It has far reaching,deep consequences, and most of what we will prove from nowon relies on this theorem.

Theorem 13.1 (Cauchys Theorem)Let D C be a domain, and f : D C be a differentiablecomplex function. Letbe a closed contour such that andits interior points lie in D.

Then,

f= 0.

Remarks 13.2(i) This is truly a great theorem. It refers to any domain in

C, any analytic function on D, and any contour with allinterior points in D. And it says that any integral aris-ing from this is zero. Thus, weak assumptions lead to astrong conclusion.

(ii) It is important to note that the interior of lies within D.If w is in the interior of but not in D, consider f(z) =1/(z w) which is analytic on D = C\{w}, but

f= dzz w = 2i n(, w)= 0,so Cauchys theorem does not apply.

(iii) The proof of theorem is too complicated for the momentand we will do it later. Note that we cant just use the FTCsince we dont know thatfhas an anti-derivative on D.

(iv) At first sight it may appear that the theorem will onlytell us about the behaviour of differentiable functions.However, it has strong implications for non-differentiablefunctions as well.

Exercise 13.3Using the contours in Exercises 2 Question 1, to which of thefollowing integrals does Cauchys theorem apply? (There is noneed to evaluate them.)

1

|z|2 dz,1

z2

z 2dz,2

z dz,

2+3

sin

1

z 1

dz.

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The calculation trick

Recall, the trick I performed where I calculated an integral

over a path, beforethe path had even been defined. How wasthis done?

The trick follows from Cauchys theorem and the funda-

mental example, (recall that the example says that

1

zdz =

2i where is a small circle round the origin).

I was trying to integrate

a

zdz, wherea was a number cho-

sen at random, andwas a path chosen at random. However,you were forced into choosing so that its winding numberwas 1.

Now, we can take a small circle C going clockwise roundthe origin, and we can make it so small that we can assumeit lies totally within the interior of. Now take another pathfrom the end ofto the start of the circle C. Call this contour.

Consider the contour = ++C. This is a closedcontour. Its interior does not include the disc encircled byC.So it doesnt contain the origin. But a/z is differentiable onD= C

\{0}

, and the interior of is a subset ofD. Thus, we canapply Cauchys theorem:

a

zdz = 0

++C

a

zdz = 0

a

zdz+

a

zdz

C

a

zdz+

a

zdz = 0

a

zdz+

a

zdz

Ca

zdz

a

zdz = 0

az

dz+ C

az

dz = 0

a

zdz =

C

a

zdz

= a

C

1

zdz

= a 2i

a

zdz = 2ai

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Thus we get the answer expected. This method of cuttingout a disc where the function is not defined will be usedagain later in more generality so make sure you understand

this simpler example.

Summary

LetD C be a domain, and f : D C be a differentiablecomplex function. Letbe a closed contour such thatand its interior points lie in D.

Then,

f= 0.

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14 Strange Consequences of Cauchys Theorem

We will prove a number of surprising theorems that can be

deduced from Cauchys theorem.

(i) Cauchys Integral Formula: The value of a function atz0is determined by the values on a contour round z0.

(ii) Liouvilles Theorem: Any differential function bounded onthe whole ofC is contant.

(iii) The Maximum Modulus Principle: The modulus of a func-tion on a domain achieves its maximum on the boundaryof the domain.

(iv) Fudamental Theorem of Algebra: Every complex polyno-mial has a complex root.

Cauchys Integral Formula

Theorem 14.1LetD C be a domain, and f : D C be differentiable. Letbe a closed contour such thatand its interior points lie inD.

IfwD\, then

f(z)z wdz= 2i n(, w)f(w).Remarks 14.2

(i) Ifw is in the interior of, then

f(w) = 1

2i n(, w)

f(z)

z wdz.

This says that the values off inside are completely de-termined by those on ! Remarkable! This contrasts withreal analysis.

(ii) This behaviour is sometimes called action at a distance.

(iii) The special case f(z) = 1 for all z D gives the windingnumber lemma.

Proof (of Theorem 14.1). Letr be the circular contourr =w+ reit,(0t2), wherer >0 is sufficiently small so thatris contained in the interior of.

[Step 1] Letbe a contour in D\{w} from the start point ofto the start point ofr.

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The contour

=++ (r) + + (r) n(,w) times

+(),

where there are n(, w) copies of (r), has winding numberzero about w, so by Cauchys theorem applied to

f(z)

z w onD\{w},

e

f(z)

z wdz = 0

+

r

r

+

f(z)

z wdz = 0

f(z)

z wdz n(, w)r f(z)z wdz = 0

f(z)

z wdz = n(, w)r

f(z)

z wdz.

[Step 2] We shall now show that

limr0

r

f(z)

z wdz= 2if(w).

We have,

rf(z)

z wdz = f(w) r

dz

z w+ r

f(z) f(w)z w

dz

= 2if(w) +

r

f(z) f(w)z w dz, by winding lemma.

Thereforer

f(z)

z wdz 2if(w) =

r

f(z) f(w)z w dz

sup

zr

f(z) f(w)z w 2r,

by the Estimation Lemma. Now f(z) f(w)

z w f(w) as z

w.

(Why does f(w) exist?) Hence RHS |f(w)|.0 = 0 as r 0.Therefore,

limr0

r

f(z)

z wdz= 2if(w).Thus we can combine these steps to get

f(z)

z wdz 2in(, w)f(w) =

n(, w)

r

f(z)

z wdz

2in(, w)f(w)

= n(, w)

r

f(z)

z

w

dz 2if(w)

.

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The RHS tends to 0 asr0, but the LHS is independent ofrand so must be 0.

Examples 14.3(i) Calculate

sin zz dz, where(t) = 3e

it, 0t4.We apply CIF with f(z) = sin zand w= . The pointw lieswithin the circle formed by, and n(, w) =2, becausewinds round w clockwise twice. (Draw a picture!)

Hence,

sin z

z dz= 2i (2) sin() =4i.

(ii) Let be a circle of radius 2 about 2, i.e. z such that|z 2|= 2.Draw the contour and indicate where the integrand is notdifferentiable.

Then,|z2|=2

ez

z2 9dz =|z2|=2

1

6

ez

z 31

6

ez

z+ 3dz

= 1

6

|z2|=2

ez

z 3dz1

6

|z2|=2

ez

z+ 3dz

= 1

62ie3 0, by 14.1,

= ie3

3 .

Liouvilles Theorem

The next theorem is also rather surprising.

Theorem 14.4Supposefis differentiable on the whole ofC, and is bounded,i.e. there exists M such that|f(z)| M for all z C. Then, fis constant.

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Proof. This proof contains a nice trick. We shall apply CIF tow Int() andf(z)k, wherek is a natural number:

f(w)k = 12in(, w)

f(z)k

z wdz.

Define the distance from w toby

dist(, w) = inf{|z w| : z}.Then, obviously|z w| dist(, w) for all z.

So,

|f(w)|k 12 |n(, w)|

Mk

dist(, w)L(), by the Estimation Lemma,

since |f(w)|k|z w|

Mk

dist(, w) on .

Therefore,

|f(w)|

L()

2 |n(, w)| dist(, w)1/k

M.

Now, we use the fact that limkx1/k = 1 for all x > 0, (this

is true because x1/k = exp( 1kln x) exp(0) = 1). Thus, lettingk gives|f(w)| M.

Fundamental Theorem of Algebra

We are now in a position to prove the Fundamental Theoremof Algebra, (which is in fact not really a theorem of algebra butof analysis!)

Theorem 14.9Every polynomialp(z) =zn +an1z

n1 +. . . a1z+a0,n1, has aroot in C.

Proof. Suppose not and derive a contradiction. Since p(z)

= 0

for all z C the function fdefined byf(z) = 1/p(z) is differen-tiable on all ofC. Now, forz= 0,p(z)zn

= 1 + an1z + + a0zn 1 as|z| .

So there exists an R such that|z| R implies thatp(z)zn

12 .This in turn means

|f(z)|=

1

p(z) 2

|zn

|

2Rn

for|z|> R.

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By the Maximum Modulus Principle applied to f with beingthe standard circle of radius R round the origin we have|z|00 x0.

Then,fis differentiable with derivative

f(x) = 2x x >00 x0.

This derivative is not differentiable at0, so f is not infinitelydifferentiable.

Some facts about power series

We now use Taylors Theorem to prove some facts about serieswhich might otherwise be difficult to demonstrate.

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Theorem 15.6Suppose f(z) =

n=0anz

n for all|z| < R1 and g(z) =

n=0bnz

n

for all

|z

| < R2. Then, f(z)g(z) = (f g)(z) =n=0cnzn for all|z|< min{R1, R2}, wherecn= nk=0akbnk.

Proof. Let R = min{R1, R2}. Then, f and g are complex dif-ferentiable for all|z| < R. The product of two differentiablefunctions is differentiable by the product rule. So, f g is dif-ferentiable. By Theorem 15.1 it has a series expansion for|z|< R, and by Corollary 10.6, we get

(f g)(z) =n=0

(fg)(n)(0)

n! zn.

Now we can apply Leibnitzs rule

(f g)(n)(0) =nk=0

n!

k!(n k)!f(k)(0)g(nk)(0)

=

nk=0

n!f(k)(0)

k!

g(nk)(0)

(n k)!

= n!nk=0

akbnk.

Therefore,

(f g)(z) =

n=0

cnzn, (|z|< R), wherecn=nk=0

akbnk.

Corollary 15.7Suppose f and g are power series with positive radii of con-vergence. Ifg(0)= 0, thenf /g has a power series expansion at0 with positive radius of convergence.

Proof. Define h(z) = 1/g(z). By continuity of g and the factthatg(0)= 0, there is a-neighbourhood of0 upon which h isdifferentiable. Thus,h has a power series expansion, and theresult follows from the theorem applied to f h.

Remark 15.8The precise value of the radius of convergence will depend onwhereg is zero.

* Moreras Theorem

There is a partial converse to Cauchys theorem. It relies onthe earlier optional material on the Fundamental Theorem ofCalculus.

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16 Zeros of functions

We have met a number of functions that are not analytic (i.e.

differentiable) at certain points. For example,1/z is not ana-lytic at0, because it is undefined when z is zero. Therefore, itseems natural to investigate zeros of functions first.

Zeros

Definition 16.1An analytic function f :D C has azero atz0 iff(z0) = 0.Examples 16.2

(i) The functionsin has zeros atk, for all k Z.(ii) By Theorem 1.35(iv) f(z) =ez has no zeros.

(iii) f(z) =z3(z2 + 1) has a zero atz=i, one atz=i, and oneatz= 0.

Exercise 16.3Find the zeros of:

(i) z2 + 9, (ii) ez2+9(z+ 4i), (iii) ez2 1.

We now know that iff :D C is differentiable atz0, it hasa Taylor series expansion aboutz0: f(z) =n=0an(z z0)n forall zwith|z z0|< Rfor someR >0.Definition 16.4We sayfhas azero of order m at z0 if

a0= a1= = am1= 0butam= 0.Remark 16.5Obviously, by Theorem 15.1, f has a zero of order m atz0 if

and only iff(j)(z0) = 0 for all j < mand f(m)(z0)= 0.

In particular, iff(z0) = 0 and f(z0)= 0, then fhas a zero of

order 1.Examples 16.6

(i) The functionf(z) =z2 has a zero of order2 at0.

(ii) The functionf(z) =z(z+ 2i)3 has a zero of order 1 at0 andone of order 3 at2i.

(iii) More generally, suppose thatfis a polynomial with a rootof multiplicitym atz0. Then,fhas a zero of orderm atz0.

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Exercise 16.7Find the zeroes and their orders of the following:

(i)(z1)3(z+1) (ii)(z2+1)2, (iii)zez2

, (iv)(2z3i)4, (v)z3+1.Exercise 16.8Suppose that f : D C is an analytic function with a zeroof orderm atz0. Then, there exists a differentiableg and anR > 0, such that f(z) = (zz0)mg(z), for all|zz0| < R, andg(z0)= 0.

Summary

An analytic functionf :D

Chas azeroatz0 iff(z0) = 0.

Suppose f(z) =n=0an(z z0)n for all z with|z z0| < Rfor someR. Thenfhas azero of order m at z0 if

a0 = a1= = am1 = 0 butam= 0.

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17 Poles

The notions of pole and residue are crucial to the application

of complex analysis to real problems, such as real integrals,or Fourier and Laplace Transforms7. We define poles in thissection and show how to locate them in the next. After thatwe define their residue.

A motivating example

Let us consider an example: Integrate

|z|=1

ez

z2dz, where|z|= 1

denotes the standard contour of a unit circle round the origin.We cant use Cauchys Integral Formula as the integrand is

not analytic at 0 and the Fundamental Theorem of Calculusdoesnt help as we cant see any obvious antiderivative.

Lets just integrate term-by-term8 and see what happens.|z|=1

ez

z2dz =

|z|=1

1

z2

1 +z+

z2

2!+

z3

3!+. . .

dz

=

|z|=1

1

z2+

1

z+

1

2+

z

6+ . . .

dz

=

|z|=11

z2dz+

|z|=11

zdz+

|z|=11

2dz+

|z|=1z

6dz+. . .

= 0 + 2i+ 0 + 0 +. . . .

Notice thatonly the 1/zterm mattered. By the Fundamen-tal Example, all the other terms were irrelevant. This is keyto understanding the terms pole and residue we are going todefine.

Laurent expansions

First we shall look at functions, like the integrand ez/z2 exam-ple above, that can be represented by a series with negative

powers.Recall for ananalytic functionwe can represent it as

f(z) =n=0

an(z z0)n for|z z0|< R.

Compare this with the following series.

7These transforms are used in Applied Mathematics. If you know what they are, then fine.If not, then dont worry about it.

8Obviously, we should check that we can do this!

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Definition 17.1Suppose thatf : D C is complex function. Then we sayfhas aLaurent expansion at z0 if there existan

C and R >0,

such that

f(z) =

n=

an(z z0)n

for all zwith 0


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Examples 17.6(i) order(1/z5) = 5at0.

(ii) order(ez/z2) = 2at0 asf(z) = 1z2

+1z

+12

+ z6

+ . . . .

(iii) order

i

(z 4i)3 + (z 4i)2 2i(z 4i)5

= 3 at4i.

(iv) f(z) = exp(1/z) has an essential singularity at0 because

f(z) =n=0

(1/z)n

n! =

n=0

zn

n! =

0n=

zn

(n)! ,

and this obviously has an infinite number of terms with

negative exponent.

Remarks 17.7(i) If such a Laurent series exists, then the coefficients are

unique. To see this look at (zz0)Nf(z), this is a powerseries and so has unique coefficients.

(ii) Suppose that f(z) has a Laurent expansion with N = 0,i.e.

f(z) =a0+a1(z z0) +a2(z z0)2 +. . . for0


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Summary

An expansion offof the form

f(z) =

n=

an(z z0)n

for allzwith0


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18 How to find poles

It is hard to give non-trivial examples of poles without using

some theory. You may have noticed that the order is somehowconnected with the order of multiplicity of the polynomial in adenominator. The following puts that intuition into a precisemathematical statement, and does it for general functions notjust polynomials.

Lemma 18.1Suppose thatf(z) =

p(z)

q(z), where

(i) p is analytic atz0, and p(z0)= 0,

(ii) qis analytic atz0, and has a zero of orderN atz0.Then,fhas a pole of orderN a

complex analysis - houston

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