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Unit IV Functions of Complex Variables

Functions of Complex Variables

This unit is mainly devoted in presenting basic concepts on Complex Numbers, Complex

Analytic Functions, the Cauchy-Riemann Equations, Laplace’s Equations, Elementary Complex

Functions (Exponential Functions, Ttrigonometric Functions and Hyperbolic Functions), Line

Integral in the Complex Plane, Cauchy’s Integral Theorems, Derivatives of Analytic Functions,

Power Sseries, Taylor Series, Laurent Series, Residue Integration and Evaluation of Real

Integrals.

4.1 Definition of Complex Numbers

The concept of complex number basically arises from the need of solving equations that has no

real solutions. Though the Italian mathematician GIROLAMO CARDANO used the idea of

complex numbers for soving cubic equation the term “complex numbers” was introduced by the

German mathematician CARL FRIEDRICH GAUSS.

Definition 4.1 A complex number z is an ordered pair (x, y) of real numbers

x and y, written z = (x, y), x is called the real part and y the imaginary

part of z, usually the real and imaginary parts of the complex number

z = (x, y) are denoted by

x = Re z and y = Im z.

Definition 4.2 Two complex numbers are equal if and only if their

corresponding real and imaginary parts are equal.

Example 4.1 Find the values of and for which the complex numbers .

Solution By definition 4.2

. Therefore, .

Definition 4.3 The complex number (0,1) usually denoted by i = (0,1) is

called imaginary unit

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4.1.1 Addition and Multiplication on Complex Numbers

Definition 4.4 For any two complex numbers and

i) = ii)

Note that: Any real number x can be written as x = (x, 0) and hence the set of complex numbers

extend the reals.

Example 4.2 Let , . Then from definition 4.4 we get: + = and = Furthermore; for any real numbers x and y,

i y = (y, 0) = (0, y) and (x, y) = (x, 0) + (0, y) = x + i y. Conequentely; for any real numbers x and y,

i y = (0, y) and (x, y) = x + i y.

Note that: 1. For any non-zero real number y, z = i y is called pure imaginary number.

2. Any point on the x-axis has coordinates of the form (x, 0) that corresponds to the

complex number x = x + 0 i, due to this reason the x-axis is called the real axis.

3. Any point on the y-axis has coordinates of the form (0, y) that corresponds to the

complex number i y = 0 + i y, and hence it is called the imaginary axis.

4.1.2 Properties of Addition and Multiplication

Let , and be complex numbers. Then

i) + = + and = ii) ( + ) + = + ( + ) and ( ) = ( ) iii) ( + ) = +

iv) 0 + = , + ( ) = 0 and =

Furthermore; for any non-zero complex number z = x + i y, there is a complex number such that

.

The complex number is usually denoted .

Consequentely;

= = = .

Therore, any non-zero complex number z = x + i y has a unique multiplicative inverse given by:

= .



The set of complex numbers form a field. However, it is not possible to define an order relation on the set of

complex numbers. Since the expressions like z > 0, < etc are meaningless unless these complex

numbers are reals.

4.1.3 Complex Plane

The concept of expressing a complex number (x, y) as a point in the coordinate plane was first

introduced by Jean Robert Argand (1768-1822), a swiss bookkeeper. The plane formed by a one

to one correspondence of complex numbers and points on the coordinate plane is called the

Argand diagram, or the complex plane or the z-plane.

In the Argand diagram the x-axis is the real axis and the y-axis is called the imaginary axis

In a complex plane any complex number

z = x + i y is represented as the point z with co-

ordinate x and ordinate y, and we say the point z in

the complex plane.

The sum of two complex numbers can be geometrically interpreted as the sum of two position vectors in the

Argand diagram.

4.1.4 Complex Conjugate

Definition 4.5 Let z = x + i y be a complex number. Then the complex conjugate

of z (or simply the conjugate of z) denoted is defined by = x i y

For any complex number z = x + i y in the complex plane, the complex conjugate of z, = x i y is obtained by reflecting z in the real axis.

Example 4.3 Let z = x + i y be any complex number. Then verify that

i) ii) Re z = (z + ) iii) Im z = (z )

Solutions Using properties of addition and multiplication on complex numbers and definition 4.5 we get:

Prepared by Tekleyohannes Negussie

O

z = 4 + 3i

4

3

x

y

O x

y

+

1z

57


i) = = = .

Therefore, = .

ii) (z + ) = = x = Re z .

Therefore, Re z = (z + ).

iii) = = y = Im z.

Therefore, z = (z ).

Example 4.4 Let and be two complex numbers. Show that:

i) = ii) = iii) , provided that 0.

Solutions Let and . Then From the properties of addition and multiplication on complex numbers and definition 4.5 we get:

i) = = = = = Therefore, = for any two complex numbers and .

ii) = = = = = . Therefore, = for any two complex numbers and .

iii) = = = = = .

Therefore, = , provided that 0.

4.1.5 Polar Form of Complex Numbers

The Cartesian coordinates x and y can be transformed into polar coordinates r and by

x = r cos and y = r sin For any complex number z = x + i y the form

z = r (cos + i sin )

is called the polar form of z, where r is the absolute value or modulus of z. The modulus of z is usually

denoted and defined by



= r = = while is called the argument of z and is denoted and defined by

arg z = = , up to multiples of 2.

The value of that lies in the interval < is called the principal value of the argument of z and

denoted by Arg z.

Note that: the value of , measured in radian, depends on the quadrant in which the complex

number z belongs.

Example 4.5 Write z = 1 + i in polar form.

Solution To write z in polar form first we need to find and Arg z. = (1 + i) (1 i) = 2 and hence =

and = arg z = = where n Z, but z lies in the second quadrant ,

hence, Arg z = .

Therefore, z = .

Example 4.6 Write z = 1 i in polar form.

Solution To write z in polar form first we need to find and Arg z. = (1 i) (1 + i) = 2 and hence =

and = arg z = = where n Z, but z lies in the third quadrant ,

hence, Arg z = .

Therefore, z = .

4.1.6 Important Inequalities

For any two complex numbers and + (Triangle Inequality)To show that this holds true, let = and = .

Then =

= + +

+ +

Therefore, + .



Furthermore; for any finite number of complex numbers , , . . . ,

(Generalized triangle inequality)

Verify! (Hint: use the principle of Mathematical induction on n)

Example 4.7 Let = and = . Find and + .

Solution = = = ,

= = =

and = = = .Therefore, + .

4.1.7 Multiplication and Division in Polar Form

Let = and = .

Multiplication

=

=

Therefore, = and arg ( ) = arg ( ) + arg ( ) up to multiplies of 2.

Division

The quotient is the number z = satisfying z = .

Thus arg (z ) = arg z + arg = arg and = = .

Hence, = and arg ( ) = arg ( ) arg ( ) up to multiplies of 2.

Therefore, = .

Example 4.8 Let = and = . Express and in polar forms.

Solution = = 2 and = = 3

and arg ( ) = = where n Z.

But lies in the quadrant, hence Arg = arg = , where n Z. But

lies in the positive imaginary axis, hence Arg = . Thus = and = .



Therefore, = 6 and = .

4.1.8 Integer powers of Complex Numbers

For any non-zero complex number

for any n Z.

In particular if = 1, then we get the De Moivre formula

for any n Z.

Example 4.9 Use the De moivre formula to show that for any angle

and

Solution If n = 2, then

and from the De Moivre formula we get:

Therefore, and

4.1.9 Roots of Complex Numbers

Suppose Z is a non-zero complex number. Now we need to solve , where n N and n 1.Note that: Each values of is called an root of z, and we write

Let z = and = .

Then = , cos = cos n and sin = sin n .

, , where k Z.

Note that: For any k Z, there exist integers m and h such that

k = m n + h, where h 0, 1, 2, 3, . . . , n 1

Let . Then =

and =

Therefore, , where, k = 0, 1, 2, 3, . . . , n 1.

Note that: These n values lie on a circle of radius with center at the origin and constitute the

vertices of a regular n-gon.

The value of obtained by taking the principal value of arg z is called the principal value of



= .Example 4.1.10 root of unity Solve the equation = 1.

Solution Now = , k = 0, 1, 2, 3, . . . , n 1.

If denotes the value corresponding to k = 1, then the n values of can be written as

1, , , . . .,

Hence let = .

Therefore, 1, , , . . ., are the roots of unity. Example 4.1.11 Solve the equation = 1.

Solution Now = , where k = 0, 1, 2, 3.

Then for k = 1 we get = .

Therefore, 1, ,1 and are the roots of unity. Note that: The n values of are:

, , , . . . ,

where = and is real.

Note that: For any complex number ,

= ,

where .

= .

Therefore, = , where

.Exercise 4.1

1. Write in the form x i y, where = 4 5 i and = 2 + 3 i

i) ii) iii)

2. Find the real and the imaginary parts of i) iii) in exercise 1.

3. Let and be complex numbers, if = 0, then show that either = 0 or = 0.

4. Compute



5. Represent in polar form.

6. Determine the principal value of the argument of

i) ii)

7. Represent each of the following in the form

i) ii)

8. Solve the equation i) ii)

9. For any two complex numbers and show that

+ = (Parallelogram equality)

4.2 Curves and Regions in the Complex Plane

4.2.1 Circles and Disks

The distance between two points z and in the complex plane is denoted by . Hence a circle C of

radius and center can be given by

=

In particular the unit circle with center at the origin is given by = 1

Furthermore; i) < represents an open circular disk.

ii) > represents the exterior of the circle C.

iii) < < represents the open circular ring (open annulus).

Example 11

i) = 9 is a circle of radius 9 centered at 2 +

ii) < 9 is an open circular disk of radius 9 centered at 2 + .

iii) ≤ 9 is a closed circular disk of radius 9 centered at 2 + .

iv) > 9 is the exterior of the circle of radius 9 centered at 2 + .

4.2.2 Half plane

i) (open) upper half - plane =



ii) (open) lower half-plane =

iii) (open) right half plane =

iv) (open) left -half plane =

4.2.2.1 Concepts Related to Sets in the Complex Plane

Now we need to define some important terms.

i) Neighborhoods

A delta, δ neighborhood of a point is the set of all points z such that < δ where δ is any

given positive number. (a deleted δ-neighborhood of is a neighborhood of in which the point

is omitted i.e. 0 < < δ).

ii) Limit points

A point is called a limit point or cluster point or accumulation point of a set S if every deleted δ-

neighborhood of contains points of S. Since δ can be any positive number, it follows that S must

have infinitely many points, and may or may not belongs to the set S.

iii) Closed sets

A set S is said to be closed if every limit point of S belongs to S.

iv) Open sets

A set S is called open if every point of S has a neighborhood consisting of points that entirely belongs

to S.

v) Connected sets

A set S is called connected if any two of its points can be joined by a path consisting of finite line

segments all of whose points belongs to S.

An open connected set is called a domain.

vi) Bounded sets

A set S is called bounded if we can find a constant M such that < M z S. A closed and

bounded set is called a compact set.

vii) Boundary points

A boundary point of a set S is a point such that every neighborhood of which contains both points that

belongs to S and points that do not belong to S.

viii) Region

A region is a set consisting of a domain plus, perhaps some or all of its boundary points.

ix) Complement of a set



A set consisting of all points which do not belong to S is called the complement of S.

4.2.2 Complex Functions; Analytic Functions

Complex Functions (Single and Multi-valued Functions)

Let S be a set of complex numbers. A function f defined on S is a rule that assigns to every z in S a complex

number called the value of f at z. We write

= f (z)

Here z varies in S and is a complex variable. The set S is called the domain of definition of f and the set of

all values of the function f is called the range of f.

Example 12 Let = f (z) = . To each complex number z there is only one value of and hence = f (z) = is a single

valued function of z.

Example 13 Let = f (z) = . To each non-zero complex number z, there are two values of and hence f (z) = is a

multi – valued function of z.

Convention: In this chapter whenever we speak of a function we shall assume single-valued function.

Let = f (z) and . If z = , then = f (z) if and only if = f ( ). Thus

u and v are real valued functions of x and y. Hence

= f (z) = .

Therefore, a complex function f (z) is equivalent to a pair of real valued functions and .

Example 14 Let = f (z) = . Find the functions and for which

f (z) =

and calculate the value of f at .

Solution Let . Then and .

Thus, = and = 2 x y. Hence, = 8 and = 6.

Therefore, .Example 15 Let = f (z) = . Find the functions and for which f (z) =

and calculate the value of f at z = .

Solution Let . Then



Thus, and . Hence, and

.

Therefore, .

4.2.3 Limit and Continuity

Definition 4.6 A function f (z) is said to have a limit ℓ as z approaches , written

if to each positive number ε there corresponds a positive number

δ such that

whenever 0 < < δ.

Note that: 1. By definition z approaches from any direction in the complex plane.

2. If a limit exists, then it is unique.

3. if and only if and .

.

Definition 4.7 A function f (z) is said to be continuous at z = if f ( ) is defined and

Note that: If , then f (z) is defined in some neighborhood of . Furthermore;

f (z) is said to be continuous in a domain if it is continuous at each point of its domain.

4.2.4 Derivative

Definition 4.8 The derivative of a complex function f at a point written

is defined by

=

Provided that this limit exists. In this case, we say f is differentiable at .

If we write , we also have

= .



Example 16 Show that f (z) = is differentiable for all complex numbers and = 2z.Solution Let be any complex number.

= = .

Therefore, for any complex number z, = 2z. 4.2.4.1 Differentiation Rules

If f and g are differentiable functions, then

i)

ii)

iii)

iv) , provided that g (z) 0.

v) .

vi) for any n N.

Note that: If f (z) is differentiable at , then f (z) is continuous at .

Let f (z) = . If f (z) is differentiable at . Then the partial derivatives of u and v at

exist.

Example 17 Show that f (z) = is not differentiable.

Solution Let be any complex number.

Now consider = = .

i) Suppose = 0, then = 1.

ii) Suppose = 0, then = 1.

Therefore, does not exist, and hence f (z) = is not differentiable.

4.2.4 Analytic Functions

Definition 4.9 (Analyticity)

A function f (z) is said to be analytic in a domain D if f (z) is differentiable at



all points of D. The function f (z) is said to be analytic at a point z = in D

if f (z) is analytic in a neighborhood of , including the point .

By analytic function we mean a function that is analytic in some domain.

A more modern term for analytic in D is holomorphic in D.

Example 18 Polynomial functions in a complex variable z are defined by

(*)

where , , , . . ., are complex constant and nN, and n is called the degree of

the polynomial.

f (z) given by (*) is analytic in the entire complex plane.

The quotient of two polynomials g (z) and h (z)

is called a rational function, and f is analytic except at those points where h (z) = 0.

Exercise 4.2

1. Find the real and imaginary parts of , where .

2. Determine whether f (z) is continuous at the origin or not, where

a) b)

3. Differentiate a) b)

4. Show that f (z) = Re z is not differentiable at any z.

5. Show that f (z) = is differentiable only at z = 0; hence it is nowhere analytic. (Hint: use = ).

4.3 The Cauchy - Riemann Equations

Let = f (z) = . Then roughly f is analytic in a domain D if and only if the first

partial derivatives of u and v satisfy the two equations

and (*)

everywhere in D.

The equations in (*) are called the Cauchy - Riemann equations.



Theorem 4.1 (Cauchy Riemann Equations)

Let f (z) = be defined and continuous in some

neighborhood of a point and differentiable at itself.

Then at , the first-order partial derivatives of u and v exist and satisfy

the Cauchy - Riemann equations.

Hence if f (z) is analytic in a domain D, those partial derivatives exist

and satisfy (*) at all points of D.

Proof Now we can approach along any curve that passes through .

i) Suppose we approach along any curve for which x = .

Then +

Therefore, = .

ii) Suppose we approach along any curve for which y = .

Then +

Thus, = .

Therefore, and .Therefore, if f (z) is analytic in a domain D, then it satisfies the Cauchy Riemann equations.

Example 19 Let = . Then and .

Now = 2x + 3 = and .

Therefore, f (z) satisfy the Cauchy Riemann equations.

Example 20 Let = . Then and .

Now = 1, = 1 and . Since , the Cauchy Riemann equations all not satisfied

Therefore, is not analytic.

Theorem 4.2 (Cauchy - Riemann Equations)

If two real -valued continuous functions and of two

real variables x and y have continuous first partial derivatives that

satisfies the Cauchy-Riemann equations in some domain D, then the

complex function is analytic in D.

Remark: The Cauchy-Riemann equations are not only necessary but also sufficient for a function



to be analytic.

Example 21 Find the most analytic function f (z) whose real part is = .

Solution f (z) is analytic implies the Cauchy- Riemann equations are satisfied.

Thus = = and hence =

and = = = . Thus, = 0 h (x) = k, constant.

Therefore, f (z) = = , in terms of z.

Example 22 If f (z) is analytic in D and = k constant in D, then f (z) is constant in D. Solution Let . Then = = .

Thus, = 0 = 0 and = 0 = 0.

Now f (z) is analytic in D implies = and = .

Hence

and .

Thus and .

i) If , then = 0 and hence . ii) If , then . Thus and hence and are constants.

Therefore, is constant.

Remark: If we use the polar form Z = r (cos + i sin ) and set f (z) = u (r, ) + i v (r, ), then

the Cauchy-Riemann equations are:

.

(Verify!)

Laplace's Equations, Harmonic Function

Theorem 4.3 (Laplace's Equation)

If is analytic in a domain D, then u and v satisfy the Lap lace's equations. i.e



respectively, in D and have continuous second partial derivatives in D.

is called the Laplacian operator and read as “nabla squared”. Proof Since f (z) is analytic in D, and is analytic in D.

Hence, and . Therefore, .Solutions of the Lap lace's equations having continuous second order partial derivatives are called

harmonic functions and their theory is called potential theory.

Remarks:

1. The real and imaginary parts of analytic functions are harmonic functions.

2. If two harmonic functions u and v satisfy the Cauchy-Riemann Equations in a domain D,

then they are the real and imaginary parts of an analytic function f in D. v is said to be

a conjugate harmonic function of u in D.

Example 23 Verify that is harmonic in the whole complex plane and find

a conjugate harmonic function v of u.

Solution Now and .

Thus and hence is a harmonic function.

Now , since v satisfies the Cauchy-Riemann Equation.

Then and hence h (x) = x + c.

Thus, .

Therefore, = .

Exercise 4.2

1. Determine whether the following functions are analytic or not.

a) b)

c) .

2. Are the following functions harmonic? If so, find a corresponding analytic function

.

a) b) c)

3. Determine a, b and c such that the given functions are harmonic and find the conjugate harmonic

functions.



a) b) c)

4. Show that if u is harmonic and v a conjugate harmonic of u , then u is the conjugate harmonic

of v.

5. Show that if f (z) is analytic and Re f (z) is constant, then f (z) is constant.

4.3 Exponential Functions

The exponential function , also written exp z, is defined by:

(*)Note that: The definition of is a natural extension of the real exponential function . i.e. i) If z = x, then = .

ii) is an entire function, i.e. an analytic function and = .

Now let and be two complex numbers. Then

In particular if = x and , then

, 0, , where n Z and hence z in Cand if x = 0, then from (*) we get:

, since

called the Euler formula .

Moreover; the polar form of a complex number can be written as

4.3.1 Periodicity of

for all z.

Hence, is periodic with pure imaginary period . Thus, all the values that w = can assume are

already in the horizontal strip of width 2.

This infinite strip is called a fundamental region of .

Example 24 Find all solutions of .

Solution .

Then = 2 cos y = and = 2 sin y = .



Hence, cos y = and sin y = , where n Z.

Therefore, z = , where n Z.

Example 25 Find all solutions of .

Solution and hence .

Then and sin 3y = 0 if and only if , where n Z.

Therefore, Z = , where n Z.

Exercise 4.3 Find all values of k such that is analytic.

4.4 Trigonometric and Hyperbolic Functions

4.4.1 Trigonometric Functions

Consider the Euler formulas

. (*)From (*) we get:

(**)

This suggests the following definitions.

Definition 4.10 For any complex number

and

Furthermore; , ,

and .

Note that: i) cos z and sin z are entire functions.

ii) tan z and sec z are analytic except where cos Z = 0

iii) cot z and csc z are analytic except where sin z = 0.

Remark: The following are immediate consequences of definition 4.10

, and .

Furthermore, the Euler's formula is valid, i.e



4.4.2 The Real and Imaginary Parts of sin z and Cos z

Example 26 For any complex number , show that

a) and

b) and

solution By definition:

a) = =

=

= = .

Therefore, = .

= =

=

= = .

Therefore, = .

b) From the result of part a) we get:

=

= = .

Therefore,

=

= =

Therefore,

Note that: i) sin z and cos z are periodic functions with period 2.

ii) tan z and cot z are periodic functions with periodic .

iii) the complex sine and cosine functions are not bounded.



Definition.4.11 A number is a zero of a complex valued function f (z) if f ( ) = 0.

Example 27 Find the zeros of sin z and cos z.

Solutions Let . Then

and .

Hence, if and only if and .

and and or .

and and .

Therefore, if and only if .

and

sin x = 0 and sinh y = 0 and y = 0, where n Z.

Therefore, sin z = 0 if and only if , where n Z.

Example 28 Solve cos z = 2.

Solution . Then cos z = 2 if and only if

Now

and and

and and .

Therefore, cos z = 2 if and only if z = , where n Z.

Exercises 4.4 For any complex numbers z, and , show that:

i) = iii) = 1

ii) = iv) =

4.5 Hyperbolic Functions

The Complex hyperbolic cosine and sine functions are defined by:

and

These functions are entire with derivatives. and .The other hyperbolic functions are defined by



, , and .

Remark: If in the definition of and replacing z by iz we get:

= and = .

similarly if we replace z by iz in the definition of and we get:

= and = .

Example 29 For any complex number , show that:

= and = Solutions Let .

Then =

=

= Therefore, = .

Similarly =

=

= Therefore, = .Exercise 4.5

1. Compute in the form and if z equals

a) b) c) d)

2. Find the real and the imaginary parts of a) b) c)

3. Write in polar form a) b) c)

4. Find all solutions of a) b) c)

5. Show that is harmonic and find its conjugate.

6. Show that are harmonic. 7. Find all solutions of the following equations

a) b) c)

8. Find all values of z for which a) and b) have real values.

9. Find .

10. Show that , and conclude that the

complex cosine and sine are not bounded in the whole complex plane.

4.6 Logarithm, General power



The natural logarithm of is denoted by and is defined by

if and only if .

Suppose that and , then

= = r and =

and Z.

Therefore, + , where n Z.

Now let . Then is called the principle value of and

, where n Z.

Note that: 1. The complex natural logarithm is infinitely many valued function having the

same real part but differing in their imaginary parts by an integral multiple of 2.

2. is single valued.

Example 30 Given with y = 0. Then find .

Solution Consider the following cases.

Case i) x 0 . Thus, = , where n Z. ii) x 0 . Thus, = , where n Z.

Example 31 Find and , where .

Solution and .

Therefore, = and = , where n Z.

Properties

In general for any non-zero complex numbers and

a) = + b) =

Note that: a) and b) need not hold true if we replace by .

Example 32 Let = = = 1.

Then = , while =

But = and + = .

Let and be non-zero complex numbers. Then



= , n Z.

Now consider where z is any non-zero complex number.

= , where n Z. = = = .Therefore, = z.But note that need not equal to z. = + , where n Z. = + = = , where n Z.Therefore, = , where n Z.

Example 33 For a fixed n and a complex number z, z is not a negative real and a non-zero complex number

and , show that = .

Solution Let = , where = and = ,

where n Z.

Then = and .

Consequently from theorem 4.1 we get:

= = = = .

Therefore, = .

General Power

General powers of a complex number are defined by

, where c is any non-zero complex number.

Since is infinitely many-valued, will, in general, multi-valued. The particular value is called the principal value of .

Remark: i) If c and c 0, then is single valued and is identical with the

integer power of z.

ii) If c = , where n N and n 1, then , where z 0, is the

root of z has n distinct values.

iii) If c = the quotient of two positive integers, then , where z 0



has only finitely many distinct values. iv) If c is irrational or genuinely complex (i.e. , b 0), then is infinitely many

valued function.

Example 34 Find the principle value of

i) ii)

Solutions i) = = = , where n N

Therefore, the principle value, n = 0, is .

ii) = , where n N

=

=

=

= .

Therefore, the principle value is .

Exercises1. Compute the principle value if z equals

i) ii) 2. Solve for z:

i) ii)

3. Find the principle value of i) ii) iii)

4.3 Complex Integration

In a complex definite integral, line integral, we integrate along a curve C in the complex plane called the

path of integration. Such a curve can be represented (parameterized) in the form

for a t b

Example 35 Find a parameterization of the line segment with endpoints

a) and b) and

Solution a) x (t) = 4 t and y (t) = 2 + 3 t

Therefore, , where 0 t 1.

b) x (t) = 7 t and y (t) =

Therefore, , where 0 t 1.



Definition 4.12 A curve C is called a smooth curve if it has a continuous and

non-zero derivative

at each point t.

Geometrically

The arrow on the curve indicates the sense of increasing t.

The Complex Line Integral

Let C be a smooth curve in the complex plane represented by

for a t b,

let f (z) be a continuous function defined at least at each point on C and let = be a

partition of , where a = .

Suppose to the subdivisions of there correspond a subdivision of C by points , where

For each m = 1, 2, 3, ….., n, let be a point on C between and , where

for some t .

Then consider the sum

, where .

as the number of partition of increases the max tends to zero and hence max tends to zero.

In this case = exists and we define this limit as the line integral of f (z) over

the oriented curve C and is denoted by


Oz(t + t))z(t)

Z (t)

O

CΔt

0z

1z 2z nz

80


If C is a closed curve for the line integral of f (z) over C we also use the notation

General assumption

All paths of integration for complex line integrals are piecewise smooth.

Theorem 4.4 (Existence of complex line integral)

If f (z) is a continuous function on a piecewise smooth curve C, then

the line integral.

exists.

Proof: Let , let and .

Then =

= +

Since f (z) is continuous, u and v are continuous. Further more u and v are real valued function implies that

as n tends to infinity the line integral over C for the real functions exist.

Hence, = + exists.

Properties of complex line integrals

1. Complex integration is a linear operation.

Let and be complex constants, and and complex continuous function on C, then

2. Decomposing C into two portions and we get:

3. Reversing the sense of integration, we get the negative of the original value.

.

Integration Methods



First method: Use of representation of a path.

Theorem 4.5 (Integration by the use of the path)

Let C be a piecewise smooth path represented by ,

where a t b and let f (z) be a continuous function on C.

Then

Example 36 Integrate on the standard unit circle oriented counter clockwise.

Solution Using Euler’s formula C: , where 0 t 2 , is the parameterization of the unit Circle oriented counterclockwise. Furthermore; and .

Thus, .

Therefore, .

Example 37 Let where m Z and is a constant. Find , where C is

a circle of radius ρ centered at and oriented counterclockwise.Solution , where 0 ≤ t ≤ 2.

Then , where 0 ≤ t ≤ 2 and hence and .

Thus, = =

If m = 1, then = 1, and = 2 and = 0.

If m 1, then = 0 = .

Therefore, = and m Z.

Remark: A complex line integral depends not only on the endpoints of the path but in general also on

the path itself.

Example 38 Let . Find along

a) b) along C consisting of and as shown below.Solutions a) : , where 0 t 1

Then and .

Prepared by Tekleyohannes Negussiex

y

1C

z = 1 + i

82


Hence =

.

b) : , where 0 t 1. Then and

and : , where 0 t 1. Then and .

Hence .

This shows that complex line integral is path dependent.Second Method: Indefinite integral

In real integration if is continuous on and if is an antiderivative of on , then

Now we need to extend this to complex integration.

Definition 4.13

1. A simple closed path is a closed path that does not intersect or touch itself.

2. A simply connected domain D in the complex plane is a domain such that

simple closed path in D encloses only points of D. A domain that is not

simply connected is called multiply connected.

Theorem 4.6 (Existence of Indefinite integral of analytic functions)

If f (z) is analytic in a simply connected domain D, then there exists an indefinite

integral F (z) of f (z) in D, which is analytic in D, and for all paths in D joining

any two points and in D, the integral of f (z) from to can be evaluated by:

= F ( ) F ( )

Example 39 Evaluate .

Solution = = .



Therefore, = .


Solution = = = .

Therefore, = .


Solution = = = = 0.

Therefore, = 0.


Solution = = .

Therefore, = .

Note that: Simply connectedness in the above theorem is essential.

Example 43 , where C is oriented counterclockwise over the unit circle.

Although is analytic in the annulus < < , this domain is not simply connected.

Bound for the Absolute Value of Integrals

Let C be a curve and f (z) be a continuous function on C. Then

≤ M L MLinequality

where L is the length of C and M is a constant for which ≤ M on C.To show that this holds true, consider the partial sum . By the generalized triangle inequality

= ≤ ≤ = ML.

Therefore, ≤ M L.



Example 43 Find an upper bound for the absolute value of the integral , where C is the line

segment from the origin to .

Solution L = 5 and ≤ . Hence, ≤ 5 .

Therefore, ≤ 5 .

Exercises

1. What curves are represented by the following functions?

i) ii)

iii) iv)

v)

2. Represent the following curves in the form .

i) = 4 ii) from (1, 1) to

iii) + = 36

Cauchy’s Integral Theorem

Theorem 4.6 (Cauchy’s Integral Theorem)

If f (z) is analytic in a simply connected domain D, then for every simple

closed path C in D,

Example 44 For any closed path C, , , and

where n W . Since these functions are entire.

Example 45 and , where C is the standard unit circle.

Now sec z is not analytic at , n Z. But these points are outside of the circle.

Similarly for the second integral, whose integrand is not analytic at outside C.

Example 46 , where C is the unit circle oriented counterclockwise.

Since is not analytic and .



Example 47 , where C is the unit circle. Now = = 0.

This result does not follow from the Cauchy’s theorem, since is not analytic at z = 0.

Note that: The condition that f (z) is analytic in D is sufficient rather than necessary for

to hold.

Example 48 Evaluate , where C is the unit circle.

Solution By using partial fraction we get:

= + .

Then = + = = .

(Since f (z) = is not analytic at z = 2 outside of C.)

Therefore, = .

Independence of Path

An integral of f (z) is said to be independent of path in a domain D if for every path from to in D

its value depends only on and but not on the choice of the path C in D.

Theorem 4.7 (Independence of path)

If f (z) is analytic in a simply connected domain D, then the integral

of f (z) is independent of path in D.

Cauchy’s Theorem for Multiply Connected Domain

We first explain this for a doubly connected domain D with outer boundary curve and inner boundary curve , refer the figure below. If f (z) is analytic in any domain that contains D as well as its boundary curves, then

=


1D

1C

2C

2D

86


Both integrals being taken counterclockwise (or both clockwise and regardless of whether

or not the full interior of belongs to ).Note that: For domains of higher connectivity the idea remains the same.

Example 49 Let C be any counterclockwise oriented simple closed path containing and m an integer, then

=

Example 50 Evaluate , where C is as shown in the figure.

Solution By using partial fraction we get:

= .

Now let and be the portion of the curve C to the

left of the origin and the right of the origin respectively.

Then = = 0 and = = .

Thus, = + = .

Therefore, = .

Existence of Indefinite Integral

If f (z) is analytic in a simply connected domain D, then by theorem 4.6 there exists an indefinite integral

F (z) of f (z) in D, which is analytic in D, and for all paths in D joining any two points and in D, the integral of f (z) from to can be evaluated by:

= F ( ) F ( )


Solution f (z) = cos z is entire and = cos z. Then

= = .

Therefore, = .




Solution f (z) = is entire and = .

Thus, = = .

Therefore, = .

Cauchy’s Integral Formula

Theorem 4.8 (Cauchy’s Integral Formula)

Let f (z) be analytic in a simply connected domain D. Then for any point

in D and any simple closed path C in D that encloses

=

the integration being taken counterclockwise.

Example 53 Evaluate , where C is any closed path enclosing = 2.

Solution = 2 and is entire. Hence by Cauchy’s integral formula, for any closed path enclosing = 2 we get:

= , since .

Therefore, = .

Example 54 Evaluate , where C is any simple closed path.

Solution There are two cases to consider

Case I C encloses

Then = . Now is entire and the region

enclosed by C is a simply connected domain and .

Therefore, = .

Case II C does not enclose



Now is analytic in the simply connected domain enclosed by C.

Therefore, by Cauchy’s integral theorem

= 0.

Example 55 Evaluate , where C is a unit circle oriented counterclockwise centered at:

i) = 1 ii) iii) iv)

Solution In i) and ii) the circle C encloses the point = 1 and does not enclose = 1.

Now let , then is analytic in the region enclosed by C, and f (1) = 1.

Therefore, = .

In iii) the circle C encloses the point = 1 but not = 1.

Now let , then is analytic in the region enclosed by C, and f (1) = 1.

Therefore, = .

In iv) the circle C does not enclose the points = 1 and = 1.

Hence let . Then is analytic in the region enclosed by C.

Therefore, = 0.

Example 56 Evaluate , where C is the circle .

Solution f (z) = tan z is not analytic at , where n Z. But all these points lie outside

of C. Hence f (z) is analytic in the region enclosed by C.

Now using partial fraction we get:

.

Thus, = = .

Therefore, = .

Remark: (Multiply Connected Domains)

If f (z) is analytic on and , and in the ring-shaped domain bounded by and , and is any point in the domain of f, then



where the outer integral (over ) is taken

counterclockwise and the inner clockwise

as shown in the figure.

Exercises

1. Integrate counterclockwise around the circle

i) ii) iii) iv) 2. Integrate the following functions counterclockwise around the unit circle

i) ii) iii)

3. Integrate the following functions over the given contour C.

i) , where C is the boundary of the triangle with vertices 1, 1 and .

ii) , where C consists of the boundary of the triangle with vertices ,

and .

iii) , where C consists of the boundary of the square with vertices

a) 3 and 3 b) 1 and 4. Integrate the following functions counterclockwise around the unit circle

i) ii) iii)

5. Integrate , where

i) , C is an ellipse with foci

ii) , C is the circle

iii) , C consists of the boundary of

a) the square with vertices 3 and 3 oriented counterclockwise b) oriented clockwise.

Derivatives of Analytic Functions

Theorem 4.9 (Derivatives of Analytic Functions)


1C2C

0z

90


If f (z) is analytic in a domain D, then it has derivatives of all orders in D,

which are then also analytic functions in D. The values of these derivatives

at a point in D are given by the formula

, where n N

here C is any simple closed path in D that encloses and whose full

interior belongs to D; and we integrate counterclockwise around C.

Example 57 Evaluate , where C is any simple closed path oriented counterclockwise

that encloses = .Solution is entire and .

Thus, =

= = .

Therefore, = .

Example 58 Evaluate , where C is any contour enclosing = oriented

counterclockwise. .Solution Let and = . Then is entire with and hence .

Thus, =

Therefore, = .

Example 59 Evaluate , where C is any contour enclosing 1 but not

oriented counterclockwise.

Solution Let and = 1. Then f (z) is analytic in the domain enclosed by C and

= .

Thus, = .



Therefore, = .

Theorem 4. 10 (Morera’s Theorem)

If f (z) is continuous in a simply connected domain D and if

= 0

for every closed path C in D, then f (z) is analytic in D.

Cauchy’s Inequality

Let f (z) be analytic in a domain D and let C be a circle of radius r and center in D. Suppose

≤ M on C. Then applying the ML-inequality

≤ .

This inequality is called Cauchy’s Inequality.

Theorem 4. 11 (Liouville’s Theorem)

If an entire function f (z) is bounded in absolute value for all z,

then f (z) must be constant.

Proof: f (z) is bounded implies there exists a constant k for which ≤ k for all z.

Let be any point. Then by Cauchy’s inequality

<

for a circle or radius any r with center , since f (z) is entire for all r.

Hence, = 0 and is arbitrary implies = 0 for all z.

Therefore, f (z) is constant.

Power Series; Taylor Series; Laurent Series

Sequence and Series, Convergence Tests

Sequences

A sequence is obtained by assigning to each positive integer n a number called a term of the sequence, and is written , , , . . . or { , , , . . . } or simply { }.



A real sequence is one whose terms are real.

Convergence

A convergent sequence { } is one that has a limit C, written

= C or → C.

By definition of limit this means that for every ε 0 we can find an N such that < ε for all n N.A divergence sequence is one that does not converge.

Example 60 The sequence

i) converges to 0. ii) is divergent. iii) is divergent.

iv) with = converges to .

Theorem 4.12 (Sequences of real and imaginary parts)

A sequence of complex numbers , where n N,

converges to if and only if the sequence of the real parts

converges to a and the sequence of the imaginary parts

converges to b.

Proof: If < ε, then is within the circle of radius ε about for all n N.

Therefore, < ε and < ε for all n N.

Suppose converges to a and converges to b.

Then for all ε 0 there exists an N such that

and for all n N.

Thus, = ≤ . Therefore, < ε for all n N. Series

Given a sequence we may form the sequence of sums

, , , . . . , , . . .called the sequence of partial sums of the (infinite) series

= (*)

, , , . . . are called terms of the series.



A convergent series is one whose sequence of partial sum converges. If = S is called

the sum or value of the series and we write

S = (1)

A divergent series is one that does not converge.

If we omit the terms of the series , there remains

.This is called the remainder of the series (1) after . If (*) converges and has sum S, then

and hence .Now = S implies = 0.

When S is known and we compute an approximation by S, then is the error.

Theorem 4.13 (Real and Imaginary Parts)

A series (*) with converges with sum

if and only if converges to and converges to .

Note that: = 0 is a necessary condition but not sufficient.

Example 61 Let = . Then , but diverges, since it is a harmonic series.

Tests for Convergence and Divergence of Series

Theorem 4.14 (Divergence)

If a series converges, then = 0.

Theorem 4.15 (Cauchy’s Convergence Principle for Series)

A series is convergent if and only if for every ε 0 (no matter

how small) we can find an N (which depends on ε, in general) such that

< ε for every n N and p N.

Or equivalently, < ε for every m, n N.

Definition 4.14 A series is called absolutely convergent if the series of the



absolute values of the terms is convergent.

If converges but diverges, then is called conditionally convergent.

Example 62 The series is conditionally convergent.

Theorem 4.16 If a series is absolutely convergent, then it is convergent.

Theorem 4.17 (Comparison Theorem)

If a series is given and we can find a convergent series with

non-negative real terms such that ≤ for m = 1, 2, 3, …, then the given series converges, even absolutely.

Theorem 4.18(Geometric series)

The geometric series converges with the sum

if < 1 and diverges if ≥ 1.

Ratio Test

Theorem 4.19 (Ratio Test)

If a series with ≠ 0 ( n = 1, 2, 3, . . . ) has the property that for

every n greater than some N,

≤ q < 1 for n N

(where q < 1 is fixed), this series converges absolutely. If for every n N,

≥ 1 for n N

the series diverges.



Theorem 4.20 (Ratio Test)

If a series with ≠ 0 ( m = 1, 2, 3, . . . ) is such that

= L

then we have the following

i) If L < 1, then the series converges absolutely

ii) If L 1, it diverges

iii) If L = 1, the test fails; that is we can not draw any conclusion

Example 63 Is the series convergent or divergent?

Solution Using ratio test with , we get:

= 0.

Therefore, is convergent.

Example 64 Is the series convergent or divergent?

Solution Using ratio test with = , we get:

= = < 1.

Therefore, is convergent.

Root Test

Theorem 4.21 (Root Test)

If a series is such that for every n greater than some N, ≤ q < 1 for n N

(where q < 1 is fixed), this series converges absolutely. If for infinitely many n, ≥ 1, then the series diverges.

Caution ! ≤ q < 1 implies < 1, but this does not imply convergence.

Example 65 Consider . = and hence < 1 for all n 1.But the series diverges.



Theorem 4.22 (Root Test)

If a series is such that = L, then we have the following

i) If L < 1, then the series converges absolutely.

ii) If L 1, it diverges.

iii) If L = 1, the test fails; that is we can not draw any conclusion.

Example 66 Consider the series and . = and = .

Hence < 1 and < 1for all n 1. But diverges, while converges.

Example 67 Is the series convergent?

Solution and . Then

and hence = 1.

Therefore, diverges.

Power Series

Definition 4.15 A power series in powers of is a series of the form

where z is a variable , , , . . . are constants, called coefficients of the series and is a constant, called the center of the series.

If = 0, then we obtain a power series in powers of z:

.

Convergence of a Power Series

A power series in powers of may converge

i) throughout the disk with center

ii) in the whole complex plane



iii) merely at the center

Consider the following examples.

Example 68 Show that the geometric series converges absolutely for < 1 and diverges

if ≥ 1.Solution Using the ratio test we get:

Hence, converges absolutely for < 1 and diverges if 1.

If = 1, then diverges.

Therefore, converges if and only if < 1.

Example 69 Show that converges on the whole complex plane.

Solution Using the ratio test we get:

= 0.

Therefore, converges on the whole complex plane.

Example 70 Show that converges only at z = 0.

Solution Using the ratio test we get:

= .

Therefore, converges only at z = 0.

Note that: Series such as are useless series.

Theorem 4.23 (Convergence of a Power Series)

If a power series

i) converges at a point z = ≠ , then it converges absolutely for

every z closer to than .

ii) diverges at a point z = , then it diverges for every z farther away

from than .



Radius of Convergence of a Power Series

Suppose converges for some ≠ but does not converge on the whole complex plane.

Let S =

Now consider the set A = .A is a bounded set, hence A has a least upper bound. Let R = ℓub A. Then

a) The series converges for all z for which < R.

b) The series diverges for all z for which R.

Definition 4.16 The number R is called the radius of convergence of the series

and the circle = R is called the circle of

convergence.

Remarks: 1. We extend R by writing

R = ∞ if the series converges for all z.

R = 0 if the series converges only at the center z = .

2. No general statement can be made about the convergence of the power series

on the circle of convergence itself.

Example 71 On the circle of convergence = 1.

a) converges every where, since converges.

b) converges at 1 but diverges at 1.

c) diverges every where.

Theorem 4.24 (Radius of Convergence R)


0z1z

x

diverges Conv.

99

y


Given the power series (with radius of convergence R)

and suppose

= L* (Cauchy-Hadamard formula)

Then,

a) if L* = 0, then R = ∞. b) if L* = ∞, then R = 0.

c) if L* ≠ 0 and L* , then R = .

Remark: If ≠ 0 and L* , then from theorem 4.24 it follows that

R =

Example 72 Determine the radius of convergence of the power series .

Solution Now . Then

R = =

= = = .

Therefore, the series converges in the open disk < , of radius and center .

Functions Given by Power Series

Consider the power series in z. i.e.

(1)

Note that: For a power series in powers of with any center we can always set = z to reduce it to the above form.

If (1) has a non-zero radius of convergence R its sum is a function of z, say f (z). Then we write

, where < R. (2)

We say that f (z) is represented by the power series or that f (z) is developed in the power series.

Example 73 For < 1, .



Therefore, represents the function for < 1.

Theorem 4.25 (Continuity of the Sum of a Power Series)

The function f (z) in (2) with R 0 is continuous at z = 0.

Proof: We must show that . i.e. we want to show that for all ε 0 there exists a

0 such that < implies < ε .

Now converges absolutely for ≤ r < R, hence

= converges.

Let S = . Then for 0 < ≤ r

= ≤ ≤ = < ε .

Thus, < ε if and only if < . Now choose < min. .

Therefore, the theorem holds.

Theorem 4.26 (Identity Theorem for Power Series)

Suppose that the power series

and

both converge for < R, where R is positive and have the same sum for all < R , then these series are identical. i.e = for all non-negative integer n.

Proof: (Use induction on n and continuity at z = 0)

Remark: If f (z) = for all < R, then the function f (z) can not be

represented by two different power series with the same center.

Power Series Representation of Analytic Functions

Term wise Addition and Subtraction

Let and be two power series with radius of convergence and respectively.

Then converges for a radius of convergence at least equal to the minimum of

and .



Term wise Multiplication

Let and . Then

=

and is called the Cauchy product.

This power series converges absolutely within the common circle of convergence of the two given series

and has sum

Term wise Differentiation and Integration

Given the power series

(1)

The power series

obtained from (1) by term wise differentiation is called the derived series of (1).

Theorem 4.27 (Term wise differentiation of a power series)

The derived series of a power series has the same radius of convergence

as that of the original.

Example 74 Find the radius of convergence R of .

Solution = = = .

Let and .

Then and . Thus,

.

Therefore, the radius of convergence R of is 1.

Theorem 4.28 (Term wise integration of a power series)

The power series , obtained by integrating the

series term by term has the same radius of convergence



as that of the original.

Theorem 4.29 (Analytic Functions and Their Derivatives) A power series with a non-zero radius of convergence R represents an analytic

function at every point interior to its circle of convergence. The derivatives of

this function are obtained by differentiating the original series term by term. All

the series thus obtained have the same radius of convergence as that of the original

series. Hence, by the first statement, each of them represents an analytic function.

Taylor Series

Let f (z) be analytic in a neighborhood of , then f (z) can be represented by

(1)

called the Taylor’s formula, is called the Taylor remainder.

If we let n approach infinity, we obtain the power series

(2)

The series in (2) is called the Taylor series of f (z) with center z = . If = 0, then we get the Maclaurin series

Theorem 4.30 (Taylor’s Theorem)

Let f (z) be analytic in a domain D and let z = be any point in D. Then there

exists precisely one power series with center that represents f (z). This series

is of the form

, where (3)

This representation is valid in the largest open disk with center in which f (z) is analytic.

Remark: In (3) can be given by

where we integrate counterclockwise around a simple closed path in D that encloses .



In (1) is given by

=

where z* is on C and z is inside C.

Definition 4.17 Singular Points

z = is called a singular point of an analytic function f (z) if f (z) is not

differentiable at z = , but every disk with center contains points at

which f (z) is differentiable.

Example 75 Let f (z) = tan z.

Then f (z) is not analytic at z = , where n Z.

Therefore, f has singularities at z = , where n Z.

Important special Taylor Series

Example 76 Geometric series

Let . Find the Maclaurin series of f (z).

Solution and . Thus = 1.

Therefore, , where and f (z) is singular at z = 1; this point lies on

the circle of convergence.

Example 77 Exponential Function

Let . Find the Maclaurin series of .

Solution and . Thus .

Therefore, = . (*)

If we set in (*) we obtain:

and recall that

Thus, = .

Therefore, . (Euler’s Formula)

Example 78 Trigonometric and Hyperbolic Functions

The Maclaurin series of sin z, cos z, sinh z and cosh z are given by:



= =

.

Therefore, = .

Similarly, = , = and = .

Example 79 Logarithm

The Maclaurin series of

(

Replacing z by z and multiplying both sides by 1 we get:

(

and adding these yields:

(

Theorem 4.31 Every power series with a non-zero radius of convergence is

the Taylor series of the function represented by that power series.

(i.e. the Taylor series of the sum)

Note that: There are real valued functions that have derivatives of all orders but can not be

represented by a power series.

Example 80 Let . If x ≠ 0 and f (0) = 0, then this function can not be represented by

a Maclaurin series since all its derivatives at zero are zero.

Power Series: Practical Methods

Methods for determining the coefficients of the Taylor series

Example 81 Substitution

Determine the Maclurian series of for < 1.

Solution Recall that: = for < 1.

Then substituting instead of z we get:

= .

Therefore, = . (*)



Example 82 Integration

Determine the Maclurian series of for < 1.

Solution . Integrating (*) term by term and using f (0) = 0 we get:

= for < 1.

Therefore, = for < 1.

This value represents the principal value of = defined as the value for which

.

Example 83 Developing by using the trigonometric series

Develop in powers of z a, where c ab ≠ 0 and b ≠ 0.

Solution Now = =

= = .

Therefore, = for < .

Example 84 Binomial Series, reduction by partial fraction

Determine the Taylor series of

with center

i) = 0 ii) = 1

Recall that: where s is any number and

= for n ≥ 0 and = 1

called the binomial coefficient, is called the Binomial series.

Solution Using partial fraction reduction

Now by the Binomial series

= .



Thus, converges for < 1.

i) = 0

= for < 2.

= = for < 3.

Therefore, for < 2.

ii) Similarly, f (z) can be written as:

.

Thus, for < 2.

Example 85 Use of Differential Equations

Determine the Maclaurin series of Solution Now and f (0) = 0. Thus, and .

Then ; ,

; ,

; ,

; , . . .


Example 86 Undetermined Coefficients

Determine the Maclaurin series of by using those of and .

Solution Since is odd the desired expansion will be = Now using = , we obtain

=

Since is analytic except at where n Z, its Maclaurin series converges

for < .



Now applying Cauchy’s product and comparing coefficients we get:

= 1, , , etc.

Hence, = 1, , , etc.


Exercises

1. Determine the center and radius of convergence of

i) ii) iii)

iv) v) vi)

vii) viii)

2. Determine the Taylor series of the given functions with the given point as center and determine the radius of convergence.

i) ; 0 ii) iii) iv)

v) vi) vii) viii)

3. Determine the Maclaurin series of the following functions and determine the radius of convergence.

i) ii) iii) iv)

v) vi) vii)

4. Determine the Taylor series of the given functions with the given point as center and determine the

radius of convergence.

i) ; 1 ii) ; iii)

iv) v) vi)

vii) viii) ix)

Laurent Series

In this section we will consider an expansion of a function f (z) around points at which it is not analytic, but

is singular. Then the Taylor series no longer applies, but we need a new type of series, known as Laurent

series, consisting of positive and negative integer powers of and being convergent in some annulus

(bounded by two circles with center at ) in which f (z) is analytic.



Theorem 4.32 (Laurent’s series)

If f (z) is analytic on two concentric circles and with center and in the annulus between them, then f (z) can be represented by the Laurent series

+

(1)The coefficients of this Laurent series are given by the integrals

and

(2)

each integral being taken counterclockwise around any simple closed path C that

lies in the annulus and encloses the inner circle.

This series converges and represents f (z) in the open annulus obtained from the

given annulus by continuously increasing the outer circle and decreasing the

inner circle until each of the two circles reaches a point where f (z) is singular.

In the important special case that is the only singular point of f (z) inside , this circle can be shrunk to the point giving convergence in a disk except at the center.Remark: 1. Instead of (1) and (2) we may write, denoting by

where

2. Uniqueness. The Laurent series of a given analytic function f (z) in its annulus of convergence is

unique. However; f (z) may have different Laurent series in two annuli with the same

center.

3. To find the coefficients of the Laurent series we do not, generally, use the integral formulas in (2).

Example 87 Use of Maclaurin series

Find the Laurent series of with center 0.

Solution for all z.

Hence = for all non-zero complex number z.

Therefore, = for 0.

Here the annulus of convergence is the whole complex plane without the origin.



Example 88 Substitution

Find the Laurent series of with center 0.

Solution for all z. Hence for 0.

Therefore, = for 0.

Example 89 Develop in

i) non-negative powers of z ii) negative powers of z

Solutions i) is singular at z = 1.

= for < 1.

Therefore, = for < 1.

ii) = = = for 1.

Therefore, = for 1.

Example 90 Laurent series in different concentric annuli

Find all the Laurent series of with center 0.

Solution is singular at z = 0 and z = 1, thus we consider two Laurent series for f (z).

I. On the annulus 0 < < 1.

= = .

Therefore, = for 0 < < 1.

II. On the annulus 1.

= = for 1.

Therefore, = for 1.

Example 91 Find all the Taylor and Laurent series of with center 0.

Solution In terms of partial fractions



.

Now = for < 1 and = = for 1.

Similarly, = for < 2 and = = for

2.

Therefore, = for < 1,

= for 1 < < 2,

and = for 2.

Example 92 Find the Laurent series of that converges in the annulus < < ,

and determine the precise region of convergence.

Solution f (z) is not analytic at .

Taylor’s series of f (z) at z = 0 is

= for < 1.

But the annulus < < contains z = 1 as its center. Hence we must develop a series in

powers of z 1.

Now =

= = for

< 2.

Therefore, = for 0 < < 2.

Note that: If f (z) in Laurent’s theorem is analytic inside the inner circle the coefficients in (2) are

zero by Cauchy’s integral theorem, so that the Laurent series reduces to a Taylor series.

Singularities and Zeros

Definition 4.18 We say that a function f (z) is singular or has a singularity at a point



z = if f (z) is not analytic (perhaps not even defined) at z = , but every

neighborhood of z = contains points at which f (z) is analytic.

The two types of singularities

i) We call z = an isolated singularity of f (z) if z = has neighborhood without singularity of f (z).

Example 93 f (z) = tan z has an isolated singularities at , where n Z.

ii) Non-isolated singularity is a singularity z = for which every neighborhood of contains another

singularity of f (z).

Example 94 f (z) = tan has no isolated singularity at z = 0.

Now consider isolated singularities only.

Suppose f (z) has an no isolated singularity at z = , then there is neighborhood of say

0 < < R for some R in which f (z) is analytic except at . Hence in the region

0 < < R f (z) has the Laurent series

+ (1)

The series is analytic at z = and the series is called the

principal part of (1).

If it has infinitely many terms, then it is of the form

+ + + . . . + , where ≠ 0.

The singularity of f (z) at z = is called a pole, m is called its order. Poles of the first order

are called simple poles.

If the principal part of (1) has infinitely many terms, we say that f (z) has an isolated singularity

at z = .

Example 95 Laurent series of f (z) = at z = 0 is

= for 0 < < 1.

Therefore, f (z) has a pole of order three at z = 0.

Example 96 Poles

The function f (z) = + has a simple pole at z = 0 and a pole of



fifth order at z = 2.

Example 97 essential singularities

Let . Then the Laurent series of is:

= .

Therefore, f (z) has an isolated essential singularity at z = 0.

Example 98 essential singularities

Let . Then the Laurent series of is:

= .

Therefore, f (z) has an isolated essential singularity at z = 0.

Theorem 4.33 (poles)

If f (z) is analytic and has a pole at z = , then

Theorem 4.34 (Picard’s Theorem)

If f (z) is analytic and has an isolated essential singularity at a point ,

then it takes on every value, with at most one exceptional value in an

arbitrary small neighborhood of .

Zeros of Analytic Functions

We say that a function f (z) that is analytic in some domain D has a zero at a point z = in D if f ( ) = 0. We also say that this zero is of order n if not only f but also the derivatives , ,

, . . ., are all zero at z = but ≠ 0.

A zero of first order is called a simple zero; for it f ( ) = 0 but . For a second order

zero f ( ) = 0 and but and so on.

Example 99 Let and .

Now if and only if , where n Z



and , but while

.

Therefore, f has second order zeros at , where n Z.

Similarly, if and only if , where n Z

and while .

Therefore, g has fourth order zeros at , where n Z.

Theorem 4.35 (Zeros)

The zeros of an analytic function f (z) ( 0) are isolated; that is, each

of them has a neighborhood

Theorem 4.36 (Poles and Zeros)

Let f (z) be analytic at z = and have a zero of order z = .

Then has a pole of order z = .

The same holds for if h (z) is analytic at z = and .

Residue Integration Method

Introduction: There are various methods for determining the coefficients of a Laurent series, without the

integral formulas, we may use the formula for to evaluate complex integrals in a very

elegant and simple fashion. will be called the residue of f (z) at z = .

Residue

Now we need to evaluate complex integrals of the form

where C is a simple closed path.

If f (z) is analytic every where on C and inside C, such an integral is zero by Cauchy’s integral theorem, and

we are done.

If f (z) has a singularity at a point z = inside C, but is otherwise analytic on C and inside C, then f (z) has a Laurent series

+



that converges for all points near z = , (except at z = itself) in some domain of the form

. The coefficient of is given by:

=

and hence

= .

Here we integrate counterclockwise around the simple closed path C that contains z = in its interior. The coefficient is called the residue of f (z) at z = and we denote it by

=

Example 100 (Evaluation of an integral by means of a residue)

Evaluate , where C is a unit circle oriented counterclockwise.

Solution = for any z.

Hence, = for 0.

Thus, this series shows that f (z) has a pole of third order at z = 0 and the residue

= .

Therefore, = .

Example 101 Integrate clockwise around the circle C: .

Solution has singularities at z = 0 and z = 1. But z = 1 lies outside of C.

Hence we need to the residue of f (z) at 0.

Thus, the Laurent series that converges for 0 < < 1 is given by:

= = for 0 < < 1.

Hence, = 1 and clockwise integration yields

= .

Therefore, = .



Two Formulas for Residue at Simple Poles

Let f (z) have a simple pole at z = . The corresponding Laurent series (with m = 1) is

0 < < R

Here, ≠ 0. Multiplying both sides by we have

Now taking on both sides of this equation we get:

Example 102 (Residue at a simple pole)

. Then .

Hence, ,

and .

Another Simpler Method for the Residue at a Simple Pole

Let where p (z) and q (z) are analytic with p ( ) ≠ 0 and q (z) has a simple zero at

z = . Hence f (z) has a simple pole at z = . By definition of a simple zero, q (z) has a Taylor series of the form

Thus, substitution yields

=

= =

Therefore, .

Example 103 Let . Then has simple pole at z = 0, z = i and z = i.



Now let p (z) = 9z + i and . Then p (0) = i ≠ 0 while q (0) = 0 and = 1.

Therefore, .

Similarly, p (i) = 10 i ≠ 0 while q (i) = 0 and = 2.

Therefore, .

and p ( i) = 8 i ≠ 0 while q ( i) = 0 and = 2.

Therefore, .

Example 104 Find all poles and the corresponding residues of the function

Solution Let and . is entire and

has simple zeros at . Hence f (z) has simple poles at these points.

Since , the residues are equal to the values of at these points.

Therefore, , , ,

and .

Formula for the Residue at a Pole of Any Order

Let f (z) be an analytic function that has a pole of any order m 1 at a point z = . Then by definition of such a pole, the Laurent series of f (z) at z = is:

+ where .

Multiplying both sides by we get:

+ (*)

Let . Then the residue of f (z) at z = is the coefficient of the power series

in the Taylor series of g (z).

Thus, = .

Therefore,



In particular, for a second order pole m = 2

.

Example 105 The function has a pole of second order at z = 1 and a simple

pole at z = 4.

Hence, =

= = 8

and = .

Example 106 Residue from partial fraction

Let .

Then

Therefore, = 8, and = 8 and f (z) has a simple pole at z = 4 and a second

order pole at z = 1.

Example 107 Integration around a second order pole

Evaluate where C is any closed path such that z + 1 is inside C and

z = 4 is outside C.

Solution =

Therefore, = .

Residue Theorem

Here we consider the residue integration method of f (z) around any closed path C containing several

singular points.

Theorem 4.37 (Residue Theorem)

Let f (z) be a function that is analytic inside a simple closed path C, except for

finitely many singular points , , , . . . , inside C. Then



= (1)

the integration being taken counterclockwise around the path C.

Example 108 Evaluate , where C is any closed path oriented counterclockwise such that

i) 0 and 1 are inside C. ii) 0 is inside C and 1 is outside C.

iii) 1 is inside C and 0 is outside C. iv) 0 and 1 are outside C.

Solutions The integrand has a simple pole at 0 and 1, with residues:

= = 4 and = = 1.

Therefore, i) ii) iii) iv) 0.

Example 109 (Poles and Essential Singularities)

Evaluate , where C is the ellipse oriented

Counterclockwise.

Solutions The first term of the integrand has a simple pole at and at .

Now is inside C with residues

= and = while lie outside C.

The second term of the integrand has an essential singularity at 0, with residue as obtained

from

= =

Therefore, = = .

Example 110 (Confirmation of an earlier result)

Evaluate , where m N and C is any simple closed path oriented

Counterclockwise enclosing .

Solutions is its own Laurent series with center z = consisting of this one term principal



part and = 1, m = 1 and = 0, m = 2, 3, 4, . . .

Therefore, = .

Evaluation of Real Integrals

We now show a very elegant and simple method for evaluating certain classes of complicated real integrals.

Integrals of Rational Functions of cos and sin

Consider integrals of the type

(1)

where is a real rational function of and and is finite on the interval of integration. Setting , we get:

= and = .

and we see that the integrand becomes a rational function of z, say f (z). As ranges from 0 to 2π, the variable z ranges once around the unit circle = 1 in counterclockwise. Since , we have

and the integral takes the form

the integration being taken counterclockwise around the unit circle.

Example 111 Show that .

Solution We use = and . Then the integral becomes

=

Thus, the integrand has two simple poles, one at and the other at . lies outside the unit circle = 1 and lies inside C where the residue is:

= .

Therefore, .

Improper Integrals of Rational Functions

We consider the real integrals of the type



(1)

The improper integral has the meaning

= (2)

If both limits exist, we may couple the two independent passages to and and write

= (3)

The expression on the right side of (3) is called the Cauchy Principal Value of the integral. It may exist even

if the limits in (2) do not exist.

Example 112 , but .

Suppose the function in (1) is a real rational function whose denominator is different from zero for

all real number x and is of degree at least two units higher than the degree of the numerator. Then

the limits in (2) exists and

= =

where S is as shown in the figure.

where the sum consists of all the residues of f (z) at the points in the upper half-plane at which f (z) has a

pole. From this we have:

=

Now as R tends to infinity the value of the integral over S approaches to zero and hence

=

where we sum over all the residues of f (z) corresponding to the poles of f (z) in the upper half-plane.

Example 113 (An improper integral from zero to infinity)

Show that .

Solution has four simple poles at the points , ,


x

y

RR

121


and .

Now and lie in the upper half-plane while and lie in the upper half-plane, and

= = =

and = = = .

Thus, = + = =

Therefore, .

Since is an even function, we get:

.

Example 114 Another improper integral

Show that .

Solution has simple poles at 2i and i in the upper half-plane and at 2i and i

in the lower half-plane. The residues at 2i and i are:

= =

and = = .

Thus, = .

Therefore, .


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