2959 complex analysis

2959 COMPLEX ANALYSIS HTML VERSION

Complex Analysis notes and interactive quizzes by PAUL SCOTT

Department of Pure Mathematics, University of Adelaide

View of River Torrens towards University footbridge

CONTENTS

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COMPLEX ANALYSIS

INTRODUCTION

These web course notes and quizzes have been used since 1999. They have evolved from many years of lecturing this course. However, the basic imprint of the book Complex Analysis and Applications by Churchill et al is still evident. I gratefully acknowledge the inspiration of this valuable book, and refer readers to it. In fact, it is difficult to conceive of a better ordering of presentation for this topic than that provided by this book. I strongly recommend it as a text or reference.

Other books which may be useful for reference are: Complex Variables, Polya and Latta (Wiley); Fundamentals of Complex Analysis, Saff and Snider (Prentice-Hall); Complex Variables, Silverman (Houghton Mifflin).

I hope you find these notes and quizzes useful.

Please send any comments or corrections to the author.

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1. COMPLEX NUMBERS

A short history

The history of the complex numbers is very interesting. By the 16th Century, although no-one understood exactly what a complex number was, it was found that complex numbers were a useful tool for solving problems. Later, mathematicians tried to understand the complex numbers. This led in turn to investigations of the real numbers, the rational numbers, the integers and finally the natural numbers. So historically there was a reverse development: the more complicated system was found to be useful early on, and the study of the simplest systems were left till later.

Leibniz (1702) described complex numbers as ‘that wonderful creation of an ideal world, almost an amphibian between things that are and things that are not’.

Definition

Complex numbers are defined as ordered pairs z = (x, y) of real numbers x and y. We shall define addition and multiplication of these numbers shortly.

We identify the pairs (x, 0) with the real numbers x. This means that the real numbers can be thought of as a subset of the complex numbers.

Complex numbers of the form (0, y) are said to be pure imaginary numbers. The numbers x and y are now called the real and imaginary parts of z respectively, and we write x = Re z, y = Im z.

Operations (I)

Equality Two complex numbers are equal if they have the same real and imaginary parts. Thus

(x1, y1) = (x2, y2) if and only if x1 = x2, y1 = y2.

Sum and Product The sum z1 + z2 and product z1.z2 of z1 = x1 + y1 and z2 = x2 + y2 are defined by:

(x1, y1) + (x2, y2) = (x1 + x2, y1 + y2),

(x1, y1).(x2, y2) = (x1x2 – y1y2, x1y2 + x2y1).

This last definition looks pretty weird, but we shall shortly see the reason for it.

Operations (II)

With these definitions of addition and multiplication, we have

(x, y) = (x, 0) + (0, 1)(y, 0).

The set of complex numbers of the form (x, 0) act just like the real numbers R. Further, setting i = (0, 1) gives z = (x, y) = x + iy, and we have

i 2 = (0, 1).(0, 1) = (–1, 0), that is, i

2 = –1.


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Addition and multiplication can now be rewritten as the more usual:

(x1 + iy1) + (x2 + iy2) = (x1 + x2) + i(y1 + y2),

(x1 + iy1).(x2 + iy2) = (x1x2 – y1y2) + i(y1x2 + x1y2).

Comments

1. Why did we not just start here? Notice that our derivation of i is firmly based on the real

numbers. There is no ‘magic’ about (–1), contrary to the use of the historical word ‘imaginary’.

2. We can now perform complex operations by treating the terms as real, and substituting i 2 =

–1.

Example:

(2 + 3i) + (4 + 5i) = (2 + 4) + (3 + 5)i = 6 + 8i,

and

(2 + 3i) . (4 + 5i) = (2.4 + 3i.5i) + (2.5i + 3.4i) = (8 – 15) + (10 + 12)i = –7 + 22i.

QUIZ 1.1A

QUIZ 1.1B

Algebraic Properties

The complex numbers behave in much the same way as the real numbers. In particular they form a field. Some of the laws satisfied are:

The Commutative Laws:

z1 + z2 = z2 + z1; z1z2 = z2z1;

The Associative Laws:

(z1 + z2) + z3 = z1 + (z2 + z3); (z1z2)z3 = z1(z2z3);

The Distributive Law:

z1(z2 + z3) = z1z2 + z1z3.

These are easily proved by writing the z s in the form x + iy and using the corresponding properties of the real numbers.

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Other Field Axioms

We now use the idea of inverse to define division of complex numbers in the following way: z1/z2 = z1 . z2

–1 (z2 0).

Try writing this out in terms of the real and imaginary parts! Later we shall give an easier way of dividing complex numbers. Clearly division by zero is not allowed, as z2

–1 is undefined when

z2 = 0.

The expressions below follow from this definition:

, , .

Example:

As mentioned earlier, complex numbers (like the reals) form a field, C. However, it is not possible to order C. Thus expressions like z > 0, z1 < z2 are meaningless unless the complex numbers are real.

QUIZ 1.2

Cartesian Coordinates

We set up the natural correspondence z = x + iy (x, y) between the complex number z and the point (x, y) in the cartesian plane. Each complex number corresponds to exactly one point in the plane, and conversely. For example, 1 + 2i is represented by the point (1,2). Notice how this corresponds to our original ‘ordered pair’ definition of a complex number. We can think of the complex number z either as the point (x, y) or as the vector from the origin to this point.

The plane of such representative points is called the Argand diagram, or the complex plane or the z-plane.

The x-axis is called the real axis and the y-axis is called the imaginary axis.

Sum and Difference; Modulus

We now have an immediate geometric interpretation of the sum and difference of two complex numbers:

Notice the direction of the arrow representing z1 – z2.

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The modulus or absolute value of a complex number z = x + iy is defined to be | z |= (x2 + y

2).

When y = 0, z is real, and we have the usual absolute value of a real number.

Length and Distance

The (real) number |z| denotes the length of the vector representing z, or the distance of the point (x, y) from the origin O. Even though z1 < z2 is meaningless, we can note that |z1| < |z2| is a valid statement, meaning that z1 is closer to the origin than z2.

We can extend this idea to see that | z1 – z2 | = {(x1 – x2)2 + (y1 – y2)}

2

is the distance between the points representing z1 and z2.

Example: What is the locus of points z satisfying | z – i | = 3 ?

We are looking at points z which are distant 3 from the fixed point i. That is, a circle of radius 3 and centre i = (0,1).

If we expressed z = x + iy, then | z – i |=3 becomes the equation of this circle:

{x2 + (y – 1)

2} = 3,

or x2 + (y – 1)

2 = 9.

Modulus and Conjugate

From the definition of | z |, we have | z |

2 = (Re z)

2 + (Im z)

2,

from which it follows that | z | |Re z| Re z, and | z | | Im z| Im z.

The complex conjugate or conjugate of z = x + iy is the number z

= x – iy. The number is represented in the complex plane by the point (x, –y) – the reflection in the x-axis of the point representing z.

We observe that for all z, = z and | | = | z |.}

More Properties of the Conjugate

The following properties are easy to establish from the definitions:

= , = . , = / .

We also have Re z = (z + ), Im z = (z – ).

Finally, z. = |z|2 = x

2 + y

2,







so for z2 0,

This gives us a nice way to find the quotient of two complex numbers.

Example

Triangle Inequality

The relation z = | z |2 quickly gives

| z1.z2 | = | z1 |.| z2 |, | z1/z2 | = | z1 | / | z2 |.

It is also not too hard to derive the Triangle Inequality:

| z1 + z2 | | z1 | + | z2 |.

Proof: |z1 + z2|2

= (z1 + z2)( )

= (z1 + z2)( + )

= z1 + z1 + z2 + z2 | z1 |

2 + 2| z1 |.| z2 | + | z2 |

2

= (| z1 | + | z2 |)2

since z1 + z2 = 2 Re(z1 )

| z1 | = | z1z2 | = | z1 |.| z2 |.

It is easy to extend this to larger finite sums:

|z1 + z2 + z3| |z1 + z2| + |z3| |z1| + |z2| + |z3|.

More on the Triangle Inequality

The triangle inequality becomes clear when seen geometrically, as in the adjacent figure. It just says that the length of any side of a triangle is not more than the sum of the lengths of the remaining two sides.

Another useful result is the ModDiff Inequality:

| |z1| – |z2| | |z1 + z2|.





We can think of this as providing a lower bound for |z1 + z2|.

Proof: |z1| = |(z1 + z2) + (–z2)| |z1 + z2| + |–z2|

so |z1| – |z2| |z1 + z2|.

Similarly, |z2| – |z1| |z1 + z2|, giving the required result.

QUIZ 1.3A

QUIZ 1.3B

Polar Coordinates

We say that z = (x, y) has polar coordinates (r, } when x = r cos , y = r sin . We write

z = r cos + i r sin = r cis .

Example

1 + 3 i = 2(cos /3 + i sin /3).

Note that r = |z|. We write = arg z – the argument of z. The argument is always expressed in

radians. We can find from the relationship tan = y/x.

The angle \theta is not unique: thus 2k is equally valid.

We use Arg z to denote the value of arg z satisfying – < arg z . This is the principal value of the argument.

If z = 0, arg z is undefined. We adopt the convention that if z is expressed in polar form, then z 0.

Using Polar Coordinates

We begin with a natural extension: z – z0 = cis . Here, r = |z – z0|,

and = arg (z – z0).

The adjacent figure illustrates the situation.

The polar form is important because it gives a very simple method of multiplying and dividing complex numbers. This is because

arg (z1.z2) = arg z1 + arg z2 (*)

We shall prove this shortly, but notice that this relation may fail for









the principal value ‘Arg’.

Example: Set z1 = –1, z2 = i.

Then Arg (z1z2) = Arg (– i) = – /2, while

Arg z1 + Arg z2 = + + /2=3 /2.

Proof of the Argument Identity

We now prove the identity (*):

arg(z1z2) = arg z1 + arg z2.

Set z1 = r1 cis 1, z2 = r2 cis 2. Now

z1z2 = r1r2 (cos 1 + i sin 1)(cos 2 + i sin 2)

= r1r2 [cos 1 cos 2 – sin 1 sin 2 + i(sin 1 cos 2 + cos 1 sin 2)]

= r1r2 [cos ( 1 + 2) + i sin ( 1 + 2)]

= r1r2 cis ( 1 + 2)

So any argument of z1 plus any argument of z2 is an argument of z1z2.

On the other hand, if arg(z1z2) = 1 + 2 + 2k , then for example, we could take arg z1 = 1,

and arg z2 = 2 + 2k .

Euler’s Formula

Euler’s Formula is:

We will justify this later. But note that

corresponding to cis . cis = cis ( + ).

In particular,

The j - Operator







We can use complex numbers to represent geometric transformations in the plane, and we shall develop this idea later.

But as an example, consider the mapping z f(z) = iz.

Here we have |i| = 1, arg i = /2. So

z = r cis iz = r cis( + /2).

That is, multiplication of z by i effects a rotation through /2.

This is the Engineers’ j - operator.

Powers and Roots

By induction, it is easy to obtain

where there are n factors on the left.

Equivalently, we have the well-known de Moivre's Theorem:

(cis )n = cis n .

It follows that

(Feel free to work with either form.)

Example

Example: Solve zn = 1.

Since rncis n = 1, we have

r = 1, cos n = 1, sin n = 0 n = 2k .

The distinct solutions are given by z = cis (2k / n), k = 0, 1, ... , n – 1.

These are called n th roots of unity. If wn = cis (2 / n), the n roots are 1, wn, wn

2, ... , wn

n – 1. Note that wn

n = 1.

The cube roots of 1 occur as the vertices of an equilateral triangle on the unit circle.

General n th Roots

We can easily extend the above method to finding n th roots of any w = cis .







For if z = r cis and zn = w, then we must have r

n = , and n = + 2k for integral k.

Thus r = (the positive nth root),and

= /n + 2k /n, k = 0, 1, ... , n–1.

Example Find all values of .

We set z = r cis , and z4 = 1 = 1.cis 0.

Then r4 = 1 implies that r = 1,

and 4 = 0 + 2k implies that = 2k /4, k = 0, 1, ... , 3.

Hence the four values of are

1.cis 0 = 1, 1.cis 2 /4 = i, 1.cis 4 /4 = –1, and 1.cis 6 /4 = –i.

QUIZ 1.4

Regions in the Complex Plane

We have already mentioned the complex plane. We look at some terms which describe certain sets in the plane.

A neighbourhood or more specifically, an - neighbourhood of z0, is the set of points satisfying |z – z0| < . That is, it is the set of points lying inside, but not on, the circle of radius centred at z0.

A point z is said to be interior to set S if there is some neighbourhood of z which just contains points of S. A point z is said to be exterior to set S if there is some neighbourhood of z which contains no points of S. A point z is said to be a boundary point of set S if every neighbourhood of z contains points in S and points not in S. All the boundary points together make up the boundary of S. Notice that a boundary point of S need not be an actual point of S.

Open, Closed and Connected

A set is open if it contains none of its boundary points. In fact, this is the same as saying that every point of S is an interior point.

A set is closed if it contains all of its boundary points. Of course a set need not be open or closed. An example would be the ring 1 < |z| 2, which includes its outer boundary, but not the inner bounding circle.

Given any set S, we can take all the points of S together with all the boundary points of S to

obtain a new closed set called the closure of S, and denoted . (Not to be confused with conjugates!)








An open set is connected if each pair of points of S can be joined by a polygonal path – a finite set of line segments joined end to end, which lies entirely within S.

Note You may wonder why we insisted on a polygonal path in the above definition. Why not any continuous path? The answer is that a ‘polygonal path’ is easily defined and understood. A ‘continuous path’ is a much more difficult concept.

Domains and Regions

An open, connected set is called a domain. A connected set is called a region.

Examples The disk |z| < 1, and the annulus 1 < |z| < 2 are domains. The disk |z| 1 and the annulus 1 |z| < 2 are regions.

A set S is bounded if every point of S lies inside some circle |z| = R; otherwise it is unbounded.

Examples The annulus 1 < |z| < 2 is bounded (it lies inside the circle |z| = 3). The straight line {z = x + iy | y = 0} is unbounded.

Points of Accumulation

We say that point z0 is an accumulation point of a set S, if each neighbourhood of z0 contains at least one point of S distinct from z0.

This is a more difficult concept, and is obviously associated with a limiting process. Thus we would expect to be able to find a sequence of points of S lying closer and closer to z0.

Lemma S is closed S contains all its points of accumulation.

Proof. ( ) Let S be closed. Then S contains all its interior and boundary points. A point of accumulation of S can not be an exterior point (see the definition of exterior point). Hence S contains all its points of accumulation.

( ) Suppose now that S contains all its points of accumulation. Does S contain all its boundary points? Let z0 be a boundary point of S not in S. By definition of boundary point,

each neighbourhood of z0 contains a point of S, so z0 is a point of accumulation of . So z0 lies in S and S contains all its boundary points. Hence S is closed.

2. ANALYTIC FUNCTIONS

Functions of a Complex Variable

Let D be a subset of C. A function f : D C is a rule that associates with each z in D a unique complex number w. We write w = f(z).

Notes

1. The set D of numbers that are mapped is called the domain of f. Notice that we now have a double use of this word. Where the domain is unspecified, we assume it to be the largest subset of C for which f(z) is defined.






2. The set of image elements {w | w = f(z)} is called the range or image of the function.

3. The above definition specifies a unique image for each z D. Later we shall extend this definition to include multivalued functions.

Examples of Functions

Example 1. f(z) = z2 + iz – 3. The domain is C.

f(z) = 1/(1 + z2). Here the domain is C \ { i}.

If f(z) only assumes real values, f is real-valued.

Example 2. f(z) = |z|, f(z) = Re z = x, f(z) = Im z = y.

We can break w = f(z) into real and imaginary parts. Thus if w = u + iv, z = x + iy, then

w = f(z) = u(x, y) + iv(x, y)

Example 3. f(z) = z2.

Here u + iv = (x + iy)2 = (x

2 – y

2) + 2ixy, so u = (x

2 – y

2), v = 2xy.

This expression in terms of real and imaginary parts may be easier said than done! In theory, it allows us to deduce properties of complex functions from our knowledge of the real numbers.

Mappings

A real function y = f(x) can be represented geometrically by a graph: the set of points {(x, y) | y = f(x)}. To represent the complex function w = f(z) geometrically, in general we need four dimensions or two planes: a plane for the domain, and a plane for the range. For simple functions, we can use the same plane twice.

Example 1. w = f(z) = z + 2.

This is a translation: each point z is translated through 2 to the point z + 2.

Example 2. w = f(z) = .

This is a reflection in the x-axis:

each point z = x + iy (or (x, y)) is mapped to the point = x – iy (or (x, –y)).





QUIZ 2.1

1. The domain of f = f(z) is the set of image elements.

(a) True ; (b) False .

2. If z = x + iy, then the function f(z) = y is real valued.


3. Geometrically, the mapping f(z) = - z is a reflection in the x-axis.


4. Geometrically, the mapping f(z) = 2z is an enlargement about the origin.


Mapping Curves and Regions

We shall be mapping curves and regions rather than just points.

Example

Find the image of the circle x2 + y

2 = c

2 (c > 0) under w = f(z) = (x

2 + y

2) – iy.

Let us set w = u + iv. Then each point (x, y) on the circle x2 + y

2 = c

2 maps to (u, v) = (c, –y),

where |y| c.

Thus the image of this circle is the line segment u = c, –c v c in the uv-plane.

The domain of f is the z-plane; the range of f is a quadrant of the w-plane.

Notice that z = (x, y) and – =(–x, y) map to the same point w.

Limits

All work on functions of two variables now carries over directly. A minor difference is that because we are dealing with complex numbers, the length or norm of a vector represented by w becomes the modulus |w| of w.

Thus the definition of limit becomes:















means

for all > 0, there exists > 0 : for all z, 0 < |z – z0| < |f(z) – w0| < .

Thus every z in the left disc has an image in the right disk. We may not obtain the whole of the right disc; for example, consider the image of f(z) = constant = w0 : this is just the central point of the second disk

A Limit Theorem (I)

Theorem 2.1 If w = f(z) = u + iv, z = x + iy, z0 = x0 + iy0, then

In brief, this theorem says: lim(u + iv) = lim u + i lim v.

Proof ( ) Let us suppose that

By definition, given , there exists > 0 :

0 < |x – x0 + i(y – y0)| < |u(x, y) – u0 + i(v(x, y) – v0)| < .

We deduce that

0 < |x – x0| < /2, 0 < |y – y0| < /2 |u(x, y ) – u0| < , |v(x, y) – v0| < .

This completes this part of the proof.

A Limit Theorem (II)

( ) Now let us suppose that

Then there exist 1 > 0, 2 > 0 such that

0 < |x – x0| < 1, 0 < |y – y0| < 1 |u(x, y) – u0| < /2,

0 < |x – x0| < 2, 0 < |y – y0| < 2 |v(x, y) – v0| < /2,

Choose = min( 1, 2). Then using the given limits,

0 < |(x – x0) + i(y – y0)| <

|(u(x, y) + iv(x, y)) – (u0 + iv0)| |u(x, y) – u0| + |i| |v(x, y) – v0| < /2 + /2 < .





More Limit Theorems

Our previous theorem quickly leads to the well-known and useful Limit Theorems. We use an easy to remember abbreviated notation.

Theorem 2.2. (Limit Theorems) If lim f, lim g exist, then

lim (f g) = lim f lim g lim (f . g) = lim f . lim g lim (f / g) = lim f / lim g (lim g 0)

Proof (a) Set f = u + iv, lim f = u0 + iv0, g = U + iV, lim g = U0 + iV0. Now

lim (f + g) = lim (u + U + i(v + V)) (substitute and rearrange) = lim (u + U) + i lim(v + V) (Thm 2.1)} = u0 + U0 + i(v0 + V0) (put in the limits) = (u0 + iv0) + (U0 + iV0) (rearrange) = lim f + lim g.

The other proofs are similar.

QUIZ 2.2A

1. Geometrically, the image of f(z) = 2 + i is

(a) a point ; (b) a line ; (c) a disk .

2. Geometrically, mapping f(z) = maps the square with vertices ( 1, i) onto itself.


3. If u + iv u0 + iv0, then we must have u u0.


4. If f(z) = 2 + i and g(z) = 2 - i then (f.g)(z) = 5.


QUIZ 2.2B

Theorem 2.1

Proof ( ) Let us suppose that { 1. } By definition, given , there exists > 0 :














0 < { 2. } |u(x, y) - u0 + i(v(x, y) - v0)| < .

We deduce that { 3. }, 0 < |y - y0| < /2 |u(x, y ) - u0| < , { 4. }.

This completes this part of the proof.

Match the above boxes 1, 2, 3, 4 with the selections:

(a) |v(x, y) - v0| < (b) |x - x0 + i(y - y0)| < (c) (d) 0 < |x - x0|

< /2

Continuity

Definition Function f is said to be continuous at z0 if f satisfies the following three conditions.

(a) f(z0) exists ; (b) lim z zo f(z) exists ; (c) lim z zo f(z) = f(z0).

Notes

1. Just writing statement (c) assumes the truth of (a) and (b).

2. Expressing (c) in terms of the limit definition, we obtain:

|z – z0| < |f(z) – f(z0)| < .

This is slightly different from the usual definition of limit, in that we allow the possibility z = z0 (omitting 0 < |z – z0| ... ).

Function f = f(z) is said to be continuous in a region if it is continuous at each point of the region.

Properties of Continuous Functions

The sum f + g, difference f – g, product f.g and quotient f/g of two continuous functions is continuous at a point z = z0, with the proviso that in the last case g(z0) 0. These results follow directly from the Limit Theorems 2.2.

Examples

1. Polynomial functions The polynomial function P(z) = ai z i is continuous forall z since it

is constructed as sums and products of the continuous constant functions (ai) and the continuous function f(z) = z.

2. Rational functions The rational function P(z) / Q(z) given by the quotient of two continuous polynomial functions P(z), Q(z) is continuous for all z : Q(z) 0.





Composition of Continuous Functions

We can also compose complex functions f, g to obtain the new function f o g defined by (f o g)(z) = f(g(z)). If f, g are continuous, will f o g be continuous also?

Theorem 2.3 The composite function f o g of two continuous functions f, g is continuous.

Alternatively, a continuous function of a continuous function is a continuous function. Formally, the proof of this theorem is exactly as for the real case, and is omitted here.

Example f(z) = sin (z2) is continuous for all z.

By Theorem 2.1, f(z) = u + iv is continuous u(x, y), v(x, y) are continuous. Thus:

Example f(z) = exy

+ i sin(x2 – 2yx

3) is continuous for all z (since the real and imaginary

parts are continuous).

Definition Say f is bounded in region R if |f(z)| M for all z R.

If f is continuous in R, then f is bounded because of the corresponding properties of u, v. Show this!

QUIZ 2.3

1. If f(z0) exists, then function f must be continuous at z = z0. (a) True ; (b) False .

2. If lim z zo f(z) exists, then function f must be continuous at z = z0. (a) True ; (b) False .

3. The function f(z) = sin(1/z) is continuous everywhere.


4. The function f(z) = cos(z3) is continuous everywhere.


The Derivative

Formally, the definition of the derivative f '(z) for functions of a complex variable is the same as for real functions.

Let f be a function whose domain contains a neighbourhood of point z0. Then















if the limit exists. In this case the function f is said to be differentiable at z0.

It is sometimes preferable to use the alternative form of the derivative obtained by setting z = z0 + z :

Note Since z0 lies in an (open) neighbourhood of the domain of f, f(z0 + z) is defined if z is small.

Examples of the Derivative

We can evaluate simple derivatives by using the basic definition.

Example f(z) = z2 .

The usual differentiation formulae hold as for real variables.

For example,

However, care is required for more unusual functions.

Example f(x) = |x|2 = x

2 f'(x) = 2x for all x.

But f(z) = |z|2 f '(z) exists only at z = 0.

This last statement is proved using the basic definition. Show it!

More on the Derivative

As in the real case, f is differentiable f is continuous.

The same rules apply for the sum, product, quotient and composite of two differentiable functions (where defined).

Example (2z2 + i)

5 = 5(2z

2 + i)

4.4z = 20z(2z

2 + i)

4.

Again, with more unusual functions, we may have to use the limit definition of differentiation.

Example Investigate (Re z).





We get

Here we get 0 approaching along x = x0, 1 along y = y0. Hence the limit does not exist.

QUIZ 2.4

1. If function f is continuous at z = z0, then f must be differentiable there. (a) True (b) False .

2. If f(z) = | z |2, then for all z, f '(z) = 2z.

(a) True (b) False .

3. If f(z) = (iz + 2)2, then f '(z) = 4i - 2z.


4. If f(z) = cos(z3), then f '(z) = - sin(z

3).


Cauchy-Riemann Equations (I)

Theorem 2.1 gives us conditions for continuity for a function of a complex variable in terms of the continuity of the real and imaginary parts. We now ask: Is there any test for differentiability?

Theorem 2.4 The derivative f '(z) of f = u + iv exists at z the first order partial derivatives ux, vx, uy, vy all exist and satisfy

ux = vy, uy = –vx (the Cauchy-Riemann equations).

Further, f '(z) = ux + ivx = vy – iuy.

Proof Since the derivative of f exists,

Cauchy-Riemann Equations (II)

By Theorem 2.1, lim Re(*) = a, lim Im (*) = b.

Set y y0 to get















Set x x0 to get

Hence all the first partial derivatives exist, ux = vy, uy = –vx, and f '(z0) = ux(x0, y0) + ivx(x0, y0) etc. as required.

Cauchy–Riemann Examples

1. Set f(z) = z2 = (x + iy)

2 = (x

2 – y

2) + 2ixy.

Now f'(z) = 2z exists for all z. So the Cauchy-Riemann equations are satisfied. We have u = x

2 – y

2, v = 2xy, and ux = 2x = vy, uy = –2y = –vx.

Also f '(z) = ux + ivx = 2x + 2iy = 2z as expected.

2. Set f(z) = |z|2. We show f '(z) does not exist for z 0.

Now f(z) = x2 + y

2, i.e. u = x

2 + y

2, v = 0, ux = 2x, uy = 2y, vx = 0 = vy.

So ux = vy x = 0, uy = –vx y = 0.

Hence f '(z) can only exist at (0, 0).

Does f '(0) exist? Yes; as suggested earlier; we must use a first principles argument.

Sufficient Conditions

Theorem 2.4 gives necessary conditions for f to be differentiable. We now seek sufficient conditions for f ' to exist: that is, a similar statement, but using .

Theorem 2.5 Let f = u + iv as before. Suppose that (i) u, v, ux, vx, uy, vy exist in the neighbourhood of (x0, y0), (ii) ux, vx, uy, vy are continuous at (x0, y0), (iii) the Cauchy-Riemann equations are satisfied at (x0, y0). Then f '(z) exists and at z0, f '(z0) = ux + ivx as before.

That is, f is differentiable at z0 the given conditions.

Proof We omit this proof. It is not hard, but rather messy.

An Example

The real functions u = ex cos y, v = e

x sin y are defined and continuous everywhere. So are ux,

vx, uy, vy and you can easily check that the Cauchy-Riemann equations are satisfied.

Hence the function f(z) = ex cos y + i e

x sin y is differentiable everywhere.

Since ux = u, vx = v,

f '(z) = f(z) ( = ex cis y = e

x + iy = e

z).







QUIZ 2.5

1. If f(z) = u + iv and the Cauchy-Riemann equations hold for u, v, then f '(z) must exist. (a) True ; (b) False .

2. For f = u + iv, the Cauchy-Riemann equations are ux = vy and vx = uy. (a) True ; (b) False .

3. If f(z) = (x2 - y

2 + 2) + 2ixy = u + iv, then the Cauchy-Riemann equations hold.


4. If f(z) is differentiable, then f '(z) = vy - i uy. (a) True ; (b) False .

Analytic Functions

Definitions f(z) is analytic at z0 if f '(z) exists not only at z0 but for all z in some neighbourhood of z0. f(z) is analytic in a domain of the z-plane if it is analytic at every point of the domain. f(z) is entire if it is analytic everywhere.

Examples

1. f(z) = |z|2 is not analytic anywhere. (It is in fact differentiable only at z = 0).

2. f(z) = 1/z is analytic (except at z = 0).

3. f(z) = a0 + a1z + ... + an zn is entire.

If f(z) is analytic throughout a domain except for a finite number of points, such points are singularities or singular points of f.

Examples

f(z) = 1/z (z = 0 is a singularity); (z = 1, 2 are singularities).

Test for Analytic Functions

Question How can we tell if a function is analytic? We can use Theorem 2.5, or

Theorem 2.6: If f = f(z) is analytic, then in any formula for f, x and y can only occur in the combination x + iy.

Proof We note that x = (1/2)(z + ), y = (

1/2i)(z – ). Hence if w = f(z) = u + iv, we can regard u, v

as functions of z, . Now, w is a function of z alone and













equating real, imaginary parts to zero.

Hence f analytic the Cauchy-Riemann equations hold as required.

Analytic Functions: Final Comments

Example

f(z) = sin(x + 3iy)

We can say immediately that this function is not analytic, as x and y do not occur in the combination x + iy. In some examples it is less clear whether or not the variables can be combined in this way.

Derivative Theorems

The theorems on derivatives quickly transfer to analytic functions. Thus the sum, product, quotient and composite of two analytic functions are formed in the obvious ways as before, and each of the resulting functions is analytic on its domain.

Harmonic Functions

Let f = u + iv be analytic in some domain of z-plane. Then the Cauchy-Riemann equations hold:

ux = vy, uy = –vx

It can be shown that for analytic function, the partial derivatives of all orders exist and are continuous functions of x, y. Hence

uxx = vyx and uyy = –vxy.

Assuming continuity of vyx, vxy, we have vyx = vxy, and hence

uxx + uyy = 0.

This is Laplace's equation, and u is called an harmonic function. In the same way we get

vxx + vyy = 0; i.e. v is an harmonic function.

If f = u + iv, u and v are conjugate harmonic functions. (Note the different use of the word ‘conjugate’ here).

Finding Conjugate Harmonic Functions

In applied mathematics (partial differential equations) we often seek an harmonic function on a given domain which satisfies given boundary conditions. If we are given one of two conjugate







harmonic functions, it is a simple matter to find the other. We use the Cauchy-Riemann equations.

Example

Let u = y3 – 3x

2y. Then u is harmonic, since uxx = – 6y = – uyy.

Now using the Cauchy-Riemann equations, ux = –6xy = vy.

Integrating v partially with respect to y gives v = –3xy2 + (x) and now

vx = –uy = –3y2 +3x

2 '(x) = 3x

2.

Hence v = –3xy2 + x

3 + c.

You can check that v is in fact harmonic! So

f(z) = (y3 – 3x

2y) + i(x

3 – 3xy

2 + c) [= i(z

3 + c) in fact].

Theorem 2.6 If f = f(z) is analytic, then in any formula for f, x and y can only occur in the combination x + iy.

Proof We note that x = (1/2)(z + ), { 1 }. Hence if w = f(z) = u + iv, we can regard u, v as

functions of z, . Now, w is a function of z alone { 2 } . So

+ i ( { 3 } ) = 0 { 4 } ux = vy, uy = -vx.



Match the above boxes 1, 2, 3, 4 with the selections

(a) y = 1/2i (z - ), (b) (c) (d)

QUIZ 2.6A

1. If f(z) is analytic, then f'(z) exists. (a) True ; (b) False .

2. Function f(z) may be differentiable at z = z0, but not analytic near z = z0. (a) True ; (b) False .

3. Function v(x, y) = -3xy2 + x

3 is an harmonic function.


4. The harmonic conjugate of u(x, y) = -2xy is (a) x

2 + y

2 + c ; (b) - x

2 - y

2 + c ; (c) x

2 - y

2 + c .










QUIZ 2.6B

Theorem 2.6 If f = f(z) is analytic, then in any formula for f, x and y can only occur in the combination x + iy.

Proof We note that x = (1/2)(z + ), { 1 }. Hence if w = f(z) = u + iv, we can regard u, v as

functions of z, . Now, w is a function of z alone { 2 } . So

+ i ( { 3 } ) = 0 { 4 } ux = vy, uy = -vx.



Match the above boxes 1, 2, 3, 4 with the selections

(a) y = 1/2i (z - ), (b) (c) (d)

3. ELEMENTARY

FUNCTIONS

Exponential and Trigonometric Functions

If z = x + iy, we define the exponential function exp z = ex cis y (= e

z).

Notes

1. We can give an alternative definition in terms of power series. Writing out a formal series for e

iy gives cis y.

2. If y = 0, then exp z = exp x = ex. Thus the complex exponential function naturally extends the

real function.

3. In this definition, y is in radian measure.

Properties of the Exponential (I)

1. The function exp is entire and (exp z) = exp z. (See Thm 2.5 and next example.)

2. If w = w(z) is analytic in some domain D, then so is exp w.

3. The function exp z = ex cis y is a complex number in polar form:

|exp z| = ex, arg(exp z) = y.

4. The range of w = exp z is the whole w-plane except O.






For, w = ex cis y with e

x > 0; to get w = cis , set x = ln ( > 0) and y = .

5. Laws of exponents exp z1 . exp z2 = exp(z1 + z2) exp z1 / exp z2 = exp(z1 – z2)

6. Powers (exp z)

m = exp(mz) m Z

+

(exp z)1/n

= exp 1/n (z + 2k i) m, n Z

+

(exp z)m/n

= exp m

/n (z + 2k i) k Z.

The proofs follow directly from the definition of the exponential. Note that (e

x cis y)

1/n = e

x/n cis ((y + 2

k)/n).

Properties of the Exponential (II)

8. We observe that exp(z + 2 i) = exp z.exp(2 i),

and that exp(2 i) = e0.(cos 2 + i sin 2 ) = 1.

It follows that exp(z + 2 i) = exp z.

Thus we can divide the z-plane into periodic strips. Each strip in the z-plane is mapped to the whole w-plane excluding the origin.

We note the further two properties of the exponential:

8. exp = .

9. cis = cos + i sin = exp(i ).

QUIZ 3.1

1. Function f(z) = 3e2z

+ 4ez is entire.


2. f(z) = exp(3 + i) = e3.


3. The range of w = f(z) = ez is the whole complex w-plane.


4. exp(2 + 3i) . exp(4 + 5i) = e6cis 8.


5. | exp(3i) | = 3. (a) True ; (b) False .

Sine and Cosine

If y is a real number, we have

















exp(iy) = cos y + i sin y, exp(–iy) = cos y – i sin y,

and so

cos y = 1/2. (exp(iy) + exp(–iy)),

sin y = 1/2i.(exp(iy) – exp(–iy)).

Thus it is natural to define cosine and sine as:

cos z = 1/2. (exp(iz) + exp(–iz)),

sin z = 1/2i.(exp(iz) – exp(–iz)).

These are Euler's relations. Again notice here how we try to generalize, or extend, a ‘real’ situation to the complex case.

Properties of Sine and Cosine

1. Both functions are entire:

(sin z) = cos z, (cos z) = – sin z.

2. Both functions are periodic, of period 2 . This follows from the periodicity of the exponential function.

The functions satisfy the usual identities, as in the real case.

3. sin2 z + cos

2 z = 1.

4. sin(z1 + z2) = sin z1 cos z2 + sin z2 cos z1 etc.

5. sin(– z) = – sin z, cos(– z) = cos z etc.

QUIZ 3.2

1. Function sin z is periodic, of period 2 i. (a) True ; (b) False .

2. (cos z) = sin z (a) True ; (b) False .

3. sin z = 0 z = n (n Z). (a) True ; (b) False .

4. If z = x + iy then | sin z |2 = sin

2 x + sinh

2 y.


5. If z = x + iy then | sin z | | sin x |.














Logarithmic Function

Does the exponential function have an inverse logarithmic function? Since the exponential function is periodic, any inverse would have to be multi-valued. Let us write

w = log z z = exp w.

If we set z = r cis , w = u + iv, then r cis = eu cis v.

From this, we deduce that

r = eu, u = ln r, v = + 2k .

That is,

w = log z = ln | z | + i( + 2k ) (k Z).

Thus there are infinitely many values of log z, the different values differing by 2k i. Each value of k gives a branch of the logarithm.

The Cut Plane

With log z = ln | z | + i( + 2k ) let us take – < .

Make a (red) cut in the complex plane along the negative x-axis. For any fixed value of k, we obtain a branch which does not cross this cut. So in the cut plane, each branch is single-valued. In particular we have the principal branch

Log z = ln r + i (– < ).

Notes

1. A path which crosses the cut moves to the next branch.

2. If z is real and positive, then Log z = ln r.

3. We can think of the branch planes interleaved together, with the x-axis as a common axis. A path drawn about the origin in one branch plane reaches the cut and then passes to the next branch plane.

4. Our choice of the positive x-axis for the cut was somewhat arbitrary. Other branch cuts are possible; but O is common to them all – O is a branch point.

Properties of the Logarithm (I)

Consider Log z = ln r + i (– < , r > 0) – that is, over the open domain excluding the

cut. There are difficulties on the cut, for is not continuous there for any branch. Hence, for example, the Log function is not continuous on the cut, and so the Log function is not differentiable there.









1. Log z is analytic over the open domain (– < < , r > 0).

Writing Log z = u + iv, we have u = 1/2 ln (x

2 + y

2), v = = arctan

y/x.

Hence

These functions are continuous on the given domain and satisfy the Cauchy-Riemann equations there. Hence by Theorem 2.5, Log z is analytic.

[Note There is a problem in defining arctan here when x = 0. We could overcome this by

defining = arccot x/y, or by taking time to develop a polar form of the Cauchy-Riemann

equations.]

Properties of the Logarithm (II)

2. Derivative

All branches have the same derivative, since they differ by a constant.

3. Inverse Property

exp(log z) = z (for any branch) log(exp z) = z (for a particular branch).

4. Sums and Differences

log z1 + log z2 = log(z1 . z2) log z1 – log z2 = log(z1 / z2)

providing we choose the appropriate logarithm branch on the right.

Examples on the Logarithm

Example 1. Evaluate Log(–1) + Log(–1).

Now –1 = 1 . cis , so Log(–1) = 0 + i .

Hence 2Log(–1) = 2 i = log 1, but not Log 1 (= 0).





Example 2. Show how to make f(z)=log z analytic on the open region A = G R.

In (green) region G, we define f(z) = Log z (the principal value). In (red) region R, we choose a different branch of the logarithm, defining f(z) = log | z | + i arg z ( < arg z < 3 ).

This definition allows a continuous transition acoss the cut.

QUIZ 3.3

1. The function log z is (a) single-valued ; (b) multiple-valued .

2. log (z1z2) = log z1 + log z2. (a) True ; (b) False .

3. Log 1 = 2n i. (a) True ; (b) False .

4. It is always true that Log(z1 / z2) = Log z1 - Log z2. (a) True ; (b) False .

5. For Log z, satisfies

(a) - < ; (b) 0 < 2 ; (c) 0 < .

Complex Exponents

Using our knowledge of real powers, we define the complex power zc (c C) by

zc = exp(c log z), (z 0).

Since zc is defined in terms of the logarithm, we expect z

c to be multivalued, so we use the cut

plane as for the logarithm. Then since log z is single-valued and analytic in the cut plane, so is z

c. Now

So

Exponent Examples

1. i 1/4

= exp(1/4 log i) = exp(

1/4 i ( /2 2k )) = exp( i/8 k i/2) – four values.


















2. i i = exp(i log i) = exp(i ( /2 2k ) i) = exp(– /2 2k ).

The principal value is exp(– /2).

3. What is the relationship between exp z and ez ?

Clearly ez = exp(z log e). Now e = e cis 0, so log e = 1 2k i, and

exp(z log e) = exp(z 2k iz).

It follows that ez = exp z . exp(2k iz). Setting k = 0 gives e

z = exp z.

Thus exp z is the principal value of the multi-valued power function ez.

QUIZ 3.4

1. (z i) = iz

i-1.


2. The principal value of i i is exp( i/2).


3. If z 0 and k is real, then | z k | = | z |

k.


4. i 1/3

(a) is single-valued ; (b) has 3 values ; (c) has infinitely many values .

5. z n, (n Z)

(a) is single-valued ; (b) has n values ; (c) has infinitely many values .













4. MAPPING BY

ELEMENTARY FUNCTIONS

There are many practical situations where we try to simplify a problem by transforming it. We investigate how various regions in the plane are transformed by elementary analytic functions.

You may find the following website helpful:

http://www.malilla.supereva.it/Pages/Fractals/Cmapper/cm.html

Linear Functions (I)

1. w = f(z) = z + c (c = (c1, c2) C) – this is a translation through c.

Each z = (x, y) maps to w = (x + c1, y + c2).

2. Let b = | b | cis be a complex constant and let w = f(z) = bz.

If z = r cis , then w = r.| b | cis( + ).

That is, each point (r, ) maps to (| b | r, + ).

Thus we have a rotation about O through and an enlargement through | b |.

Linear Functions (II)

The general linear function is the mapping w = f(z) = bz + c. This is the rotation and enlargement (2), followed by the translation (1) previously discussed. That is,

z bz bz + c.

Example Consider the transformation w = iz + (i + 1).

This maps square A to square B in the adjacent figure. Notice that S is an invariant point of this transformation. Geometrically, we have:

rotation translation = rotation.

The Function f(z) = z

2 (I)

Let w = f(z) = z 2. Then if z = r cis and w = cis , we have

cis = r2 cis 2 and (r, ) ( , ) = (r

2,2 ).

Example The first quadrant of the complex plane maps to the top half-plane under this

mapping, since the angle range 0 /2 maps to the range 0 .

http://www.malilla.supereva.it/Pages/Fractals/Cmapper/cm.html








We note that in general, the mapping w = f(z) = z2 will not be 1–1. For example, the points z

both map to the same w.

The Function f(z) = z

2 (II)

Let us now write w = f(z) = z 2 in cartesian coordinates. Setting w = u + iv, z = x + iy, we obtain u

+ iv = (x2 – y

2) + 2xyi. That is, (x, y) (x

2 – y

2, 2xy).

So z-points on the hyperbola x2 – y

2 = k map to points on the straight line u = k.

Similarly z-points on the hyperbola 2xy = k' map to points on the straight line v = k'.

Quiz 4.1

The Function f(z) = 1/z

The mapping w = f(z) =1/z (equivalently z =

1/w) sets up a 1–1 correspondence between points in

the z- and w-planes excluding z = 0, w = 0.

In polar coordinates w = 1/z becomes cis =

1/r cis (– ).

This transformation is the product of two simpler transformations:

z = r cis z' = 1/r cis and z' =

1/r cis w = '

– inversion in the unit circle followed by reflection in the x-axis.

Notes

1. Under inversion, the points on remain invariant.

2. Under w = 1/z, (and under inversion),

the centre of remains invariant;

lines through O map to lines through O;








the interior of the exterior of ; (in fact, circles centred at O map to circles centred at O).

Mapping Circles and Lines under f(z) = 1/z (I)

Question What is the effect of f(z) = 1/z on more general lines and circles?

Expressing w = f(z) = 1/z in terms of cartesian coordinates, we obtain

Thus

and inversely

Mapping Circles and Lines under f(z) = 1/z (II)

Now consider the equation

a(x2 + y

2) + bx + cy + d = 0 (a, b, c ,d R). (*)

If a 0, this is the equation of a circle; if a = 0, the equation of a straight line.

Substituting for x, y in terms of u, v we get

d(u2 + v

2) + bu – cv + a = 0 (†)

Points (u, v) satisfying (†) correspond to points (x, y) satisfying (*). Hence we have the four cases:

(a) a 0, d 0. Circle not through O circle not through O.

(b) a 0, d = 0. Circle through O straight line not through O.

(c) a = 0, d 0. Straight line not through O circle through O.

(d) a = 0, d = 0. Straight line through O straight line through O.

Example of a Mapping under f(z) = 1/z

What is the image under w = f(z) = 1/z of line x = c?

The line x = c maps to the set of points satisfying

that is, u2 + v

2 –

u/c = 0, or (u –

1/2c)

2 + v

2 = (

1/2c)

2.

This is the circle with centre (1/2c, 0), passing through the origin O.

The half plane x > c maps to the interior of the disc.







QUIZ 4.2

Bilinear Transformations

Let T be the transformation , (ad – bc 0, a, b, c, d C).

This is called a bilinear transformation as it can be rewritten as

cwz + dw – az – b = 0

– an equation which is linear in each of its variables, z and w.

Solving for w in terms of z, we see that the inverse of T is another bilinear transformation:

.

Notes 1. The singular points for T, T

–1 are z = –d/c, w =

a/c respectively.

As each mapping has just one singular point, we write

z = –d/c ‘w = ’, w =

a/c ‘z = ’.

2. The set of bilinear transformations forms a group.

Understanding the Bilinear Transformation

In the formula for a bilinear transformation, if c 0, we can write

Setting z' = cz + d, z'' = 1/z', we obtain

(*)








If c = 0, we get an expression of type (*) immediately.

It follows that any bilinear transformation can be obtained as the composition of linear transformations and the mapping f(z) =

1/z , all of which map lines and circles to lines and

circles.

Hence any bilinear transformation maps the set of lines and circles to itself.

Images under a Bilinear Transformation

The following reult will be useful.

Theorem 4.1 There exists a unique bilinear transformation which maps three given distinct points z1, z2, z3 onto three distinct points w1, w2, w3 respectively.

Proof The algebraic expression for the bilinear transformation can be written as cwz + dw – az – b = 0. This is an equation in four unknowns a, b, c, d. Hence the three ratios a : b : c : d of these numbers are determined by substituting three pairs of corresponding values of zi, wi (1 i 3).

In practice, this means that in general if we allocate image points to three points in the z-plane, then the bilinear transformation will be completely determined. Conversely, if we want to find the image of a circle (say) under a given bilinear transformation, then it is sufficient to find the images of three points on the circle.

Bilinear Transformation: Example

Let a C be constant with Im a > 0. Find the image of the upper half plane (y 0) under the bilinear transformation

.

We first consider the boundary. For z on x-axis,

we have | z – a | = | z – |.

Hence for such points, .

That is, the x-axis maps to the unit circle having centre O.

Also, z = a (a point in the upper half plane) maps to w = 0 (a point interior to the unit circle). Observing that the mapping is continuous, we deduce that the image of the upper half plane is the interior of the disc.

QUIZ 4.3

The Transformation w = exp z

As before, we set z = x + iy, w = cis and w = exp z gives = ex and = y.










The line y = c maps onto the ray = c (excluding the end-point O) in a 1–1 fashion. Similarly,

the line x = c maps onto the circle = ec. However, here, an infinite number of points on the

line map to each image point.

Combining these results, we see that the rectangular region

a x b, c y d

is mapped to the region

ea e

b, c d

bounded by portions of circles and rays.

This mapping is 1–1 if d – c < 2 .

A Special Case of w = exp z

As a particular case, the strip 0 < y < maps to the upper half plane > 0, 0 < < of the w-plane.

It is interesting to map the boundary points here. This mapping is useful in applications.

5. COMPLEX

INTEGRATION

(TO PART B)

Definite Integrals

http://web.me.com/paulscott.info/CA2/ca5B.html






Integrals are extremely important in the study of functions of a complex variable. The theory is elegant, and the proofs generally simple. The theory is put to much good use in applied mathematics.

We shall study line integrals of f(z). In order to do this we shall need some preliminary definitions.

Let F(t) = U(t) +iV(t) (a t b) where U,V are real-valued, piecewise-continuous functions of t on [a, b], i.e. continuous except for at most a finite number of jumps. The definite integral of F on the interval a t b is now defined by:

Definition F(t) dt = U(t) dt + i V(t) dt.

Properties of the Definite Integral

1. Re ( F(t) dt) = U(t) dt = Re (F(t)) dt. [Real part property]

2. kF = k F ;(k C, const). [Scalar multiple property]

3. (F + G)= F + G. [Addition property]

4. F = – F. [Interchange endpoints property]

5. | F | | F | (a b) [Modulus property]

Property (1) follows immediately from the definition. The proofs of (2), (3), (4) are trivial, and follow from the properties of real integrals. We see from (2) and (3) that the integral behaves in a linear way.

Proof of Property (5)

By definition, F dt is a complex number. So we can set r0 cis 0 = F dt.

By Property (2), r0 = F cis (– 0) dt.

Each side is real, and when a complex number is real, it is the same as its real part.

So by Property (1), r0 = Re (F cis (– 0)) dt.

But Re (F cis (– 0)) | F cis (– 0) | = | F |.| cis (– 0)| = | F |.

So r0 | F | dt, providing a b.

Hence | F dt | | F | dt.





Arcs

Definition A continuous arc is a set of points (x, y): x = (t), y = (t) (a t b), where , are real continuous functions of the real parameter t.

Notes

1. The definition gives a continuous mapping of [a, b] to the arc with a corresponding ordering of points.

2. If no two distinct values of t map to the same point (x, y), the arc is a simple arc or a Jordan arc.

Examples

Jordan arc Not simple Jordan curve Non-simple closed curve

3. If Note (2) holds except that (a) = (b) and (a) = (b), the arc is a simple closed curve or Jordan curve.

Contours

If functions , have continuous derivatives '(t), '(t) not simultaneously zero, we say the

arc (x, y): x = (t), y = (t) (a t b) is smooth (has a continuously turning tangent).

Definition A contour is a continuous chain of a finite number of smooth arcs joined end to end.

Examples

Contour Simple closed contour

Length of an Arc

Definition For a smooth arc, the length exists and is given by

, providing a b.







Now, where does this strange formula come from? From Pythagoras' Theorem, the length s of a small portion of the arc is given by

.

Dividing through by t gives

.

Integrating this expression with respect to t gives the required result.

Note It can be shown that this formula is independent of the choice of parameter used.

QUIZ 5.1A

QUIZ 5.1B

Line Integrals

Let C be a contour extending from z = to z = , and set z = x + iy. Thus for z on C, x = (t), y

= (t) (a t b) say, where , are continuous and ', ' are sectionally continuous. Also t

= a when z = , and t = b when z = .

Let f(t) = u(t) + iv(t) be a (sectionally) continuous function on C (that is, the real functions u = u(t) and v = v(t) are sectionally continuous over a t b).

Definition We define

f(z) dz = f [ (t) + i (t)].[ '(t) + i '(t)] dt.

Here, f(z) dz is a line integral, or contour integral.

Note that the line integral exists because the integrand on right is sectionally continuous.

Expanding the Line Integral

Suppose f(z) = u + iv = u( (t), (t)) + iv( (t), (t)).

Then substituting f(z) = u + iv in the defining expression











f(z) dz = f [ (t) + i (t)].[ '(t) + i '(t)] dt.

gives

f(z) dz = (u ' – v ' )dt + i (u ' + v ' )dt,

or simply

f(z) dz = (u dx – v dy) + i (u dy + v dx).

Note In summary, f(z) dz = (u + iv)(dx + i dy).

Thus we have expressed the complex line integral in terms of two real line integrals.

Properties of the Line Integral

1. (taken over the same contour).

2. kf(z)dz = k f(z)dz (k C , constant).

3. [f(z) + g(z)] dz = f(z) dz + g(z) dz

4. If C1 is a contour (with parameter t from to ), C2 a contour from to , and C = C1 C2 a

contour from to , then .

5. If C has length L and |f(z)| M on C, then | f(z)dz | ML.

Properties (2) and (3) express the linearity of the integral. Properties (1) to (4) follow easily from known properties of the real integral.

Proof of Property 5

Proof We use Property 5 of the Definite Integral:

| F | | F | (a b) (*)

Now,

| f(z)dz | = | f ( (t) + i (t)].[ '(t) + i '(t)) dt.| (a b)

| f ( (t) + i (t)].[ '(t) + i '(t)) | dt (by inequality (*))

M | '(t) + i '(t) | dt ( since | f(z) | M)





= M {( '(t))2 + ( '(t))

2 }dt

= ML.

Line Integral Example I

Find I1 = z2 dz, where C1 is the illustrated path OB.

Line OB has equation x = 2y. Taking y as parameter we obtain the set of points (2y, y) (0 y 1).

Now, z2 is continuous and on C1,

z2 = (x

2 – y

2) + 2ixy = 3y

2 + 4y

2i ,

dz = dx + i dy = (2 + i) dy

[strictly ( '(y) + i '(y)) dy].

Hence

I1 = (3y2 + 4y

2i)(2 + i)dy = (3 + 4i)(2 + i) y

2 dy =

1/3.(2 + 11i).

Line Integral Example II

Find I2 = z2 dz, where C2 is the illustrated path

OAB.

Now

I2 = OA z2 dz + AB z

2 dz

= x2 dx + [(4 – y

2) + 4iy] i dy

= 8/3 + [4 –

1/3 + 2i] =

1/3(2 + 11i).







Some Interesting Questions

Question 1 We see that in the previous examples, the integral from O to B is the same for both contours. Is this a coincidence? Or does it always happen? If it doesn't always happen, when does it?

We can express this in a different way. We see that

OABO = OAB + BO = I2 – I1 = 0.

So we ask: Is the integral around a closed contour always zero?

Question 2 We observe that if we forget the contour altogether, and simply integrate, then we obtain:

, again!

Does this always happen? This would say that C f is independent of the contour C.

Line Integral Example III

Find I3 = dz, where the contour C3 is shown in red.

Now C3 is the contour of points

(x, y) | x = cos , y = sin , 0

and dz = (– sin + i cos ) d .

Hence we have

I3 = dz

= (cos – i sin )(– sin + i cos )d

= i d = – i.





Line Integral Example IV

Find I4 = dz, where the contour C4 is shown in red.

In this case we obtain

I4 = dz = i d = i.

We observe that this is different from I3!

And if we set C = (– C3) C4, then dz = 2 i. (not 0) where the contour C is traversed in an anti-clockwise direction.

Note On the contour C, |z|2 = z = 1. That is, =

1/z. This suggests that there might be some

special significance about the singular point z = 0 being inside the contour.

Line Integral Example V

Without evaluating the integral, find an upper bound for | dz

/z4 | on the given contour C5.

We make use of Property 5:

| f(z)dz | ML.

In this case, the length L = 2 . Also, the contour C has equation y = 1 – x (0 x 1).

Therefore for z on C

| z4 | = (x

2 + y

2)2 = [ x

2 + (1 – x)

2 ]2 = [2x

2 – 2x + 1]

2.

That is, | z4 | = [2(x –

1/2)

2 +

1/2]

2 1/4.

(In retrospect, this is obvious from the figure! Why?)

Hence | 1/z

4 | 4 and | I | 4 2 .

QUIZ 5.2

Green’s Theorem

Let R be a closed region in the real plane made up of a closed contour C and all interior points.










If P(x, y), Q(x, y) are real continuous functions over R and have continuous first order partial derivatives, then Green’s Theorem says:

(P dx + Q dy) = R (Qx – Py) dxdy

where C is described in the anti-clockwise direction.

Outline Proof

Py dxdy = [ Py dy] dx

= [ P(x, u(x)) – P(x, l(x)) ] dx

= – P dx etc.

Cauchy’s Theorem

Now consider a function f(z) = u(x, y) + iv(x, y) which is analytic at all points within and on the closed contour C, and is such that f '(z) is continuous there.

We show that f(z) dz = 0.

The given conditions tell us that u, v and their first order partial derivatives are continuous. Now

f(z) dz = (u dx – v dy) + i (u dy + v dx)

= – R (vx + uy) dxdy + i R (ux – vy) dxdy

= 0, using Green's Theorem and the Cauchy-Riemann equations.

This result was discovered by Cauchy.

Examples and the Cauchy-Goursat Theorem

As a consequence of Cauchy's Theorem we have the following:

Examples dz = 0, z dz = 0, z2 dz = 0,

for all closed contours C, since these functions are analytic everywhere and their derivatives are everywhere continuous.

Goursat showed that Cauchy’s condition ‘f '(z) is continuous’ can be omitted. This discovery is important, because from it we can deduce that all derivatives of analytic functions are also analytic. But, proving this stronger result takes much more effort!





Theorem 5.1 (Cauchy-Goursat Theorem) If a function f is analytic at all points interior to and on

a closed contour C, then f(z) dz = 0.

Cauchy-Goursat Theorem, Example I

Find where the contour is the unit circle | z | = 1.

Now the integrand z2/(z–3) is analytic everywhere except at the point z = 3. This point lies

outside the circular disk | z | 1. Hence by the Cauchy-Goursat Theorem

Most of the useful applications of the Cauchy-Goursat Theorem are as yet beyond us, but the next example gives us a glimpse of what can be done.

Cauchy-Goursat Theorem, Example II

Assuming Laplace’s Integral: show that

[We note that Laplace's integral can be evaluated by writing

and expressing this second integral in polar coordinates.]

We integrate f(z) = exp(–z2) around the rectangle C defined by | x | R, 0 y b. Later we let R

.

Since exp(–z2) is entire, the Cauchy-Goursat Theorem applies, that is,

[Continued]

Example II, continued

Hence

Taking out the factor from the second and fourth integrals and letting R , we get a

zero contribution (using the | dy | ML. formula). Hence letting R ,







Equating the real parts,

or

since the integrand is even.

Indefinite Integrals

Let f be analytic in D and z0, z D. Then for contours C1, C2 lying in D, joining points z0, z, and being traversed from z0 to z,

by the Cauchy-Goursat Theorem.

Theorem 5.2 (‘Primitive’ Theorem) For all such paths in the domain D

has the same value, and F'(z) = f(z).

Note This will allow us to evaluate some line integrals by straight integration. We shall see that the name of the theorem comes from the fact that the function F satisfying F'(z) = f(z) is called a primitive of f.

Proof of the ‘Primitive’ Theorem

Proof Let z + z be a point in D. Then

F(z + z) – F(z) = ,

where since z is very small, we may take the path from z to z + z to be a line segment.

Now clearly on the segment, So

.

Then





[Continued]

Proof of the ‘Primitive’ Theorem (Continued)

Now f is continuous at z. Hence for all > 0, there exists > 0:

| z | < | f(z ') – f(z) | < .

Hence, when | z | < ,

(using Property 5).

That is,

and F(z) exists at each point of D, and F'(z) = f(z).

Notes on the ‘Primitive’ Theorem

We say that F is an indefinite integral or primitive (anti-derivative) of f and write F(z) = f(z) dz.

That is, F is an analytic function whose derivative is f(z).

Since

we can use this as a means of evaluating line integrals. (All paths here are assumed to be in the domain D.)





COMPLEX

INTEGRATION PART B

(BACK TO PART A)

The ‘Primitive’ Theorem: Example I

The function f(z) = z2 is entire.

It has primitive F(z) = 1/3.z

3.

So by our theorem,

along any contour joining 0 and 1 + i.

The ‘Primitive’ Theorem: Example II

We recall that 1/z dz = – i,

1/z dz = i, where C1, C2 are upper and lower semicircles of

the unit circle. Now we can easily choose our domain D to avoid O and contain the given contour Ci. Mysterious question! So why does the difference occur?

The answer is that for the different contours, we need different primitives!

For C1, we might choose as primitive

log z = ln | z | + i arg z (– /2 < arg z < 3 /2).

Then

But for C2, our choice of primitive could be

log z = ln | z | + i arg z ( /2 < arg z < 5 /2)

and

Simply and Multiply Connected Domains

Defintion A simply connected domain D is an open connected region such that every closed contour within it encloses only points of D. Otherwise the domain is multiply connected.

Examples










simply connected multiply connected

The Cauchy-Goursat Theorem has been stated for simply connected domains. i.e. if f(z) is analytic throughout a simply connected domain D, then for every closed contour C within D,

f(z) dz = 0.

Extending the Cauchy-Goursat Theorem

It will be useful to extend the Cauchy-Goursat Theorem to certain multiply-connected domains. Consider the illustrated (red) region D and suppose that f(z) is analytic over this (closed) region.

We assert that f(z) dz = 0, where B is the total directed boundary (C C1 C2), with all components traversed so that the region is on the left.

This is easy to prove. We insert the indicated green links partitioning D into two simply-connected domains. We apply the Cauchy-Goursat Theorem to the boundaries of the left and right regions, obtaining two line integrals having value 0. Putting the two circuits together, the integrals along the introduced links cancel, giving the required result.

Multiply-Connected Domains: Examples

Example 1 = 0 where B is the two-circle contour shown.

For, the integrand has singularities 0, 3i, and is analytic over the enclosed domain.

Example 2 Find where T is the illustrated triangle. Now

1/z is analytic

inside T, except at O. Consider the unit circle C, centre O. Since 1/z is analytic in the domain bounded by C and T, noting the

direction of C, – = 0, that is, = .

Hence it is sufficient to evaluate the integral around C. So although we have not solved this problem, we have simplified it.

The Cauchy Integral Formula







Theorem 5.3 Let f be analytic everywhere within and on a closed contour C. If z0 is any point interior to C, then

where the integral is taken in the positive sense around C.

Notes

(1) The formula is the Cauchy integral formula. It is remarkable because it gives the value of f at z0 in terms of the values of f on the boundary. That is, for an analytic function, fixing the boundary values completely determines f at points inside C.

(2) The theorem can also be used to evaluate certain integrals.

Cauchy Integral Formula Example

Evaluate where C is | z | = 2.

Take f(z) = z/(9 – z

2), z0 = – i. Then we observe that f(z) is analytic

in and on C. So using the Cauchy integral formula,

Proof of the Cauchy Integral Formula (I)

Let C0 be a circle centre z0, radius r0 interior to C. Now the function f(z)

/(z–z0) is analytic at all points in and on C except at z = z0, in particular in the region D lying between C and C0. Hence as in Example 2 above





where both contours are traversed in the positive direction. Hence

or I = f(z0) I1 + I2 say.

[Continued]

Proof of the Cauchy Integral Formula (II)

For z on C0, we can write z – z0 = r0 cis = r0 exp(i ).

Hence dz = i r0 exp(i ) d , and

for every r0 > 0.

Also, f is continuous at z0, so given > 0, there exists > 0: | z – z0 | | f(z) – f(z0) | < .

Take r0 = . Then | z – z0 | = , and

Hence I2 can be made arbitrarily small by taking r0 sufficiently small. Since I and I1 are independent of r0, I2 must be too. Therefore I2 = 0, and

I = f(z0) I1 = 2 i f(z0).

Derivatives of Analytic Functions

Theorem 5.4 Suppose f is analytic inside and on a closed contour C, and z0 lies inside C. Then

That is, we can take the Cauchy integral formula and formally differentiate with respect to z0.

Proof We omit this proof. It is not unlike the proof of the Cauchy integral formula.

Note We can similarly prove:





A Glimpse and an Example

One of most beautiful parts of course on Complex Functions is the development of Taylor series. We look at this a little later, but for now, note the appearance here of the terms

f(n)

(z0)/n!.

Example Evaluate (a) , (b) where C is the unit circle | z | = 1 taken in the anti-clockwise direction.

(a) Take f(z) = cos z, z0 = 0 (within C). Since cos z is entire,

I(a) = 2 i .cos z0 = 2 i (the Cauchy integral formula).

(b) Take f(z) = sin z, z0 = 0, f '(z) = cos z. Then

I(b) = 2 i f '(z0) = 2 i cos 0 = 2 i (the Derivative formula).

Another Example

Let us take f(z) = 1 in Cauchy's integral formula and the Derivative formulae, and let C be any contour about z0.

(a) By Cauchy's integral formula, we have

This can be rewritten as

(b) By the Derivative formulae,

This can be rewritten as





Corollary of the Derivative Formulae

If a function f is analytic at a point, then by the Derivative formulae, its derivatives of all orders are also analytic functions at that point.

Now if f(z) = u + iv, then

f '(z) = ux + ivx = vy – iuy.

So if f '(z) is analytic, then ux, vx, uy, vy are all differentiable and so continuous.

In the same way, using f "(z) etc., we see that all partial derivatives of u, v of all orders are continuous at any point where f(z) is analytic. Thus we have:

Corollary If f = u + iv is analytic at a point, then all partial derivatives of u, v of all orders are continuous there.

Note Cauchy's integral formula and the Derivative formulae easily extend to the boundaries of multiply connected domains.

A Useful Lemma

Here is an interesting little result.

Lemma If f is analytic, and | f | is constant, then f is constant.

Proof Let f = u + iv. Then we are given that u2 + v

2 = c.

So

2uux + 2vvx = 0, 2uuy + 2vvy = 0,

leading to

ux/uy = vx/vy.

Also, by the Cauchy-Riemann equations, ux = vy, uy = –vx.

So, eliminating the u terms, we get vx2 + vy

2 = 0

and so vx = 0 = vy = ux = uy.

Hence f '(z)=0 and so f(z) is constant.

Maximum Modulus

Let f be analytic and not constant on the open disk | z – z0 | < r0, centred at z0. If C is any circle | z – z0 | = r (0 < r < r0) then by the Cauchy integral formula,

(*)

Path C is in the positive sense and parametrizes: z( ) = z0 + r exp(i ) (0 2 ).







So (*) becomes (i.e. the value at the centre is the arithmetic mean of the values on the circle). Thus

(0 r < r0). (+)

Now suppose | f(z0) | is a maximum. Then | f(z) | | f(z0) | for all z: | z – z0| < r0.

[Continued]

Maximum Modulus Principle

So, (0 r < r0). (+)

Combining the two inequalities (+), we have

It follows that, | f(z) | = | f(z0) | for all z : | z – z0 | < r0.

For, write the above equation as

Since the integrand is non-negative, we deduce it must be 0 for all z : | z – z0 | < r0.

This shows that f(z) = f(z0) for all z in the disk. So f is constant in the disk. This is a contradiction.

Theorem 5.5 (Maximum Modulus Principle) If f is analytic and not constant in the interior of a region then | f(z) | has no maximum value in that interior.

Some Observations

We deduce that if a function f is continuous in a closed bounded region R and is analytic and not constant in the interior of R, then | f(z) | assumes its maximum value on the boundary of R and never in its interior.

Now let f be analytic in and on the circle C0 defined by |z – z0| = r0, and traversed in a positive sense. Then by the Derivative formulae

, n = 0, 1, 2, ... .

If | f(z) | M on C0, then

| f (n)

(z0) | n! M

/ r0n (n = 0, 1, 2, ...),

and for n = 1





| f '(z0) | M / r0.

Liouville’s Theorem

The preceding observations lead to the following theorem.

Theorem 5.6 (Liouville) If f is entire and bounded for all values of z in the complex plane, then f(z) is constant.

Proof By assumption | f(z) | M for all z.

Therefore, as before, | f '(z0) | M / r0 for each z0 in the plane, and for any positive r0, no matter how large.

It follows that f '(z0) = 0.

But z0 is arbitrary. Thus for all z, f '(z) = 0, and so f(z) = constant.

The Fundamental Theorem of Algebra

Theorem 5.7 (Fundamental Theorem) Any polynomial

P(z) = a0 + a1z + ... + anzn (an 0)

where n 1, has at least one zero. That is, there is at least one point z1 : P(z1) = 0.

It is curious that this important result has no easy algebraic solution.

Proof Suppose P(z) 0 for any z. Then f(z) = 1/P(z) is entire.

In fact f(z) is also bounded. For f is continuous and so bounded in any closed disk centred at the origin. Further, if R is large and z is exterior to the disk | z | R, then

| f(z) | = 1/| P(z) |

1/R

n (or worse!)

so f is bounded for all values of z in the plane. (We can tidy up this last argument, but the idea is that when | z | is large, so is | P(z) |.)

Now by Liouville's Theorem, f(z) and so P(z) is constant. This contradiction shows that P(z) has at least one zero.

Factorization of Complex Polynomials

Let z1 be the zero guaranteed by the Fundamental Theorem.

Then P(z) = (z – z1)Q(z), where Q(z) is a polynomial of degree n – 1.

We deduce (by induction) that

P(z) = c(z – z1)(z – z2) ... (z – zn).

Corollary Any polynomial of degree n, where n 1 can be expressed as a product of n linear factors. That is,

P(z) = c(z – z1)(z – z2) ... (z – zn),







where c and the zk are complex constants.

You might like to compare this result with what happens for polynomials over the reals R.

6. SERIES

Convergence of Sequences

A sequence z1, z2, ... , zn, ... has limit z if for all > 0 there exists N : n > N | zn – z | < .

That is, we can make zn arbitrarily close to z by taking n sufficiently large.

This definition is formally the same as for the real case. Of course, here the points of the sequence lie in the complex plane, rather than on the real line.

The limit, if it exists, is unique.

We say the sequence converges to z0 and write zn z or zn = z.

If there is no limit, we say that the sequence diverges.

A First Convergence Theorem

Theorem 6.1 If zn = xn + iyn (n = 1, 2, ...) and z = x + iy, then

zn = z xn = x and yn = y.

Proof

( ) Given > 0 there exists N : n > N | xn + iyn – (x + iy) | < . Hence n > N | xn – x | < and | yn – y | < .

i.e. xn = x and yn = y.

( ) Given > 0 there exist N1, N2 :

n > N1 | xn – x | < /2 , n > N2 | yn – y | < /2.

So n > max(N1, N2) | xn – x | + | yn – y | < . Now | xn + iyn – (x + iy) | | xn – x | + | yn – y | < .

So zn = z.

Infinite Series and Partial Sums

The expression

z1 + z2 + z3 + ...

is an infinite series.

The set

SN = z1 + z2 + ... + zN

is a partial sum of the series.








If the sequence S1, S2, ... , SN, ... converges to limit S we write zn = S, and S is the sum of the series.

The sum, when it exists, is unique.

When a series does not converge, it diverges.

Convergence of Complex Series

Theorem 6.2 Suppose that zn = xn + iyn (n = 1, 2, ...) and S = X + iY. Then

zn = S xn = X and yn = Y.

Proof Let SN = XN + iYN denote the Nth partial sum, where XN = xn and YN = yn.

Now

zn = S SN = S XN = X and YN = Y

by Theorem 6.1. Since XN, YN are the partial sums of xn and yn, the result follows.

Remainder and Power Series

In establishing that a given series has sum S, we define the remainder after N terms to be:

RN = S – SN.

Since | S – SN | = | RN – 0 |, SN S iff RN 0 as N .

Hence, a series converges to sum S the sequence of remainders converges to 0.

We shall be particularly concerned with power series. A power series is a series of the form

a0 + an(z – z0)n = an(z – z0)

n

where z0 and the an are complex constants, and z is any number (variable) in a stated region. We will use the notation S(z), SN(z), RN(z) for the sum, partial sum and remainder respectively.

Taylor Series

Theorem 6.3 (Taylor) Let f be analytic everywhere inside the circle C0 : | z – z0 | = r0. Then at each point z inside C0

f(z) = f(z0) + f '(z0)(z – z0) + f ''(z0)/2! .(z – z0)2 + ... + f

(n)(z0)/n! .(z – z0)

n + ... .

That is, the power series converges to f(z) when | z – z0 | < r0.







Notes

(1) This is the Taylor series expansion about point z0.

(2) If all terms are real, we get the real Taylor series.

(3) The proof of Taylor's Theorem we give is remarkably ‘natural’, and is one of the rewards in our study of complex functions.

Proof of Taylor's Theorem (I)

Proof Let z be any fixed point inside C0 and set | z – z0 | = r (so r < r0). Let C1 be a circle centred at z0 and having radius r1 : 0 < r < r1 < r2.

Let be any point on this circle; i.e. | – z0 | = r1.

Now z lies inside C1, and f is analytic in and on C1, so by the Cauchy

integral formula where C1 is taken in the positive sense. Now

Also for any complex c 1,

[Continued]

Proof of Taylor's Theorem (II)

So

Hence

We next integrate each term anticlockwise around C1, divide by 2 i, and substitute the Cauchy integral formula and the Derivative formulae:

, n = 0, 1, 2, ... .

[Continued]

Proof of Taylor's Theorem (III)

http://web.me.com/paulscott.info/CA2/ca6.html#anchorPofTT

http://web.me.com/paulscott.info/CA2/ca6.html#anchorPTT2

http://web.me.com/paulscott.info/CA2/ca6.html#anchorPTT3







So

where (*)

Recall that | z – z0 | = r, | – z0 | = r1 (> r), and | – z | | – z0 | – | z – z0 | = r1 – r.

Thus if M denotes the maximum value of | f( ) | on C1, (*)

Since r/r1 < 1, RN(z) = 0.

So for each z interior to C0, the Taylor series for f converges to f(z).

Special Case and Observations

Let us seek the Maclaurin expansion for f(z) = ez.

We have f (n)

(z) = ez, so f

(n)(0)=1. Also, e

z is analytic for all z, so

ez = 1 + z + z

2/2! + ... + z

n/n! + ... = z

n/n! , |z| < .

Similarly

sin z = z – z3/3! + z

5/5! – ... , |z| <

cos z = 1 – z2/2! + z

4/4! – ... , |z| <

sinh z = z + z3/3! + z

5/5! + ... , |z| <

cosh z = 1 + z2/2! + z

4/4! + ... , |z| <

Geometric Series

Let us try to find the Maclaurin series for the function

Now

This is not a Maclaurin series: the first two terms are unexpected, and function f has a singularity at z = 0.





Question Perhaps there are other interesting series to investigate?

Laurent’s Theorem

Let C1, C2 be concentric circles, centre z0, with radii r1, r2 (r1 > r2).

Theorem 6.4 (Laurent) If f is analytic on C1 and C2 and throughout the annulus between these two circles, then at each point in this domain, f(z) is represented by the expansion

(*)

where

(**)

each path of integration taken counter-clockwise.

The series (*) is a Laurent series.

Notes on Laurent's Theorem (I)

(1) If f is analytic at all points in and on C1 except at z0, we can take r2 to be arbitrarily small. Then (*) is valid for 0 < | z – z0 | < r1.

(2) If f is analytic at all points in and on C1, then f(z)

/(z – z0)– n+1

is analytic in and on C2 (since –n + 1 0). So integral (**) is zero, and (*) reduces to the Taylor series.

Notes on Laurent's Theorem (II)







(3) Since f(z)

/(z – z0)n+1

and f(z)

/(z – z0)– n+1

are analytic throughout the annular region r2 | z – z0 |

r1, we can replace and by where C is any closed contour around the annulus in the positive direction.

This means that (*) can be written

r2 < | z – z0 | < r1

where

(***)

Notes on Laurent's Theorem (III)

(4) In practice, some, or even many of the coefficients may be zero.

Example Consider f(z) = 1/(z – 1)

2 where | z – 1| > 0.

Here z0 = 1, c– 2 = 1, and all the other coefficients are zero.

Using (***), we observe that

Notes on Laurent's Theorem (IV)

(5) For particular examples we usually do not find the coefficients of an expansion using the formula. In other words, the general formula is useful more as an existence formula.

Example Find the Laurent series (z0 = 0) for f(z) = ez/z

2.

Using the Maclaurin expansion for ez, we obtain:

Example Find the Laurent series (z0 = 0) for f(z) = e(1/z)

.

Using the Maclaurin expansion for ez with a change of variable,we obtain:





Proof of Laurent's Theorem (I)

If z lies in the annular region, then

(†)

This is the Cauchy integral formula extended to a multiply-connected domain.

To show this explicitly in this special case, we take a small anti-clockwise directed circle K with centre z, lying within domain. Then

Notice that by the Cauchy integral formula,

Comparing this equation with equation (†), the validity of (†) is confirmed.

[Continued]

Proof of Laurent's Theorem (II)

The first integral of (†) will give the Taylor series part, so as before

For the second integral of (†) we note that

Multiplying through by f( ), and expanding the last quotient as a geometric series gives

[Continued]

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Proof of Laurent's Theorem (III)

So from (†), where an, bn are as given in the statement of the theorem, RN(z) is as before, and RN(z) 0 as N .

Also,

If r = | z – z0 |, and r2 is the radius of C2, then r2 < r.

Let M be the maximum of | f( )| on C2. Then

0 as N .

This completes the proof of the theorem.

Further Properties of Series

There are many parallels with real series. This follows from the fact that zn is convergent if

and only if xn and yn are convergent.

Now if xn and yn are convergent, then xn 0, yn 0. We deduce that

zn convergent zn 0. Of course, the converse is false!!

So the terms of a convergent complex sequence are bounded:

that is, there exists M: | zn | < M for all n.

We say that zn is absolutely convergent (AC) if the series

| zn | = (xn2 + yn

2) is convergent. So by the Comparison Test for real series, | xn | and

| yn | are convergent. Thus xn, yn are AC and so convergent. It follows that zn is convergent.

Thus zn absolutely convergent zn is convergent.

Absolute Convergence of Power Series

We now prove an important result for power series. Analogues of the next results hold for an(z – z0)

n, but we give this proof for z0 = 0.

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Theorem 6.5 If a power series anzn converges when z = z1 ( 0) then it is A.C. for all z : | z | <

| z1 |.

Proof Since anz1n is convergent, for some M we have | anz1

n | < M for all n.

We write | z |

/ | z1 | = k ( < 1 ).

Then

| anzn

| = | anz1n

|.| z / z1 |

n < Mk

n.

Now the series with terms Mkn (k < 1) is a real, convergent, geometric series.

So by the Comparison Test,

anzn is convergent.

Circle of Convergence

Our previous result shows that the set of all points inside some circle centred at the origin is a

region of convergence for anzn. The largest such circle is the circle of convergence.

We note that by the theorem, the series cannot converge at any point z2 outside this circle.

Similarly, if the series bn / zn converges for z = z1, then it is absolutely comvergent at every

point z exterior to the circle centre O passing through z1. The exterior of some circle centred at O is therefore a region of convergence.

Some Stated Results (I)

The theory starts to get a bit solid (boring!) here, so we settle for some stated results. Nevertheless, these results are important.

Let S(z) = anzn over some circle of convergence C1. Thus S is the function defined by the

convergent power series. Then

(1) Function S(z) is continuous at each z interior to C1.

(2) Function S(z) is analytic at each z interior to C1.

(3) If C is any contour interior to C1, then the power series can be integrated term by term, i.e.

S(z) dz = an zn

dz.

[If C is a closed contour, then of course we get the value 0 on both sides.]

Some Stated Results (II)

(4) The power series can be differentiated term by term.

Thus for each z inside C1,

S'(z) = nanz n –1.

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(5) The Taylor/Laurent series about z0 for a given function is unique.

Illustration

sin(z2) = z

2 – z

6/3! + z

10/5! – ... .

Even though this series is obtained by substituting z2 in the series for sin z, it will be

the same as the Maclaurin series for sin (z2).

Some Stated Results (III): Cauchy Product

(6) Let f(z) = an zn, g(z) = bn z

n.

If we formally multiply these series together and collect the coefficients of like powers of z, we get the Cauchy Product of the two series:

f(z).g(z) = a0b0 + (a0b1 + a1b0)z + (a0b2 + a1b1 + a2b0)z2 + ...

+( akb

n–k ) z

n + ... .

We now have:

The Cauchy Product of two power series converges to the product of their sums at all points interior to their circles of convergence.

Laurent’s Theorem: Example I

For the next examples, set

This function is analytic everywhere except at z = 1 and z = 2.

Find the Maclaurin series for f(z) valid in | z | < 1.

Now

Also | z | < 1 | z/2 | < 1. Hence

valid for | z | < 1.

Thus this is the Maclaurin series for f(z) over the given domain.





Laurent's Theorem: Example II

We are given

Find the Laurent expansion for f(z) in 1 < | z | < 2.

In this case we have | 1/z | < 1 and | z / 2 | < 1.

So

valid for 1 < | z | < 2.

This is the required Laurent expansion.

We observe here that c-1 (or b1) = 1.

Evaluation of Integrals

We noted in the previous calculation that c–1 (or b1) = 1.

Recall that

Thus

where C is any simple closed contour around the annulus (taken in the positive direction). That is,

f( ) d = 2 i.

This suggests the use of the Laurent series for the evaluation of integrals.

Laurent's Theorem: Example III

Obtain the Laurent expansion in | z | > 2 for

Here | 2/z | < 1 and so |

1/z | < 1.







So

valid for | z | > 2.

Note that the coefficient of z-1

is zero. Hence f( ) d = 0 for any small contour C about 0 and exterior to circle | z | = 2.

Note It is possible to develop the whole theory of analytic functions beginning with series, but there is no great advantage. The proofs are not easy and motivation is lacking.

Zeros of an Analytic Function (I)

If f is analytic at z0, there exists a circle centre z0 within which f is represented by a Taylor series:

f(z) = a0 + an(z – z0)n (| z - z0 | < r0)

where a0 = f(z0) and an = f (n)

(z0)/n! . If z0 is a zero of f, then a0 = 0. If in addition

f '(z0) = 0 = f ''(z0) = ... = f (m -1)

(z0)

but f (m)

(z0) 0, then z0 is a zero of order m. In this case,

f(z) = (z – z0)m

am+n(z – z0)n

(am 0, | z – z0 | < r0)

= (z – z0)m

g(z) say,

where g(z0) = am 0.

Zeros of an Analytic Function (II)

Thus f(z) = (z - z0)m

g(z) where g(z0) = am 0.

Since g(z) is represented by a convergent power series, g is continuous at z0. That is, given >

0, there exists a > 0 such that | z - z0 | < | g(z) – g(z0) | < .

Now choose = | am | /2.

Then there exists : | z - z0 | < | g(z) – am | < | am | /2.

It follows that g(z) 0 at any point in neighbourhood | z - z0 | < .

We have proved:

Theorem 6.6 Let f be analytic at point z0 which is a zero of f. Then there exists a neighbourhood of z0 throughout which f has no other zeros, unless f 0. That is, the zeros of an analytic function are isolated.





7. RESIDUES

AND POLES

Introduction

Definition If there exists a neighbourhood of z0 throughout which f is analytic except at z0 itself, then z0 is an isolated singularity of f.

Example has three isolated singularities: z = 0, z = i.

Contrast Log z which has a continuous ray of singularities.

Now from the Cauchy integral formula and the Derivative formulae,

where the integrals are in an anti-clockwise direction about a simple closed contour containing z0. We observe that the values of these integrals are of the form 2 i.K.

Residues

We now change our notation, replacing f(z)

/ (z – z0) by f(z). So denote by f(z) a function which is analytic on and inside C except at an isolated singular point z0 inside C.

Then f(z) dz = 2 i.K, where K is a constant and the integral is once anti-clockwise round C.

Definition is the residue of f at the isolated singular point z0.

Theorem 7.1 (Residue Theorem) Let C be a closed contour within and on which function f is analytic except for a finite number of singular points z1, z2, ... , zn interior to C. If Ki denotes the residue of f at zi, then

f(z) dz = 2 i.(K1 + K2 + ... + Kn),

where the integral is around C in positive sense.

Proof of the Residue Theorem

Proof Let each zi be enclosed as shown in a circle Ci with radius small enough so that C and Ci are all separated. Now f is analytic in the remaining region within and including C, so by Cauchy's Theorem:

f(z) dz – f(z)dz – ... – f(z)dz = 0,








so f(z) dz = f(z)dz – ... – f(z)dz = 2 i.(K1 + K2 + ... + Kn),

using the definition of Ki.

It follows that evaluation of such integrals depends on our ability to evaluate residues.

Example on the Residue Theorem

Evaluate , where C is | z | = 2 taken in the positive sense.

The singularities inside C are z = 0, 1. So I = 2 i.(K0 + K1) (say).

Find K0. Set g(z) = (5z – 2)

/(z – 1) – analytic in a small circle C0 centred at 0.

By the Cauchy integral formula, g(z)

/z dz = 2 i.g(0) = 2 i.2 = 2 i.K0.

Find K1. Set h(z) = (5z – 2)

/z – analytic in a small circle C1 centred at 1.

Again by the Cauchy integral formula, h(z)

/(z – 1) dz = 2 i.h(1) = 2 i.3 = 2 i.K1.

Hence I = 2 i.(2 + 3) = 10 i.

Alternative Solutions

(1) Use partial fractions:

(2) Use the Laurent expansion with | z | > 1, and find the coefficient of 1/z. So

and noting that the coefficient of 1/z is 5, we obtain the result as before.

QUIZ 7.1

Singularities and Poles

Singularities are obviously important in the theory. The nature of a singularity can be determined from the Laurent expansion

(over some domain)










and in particular from the portion involving negative powers, known as the principal part of f at z0.

Suppose f(z) = (z) / (z – z0)n, where is analytic at z0 and (z0) 0. (This certainly means that

has no (z - z0) factors).

Then f has a pole of order m at z0 (or z0 is a pole). A pole of order 1 is a simple pole.

Examples of Poles (I)

(1) Consider f(z) = (z2 – 2z + 3) / (z – 2).

Now (z) = z2 – 2z + 3 is entire and (2) 0, so z = 2 is a simple pole.

(2) Consider f(z) = (z + 1)/(z2 + 1)

2 and z = –i.

Set (z) = (z + 1) / (z – i)2.

Then f(z) = (z) / (z + i)2, where is analytic at z = –i and (–i ) 0.

So z = –i is a pole of order 2.

Examples of Poles (II)

(3) Consider f(z) = sin z

/ z .

Here (z) = sin z is analytic, but (0) = 0. Hence this is not a simple pole. In fact

We say z = 0 is a removable singularity. We could define f(0) = 1. (This would give a new function, coinciding with f for z 0, but also defined at z = 0 and continuous there.)

(4) Consider f(z) = e1/z

. There is a problem here at z = 0. Assuming

it is not possible to put this in the form (z)/zm

for any m. We call this an essential singularity.

Examples of Poles (III)

(5) Essential singularities often exhibit strange behaviour. When f has a pole at z0, we expect f(z) as z z0. This occurs. But it may not occur for an essential singularity.

Example







Consider

For this function there is an essential singularity at z = 0.

Take z = r cis , so exp(1/z) = exp[(1/r)cis(- )].

Now take = /2 and let r 0 (i.e. approach O along the positive imaginary axis.) Approaching 0 along this path, | e

1/z | = exp[(1/r) cos /2] = e

0 = 1;

that is, e1/z

does not tend to .

Notes on Poles

(1) We can give general formulae for the residues for poles of order m – essentially using Theorems 6.3, 6.4.

(2) Work on series is useful here.

(3) Most examples treat poles of low order. It is suggested that you learn the Cauchy integral formula and the Rules on Differentiation with respect to z0. Thus:

More Examples on Poles and Residues

(1) Consider .

This function has a pole of order 3 at z = 0. Hence the residue there is

1/2! . f ''(0) =

1/2! . 4 . e

0 = 2.

(2) Consider .

This function has a simple pole at z = 3i. The residue there is





QUIZ 7.2

Improper Real Integrals: Cauchy Principal Value

In the case of real improper integrals, we make the definition:

(1)

where both integrals on the right exist. Notice that the variables R, R' tend to infinity independently in the two integrals.

It is useful to define the Cauchy Principal Value (Cauchy P.V.) in the following way:

(2)

So with the Cauchy P.V., we are insisting that the upper and lower infinite limits are approached at the same rate.

If the improper integral defined by (1) converges, then the value obtained is the Cauchy P.V. On the other hand, the Cauchy P.V. may exist and integral (1) not converge.

Using the Cauchy Principal Value

Example Let f(x) = x. Here the Cauchy P.V. is 0, but the integral is not convergent.

Special case If f is even and the Cauchy P.V. exists, then converges.

For in this case

and the existence of the last Cauchy P.V. guarantees the existence of the first two integrals.

Example If f(x) = p(x)

/ q(x) where p, q are real polynomials with no common factors, q(x) has no real zeros, and the degree of q(x) is greater than or equal to the degree of p(x) + 2, then

converges. Its value can easily be found using residues.

Cauchy Principal Value: Example (I)

Evaluate Consider

This has simple poles z = 3i; poles of 2nd order z = 2i. We find the residues for the poles lying inside the illustrated contour C.










For z = 3i:

For z = 2i:

Hence

[Continued]

Cauchy Principal Value: Example (II)

Now on CR,

and the length of CR is R. Thus

So = /100 (noting that f is even), or I = /200 .

Question Is this easier than factorizing and using partial fractions in the real case?!

Probably yes, especially as the integral around CR will fairly clearly always vanish when the difference in degree is 2 or more.

Improper Integrals involving Trigonometric Functions

Residue theory is also useful for evaluating integrals of the form

(*)

where p, q are real polynomials and q(x) has no real zeros.

Note that the previous method cannot be used here. For we have

| sin z |2 = sin

2 x + sinh

2 y and | cos z |

2 = cos

2 x + sinh

2 y,

so | cos z | and | sin z | increase like sinh y as y .

However, the integrals (*) can be combined to give

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,

and | e iz | = e

– y, which is bounded in the upper half plane.

Trigonometric Function Integral: Example (I)

Show that

This integral is the real part of and we obtain this by integrating

along the real axis. The function f is analytic except for poles of order 2 at z = i. The pole z = i lies inside the illustrated semicircular contour.

So

Now (calculating).

[Continued]

Trigonometric Function Integral: Example (II)

We show that the second integral tends to 0 as R . For z in CR,

| z2 + 1 |

2 (R

2 – 1)

2 and | e

iz | = | e

–y | 1 (y 0)

so as R .

Hence

So, taking real parts,

Since the integrand is even here, this Cauchy P.V. is the required integral.

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Definite Integrals of Trigonometric Functions

We can use residues to evaluate certain definite integrals of the type .

The variation of from 0 to 2 suggests that we use as the argument of a point z on the unit

circle C. That is, z = exp(i ) (0 2 ). Then

and the integral becomes

That is, a contour integral of a function of z around the circle C in the positive sense.

Integrals of Trigonometric Functions: Example (I)

Show that

The formula is valid for a = 0. Suppose that a 0. Then

where C is the circle | z | = 1 traversed in the positive direction.

The denominator has zeros:

Hence the integrand is .

[Continued]

Integrals of Trigonometric Functions: Example (II)

Also, noting that | a | > 1, we have | z2 | = (1 + (1 – a2)) / |a| > 1;

that is, z2 lies outside C.

Further, | z1z2 | = 1, so | z1 | < 1, – a simple pole inside C.

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The corresponding residue K1 is:

Hence I = 2 i.K1 = 2 / (1 – a2) .

QUIZ 7.3

Final Comment

Because of time constraints, the course finished here. With a little more time, we would have showed that analytic mappings are conformal (preserve angle measure), and worked through a few boundary value problems. These would have demonstrated again the practical nature of complex analysis, and given us practice in the use of complex mappings.

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2959 complex analysis

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