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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 3 , 4 12, 2 .
Solution By definition 4.2
3 , 4 12, 2 3
12 and 2 4 4 and 2 .
Therefore, 4 and 2 .
Definition 4.3 The complex number (0,1) usually denoted by i = (0,1) is
called imaginary unit
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Unit IV
Functions of Complex Variables
4.1.1 Addition and Multiplication on Complex Numbers
Definition 4.4 For any two complex numbers z1 ( x1 , y1 ) and z 2 ( x2 , y 2 )
i) z1 z 2 = ( x1 x2 , y1 y 2 )
ii) z1 z 2 ( x1 x2 y1 y 2 , x1 y 2 x2 y1 )
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 x1 , x 2 . Then from definition 4.4 we get:
(x1 , 0) + (x2 , 0) = ( x1 x2 , 0) and (x1 , 0) (x2 , 0) = ( x1 x2 , 0)
Furthermore; for any real numbers x and y,
i y = (0, 1) (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 z1 , z 2 and z 3 be complex numbers. Then
i) z1 + z 2 = z 2 + z1 and z1 z 2 = z 2 z1
ii) ( z1 + z 2 ) + z 3 = z1 + ( z 2 + z 3 ) and ( z1 z 2 ) z 3 = z1 ( z 2 z 3 )
iii) z1 ( z 2 + z 3 ) = z1 z 2 + z1 z 3
iv) 0 + z1 = z1 , z1 + ( z1 ) = 0 and z1 1 = z1
Furthermore; for any non-zero complex number z = x + i y, there is a complex number
z
1
such that
z
1 1
z 1.
z z
The complex number
1
1
is usually denoted z .
z
Consequentely;
z
1
=
1 x iy
1
x
y
= 2
=
.
i 2
2
x i y x iy x iy x y
x y2
Therore, any non-zero complex number z = x + i y has a unique multiplicative inverse given by:
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Unit IV
Functions of Complex Variables
z
1
=
x
x2 y2
y
i
x2 y2
.
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, z1 < z 2 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
y
3
z = x + i y is represented as the point z with co-
z = 4 + 3i
ordinate x and ordinate y, and we say the point z in
4
O
x
the complex plane.
The sum of two complex numbers can be geometrically interpreted as the sum of two position vectors in the
Argand diagram.
y
z2
z1 + z 2
z1 z 2
z1
O
x
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 z is defined by
z =xiy
For any complex number z = x + i y in the complex plane, the complex conjugate of z, z = x i y
is obtained by reflecting z in the real axis.
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Example 4.3 Let z = x + i y be any complex number. Then verify that
i) z z x 2 y 2
ii) Re z =
1
(z + z )
2
iii) Im z =
1
(z z )
2i
Solutions Using properties of addition and multiplication on complex numbers and definition 4.5 we get:
i) z z = ( x i y ) ( x i y ) = ( x 2 i 2 y 2 ) = x 2 y 2 .
Therefore, z z = x 2 y 2 .
1
1
(z + z ) = ( x i y) ( x i y) = x = Re z .
2
2
1
Therefore, Re z = (z + z ).
2
1
iii)
( x i y) ( x i y)= 1 2i y= y = Im z.
2i
2i
1
Therefore, z =
(z z ).
2i
Example 4.4 Let z1 and z 2 be two complex numbers. Show that:
ii)
i) z1 z 2 = z1 z 2
ii) z1 z 2 = z1 z 2
iii) 1 1 , provided that z 2 0.
z
z
z
2
z
2
Solutions Let z1 x1 i y1 and z 2 x 2 i y 2 . Then From the properties of addition and
multiplication on complex numbers and definition 4.5 we get:
i) z1 z 2 = ( x1 i y1 ) ( x 2 i y 2 ) = ( x1 x 2 ) i ( y1 y 2 )
= ( x1 x2 ) i ( y1 y2 ) = ( x1 i y1 ) ( x2 i y2 ) = z1 z 2
Therefore, z1 z 2 = z1 z 2 for any two complex numbers z1 and z 2 .
ii) z1 z 2 = ( x1 i y1 ) ( x 2 i y 2 ) = ( x1 x 2 y1 y 2 ) i ( x1 y 2 x 2 y1 )
= ( x1 x2 y1 y2 ) i ( x1 y2 x2 y1 ) = ( x1 i y1 ) ( x2 i y2 ) = z1 z 2 .
Therefore, z1 z 2 = z1 z 2 for any two complex numbers z1 and z 2 .
z
z
z
1
1
z
iii) 1 = z1 = z1 = z1 2 = z1 2 = 1 .
z
z
z
z z
z z
z2
2
2
2
2 2
2 2
z
z
Therefore, 1 = 1 , provided that z 2 0.
z
z2
2
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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
z =r=
x2 y2 =
zz
while is called the argument of z and is denoted and defined by
1 y
, if z 0 and z is not a pure imaginary
tan x
arg z = =
2
2
, if z is pure imaginary and y 0
, up to multiples of 2.
, if z is pure imaginary and y 0
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 z and Arg z.
2
z = (1 + i) (1 i) = 2 and hence z = 2
1
and = arg z = tan (1) = n where n Z, but z lies in the second quadrant ,
4
3
hence, Arg z = .
4
3
3
Therefore, z = 2 cos i sin .
4
4
Example 4.6 Write z = 1 i in polar form.
Solution To write z in polar form first we need to find z and Arg z.
2
z = (1 i) (1 + i) = 2 and hence z = 2
1
n where n Z, but z lies in the third quadrant ,
and = arg z = tan (1) =
4
3
hence, Arg z = .
4
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3
3
2 cos i sin .
4
4
Therefore, z =
4.1.6 Important Inequalities
For any two complex numbers z1 and z 2
z1 z 2 z1 + z 2
(Triangle Inequality)
To show that this holds true, let z1 = x1 i y1 and z 2 = x2 i y 2 .
Then z1 z 2
2
= z1 z 2 z1 z 2
= z1 + 2 x1 x2 y1 y 2 + z 2
2
2
z1 + 2
x1 x2
2
y1 y 2 x1 y 2 y1 x2
2
2
2
+ z2
z1
z2
2
Therefore, z1 z 2 z1 + z 2 .
Furthermore; for any finite number of complex numbers z1 , z 2 , . . . , z n
n
zk
k 1
n
zn
k 1
(Generalized triangle inequality)
Verify! (Hint: use the principle of Mathematical induction on n)
Example 4.7 Let z1 = 1 i and z 2 = 2 3 i . Find z1 z 2 and z1 + z 2 .
1 4 i 1 4 i =
Solution z1 z 2 = 1 4 i =
z1 = 1 i =
1 i 1 i =
and z 2 = 3 2 i =
Therefore,
17
17 ,
2
2 3i 2 3i =
13 .
2 + 13 .
4.1.7 Multiplication and Division in Polar Form
Let z1 = r1 cos 1 i sin 1 and z 2 = r2 cos 2 i sin 2 .
Multiplication
z1 z 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 )
Therefore, z1 z 2 = z1 z 2 and arg ( Z1Z 2 ) = arg ( z1 ) + arg ( z 2 ) up to multiplies of 2.
Division
The quotient
z1
z2
is the number z =
z1
z2
satisfying z z 2 = z1 .
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Thus arg (z z 2 ) = arg z + arg z 2 = arg z1 and z1 z 2 = z
z1
Hence, z =
Therefore,
z2
z1
z2
z 2 = z1 .
z
and arg ( 1 ) = arg ( z1 ) arg ( z 2 ) up to multiplies of 2.
z2
r
= 1 cos (1 2 ) i sin (1 2 ) .
r2
Example 4.8 Let z1 =
Solution z1 =
3 i and z 2 = 3 i . Express z1 z 2 and
3i
3 i = 2 and z 2 =
z1
z2
in polar forms.
3i 3i = 3
1
= n where n Z.
6
3
2 n , where n Z. But z 2
But z1 lies in the 4 th quadrant, hence Arg z1 =
arg z 2 =
2
6
2
lies in the positive imaginary axis, hence Arg z 2 = . Thus 1 2 = and 1 2 = .
3
2
3
z
2
2
2
i sin and 1 = cos i sin .
Therefore, z1 z 2 = 6 cos
3
3
3
3
z2 3
4.1.8 Integer powers of Complex Numbers
and arg ( z1 ) = tan
1
For any non-zero complex number z r cos i sin
z n r n cos n i sin n for any n Z.
In particular if z = 1, then we get the De Moivre formula
cos
i sin n
cos n i sin n for any n Z.
Example 4.9 Use the De moivre formula to show that for any angle
cos 2 cos 2 sin 2 and sin 2
Solution If n = 2, then
cos
i sin 2
cos 2
2
cos sin
sin 2 i 2 sin sin
and from the De Moivre formula we get:
cos
i sin 2
cos 2 i sin 2
Therefore, cos 2 cos sin 2 and sin 2 2 cos sin
2
4.1.9 Roots of Complex Numbers
Suppose Z is a non-zero complex number. Now we need to solve z n , where n N and n 1.
Note that: Each values of is called an n th root of z, and we write
n
z
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Let z = r cos i sin and = R cos i sin .
Then z n r cos i sin = R n cos n i sin n
R n r , cos = cos n and sin = sin n .
2k
Rnr ,
, where k Z.
n
n
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
2k
2 h
2m = cos h
. Then cos k cos
n
n
n
n
2 h
2m = sin h
and sin k sin
n
n
2k
2k
Therefore, k n r cos
i sin
, where, k = 0, 1, 2, 3, . . . , n 1.
n
n
n
n
Let k
Note that: These n values lie on a circle of radius n r with center at the origin and constitute the
vertices of a regular n-gon.
The value of n z obtained by taking the principal value of arg z is called the principal value of
= n z .
Example 4.1.10 n th root of unity
Solve the equation z n = 1.
2k
2k
Solution Now n 1 = cos
i sin
, k = 0, 1, 2, 3, . . . , n 1.
n
n
If denotes the value corresponding to k = 1, then the n values of n 1 can be written as
1, , 2 , . . .,
n 1
2
2
i sin
n
n
Hence let = cos
Therefore, 1, , 2 , . . .,
n 1
.
are the n th roots of unity.
Example 4.1.11 Solve the equation z 4 = 1.
k
k
Solution Now 4 1 = cos
i sin
, where k = 0, 1, 2, 3.
2
Then for k = 1 we get = i .
2
Therefore, 1, i ,1 and i are the 4 th roots of unity.
Note that: The n values of n z are:
1 , 12 , 13 , . . . , 1n
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2
2
where 1 = n z cos
i sin
and n z is real.
n
Note that: For any complex number z x i y ,
1
2
z x i y=
z x i ( sign y )
x2
1
z x ,
2
2
1
z x .
2
1
z x , where sign y 1
z x i ( sign y )
2
1
1
z x i (sign y)
2
1
z =
2
Therefore,
2
z
if y 0
.
if y 0
1
1
where sign y
=
n
if y 0
.
if y 0
Exercise 4.1
1. Write in the form x i y, where z1 = 4 5 i and z 2 = 2 + 3 i
i)
ii) z1 z 2
1
2
z
iii) z1 2 z 2
1
2. Find the real and the imaginary parts of i) iii) in exercise 1.
3. Let z1 and z 2 be complex numbers, if z1 z 2 = 0, then show that either z1 = 0 or z 2 = 0.
4. Compute
1 i 6
i 3 1 4i 2
5. Represent
2i
in polar form.
4 4i
6. Determine the principal value of the argument of
ii) 2 2 i
i) 10 i
7. Represent each of the following in the form x i y
i) 4 cos
3
i sin
3
ii) 2 2 cos
3
3
i sin
4
4
8. Solve the equation
i) z 2 z 1 i 0
ii) z 4 3 (1 2 i ) z 2 8 6 i 0
9. For any two complex numbers Z1 and Z 2 show that
z1 z 2 2 + z1 z 2 2 = 2 z1 2 z 2 2
(Parallelogram equality)
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4.2 Curves and Regions in the Complex Plane
4.2.1 Circles and Disks
The distance between two points z and z 0 in the complex plane is denoted by z z0 . Hence a circle C of
radius and center z 0 can be given by
z z0 =
In particular the unit circle with center at the origin is given by z = 1
Furthermore;
i) z z0 < represents an open circular disk.
ii) z z0 > represents the exterior of the circle C.
iii) 1 < z z0 < 2 represents the open circular ring (open annulus).
Example 11
i)
z 2 i = 9 is a circle of radius 9 centered at 2 + i
ii)
z 2 i < 9 is an open circular disk of radius 9 centered at 2 + i .
iii)
z 2 i ≤ 9 is a closed circular disk of radius 9 centered at 2 + i .
iv)
z 2 i > 9 is the exterior of the circle of radius 9 centered at 2 + i .
4.2.2 Half plane
i) (open) upper half - plane = z : z x i y and y 0
ii) (open) lower half-plane = z : z x i y and y 0
iii) (open) right half plane = z : z x i y and x 0
iv) (open) left -half plane = z : z x i y and x 0
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 z 0 is the set of all points z such that z z0 < δ where δ is any
given positive number. (a deleted δ-neighborhood of z 0 is a neighborhood of z 0 in which the point
z 0 is omitted i.e. 0 < z z0 < δ).
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ii) Limit points
A point z 0 is called a limit point or cluster point or accumulation point of a set S if every deleted δneighborhood of z 0 contains points of S. Since δ can be any positive number, it follows that S must
have infinitely many points, and z 0 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 z < 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) = z 2 .
To each complex number z there is only one value of and hence = f (z) = z 2 is a single
valued function of z.
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Example 13 Let = f (z) =
z.
To each non-zero complex number z, there are two values of and hence f (z) =
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 u i v . If z = x i y , then = f (z) if and only if u i v = f ( x i y ). Thus
u and v are real valued functions of x and y. Hence
= f (z) = u ( x, y ) i v ( x, y ) .
Therefore, a complex function f (z) is equivalent to a pair of real valued functions u ( x, y ) and v ( x, y ) .
Example 14 Let = f (z) = z 2 . Find the functions u ( x, y ) and v ( x, y ) for which
f (z) = u ( x, y ) i v ( x, y )
and calculate the value of f at z 1 3 i .
Solution Let z x i y . Then u Re f ( z ) x 2 y 2 and v Im f ( z ) 2 x y .
Thus, u ( x, y ) = x 2 y 2 and v ( x, y ) = 2 x y. Hence, u (1, 3) = 8 and v (1, 3) = 6.
Therefore, f (1 3 i ) 8 6 i .
Example 15 Let = f (z) = 2 i z 6 z . Find the functions u ( x, y ) and v ( x, y ) for which
f (z) = u ( x, y ) i v ( x, y )
1
4i .
2
Solution Let z x i y . Then f ( x i y ) 2 i ( x i y ) 6 ( x i y ) (6 x 2 y ) i (2 x 6 y )
and calculate the value of f at z =
1
2
1
2
Thus, u ( x, y ) 6 x 2 y and v ( x, y ) 2 x 6 y . Hence, u ( , 4) 5 and v ( , 4) 23 .
1
2
4.2.3 Limit and Continuity
Therefore, f ( 4 i ) 5 23 i .
Definition 4.6 A function f (z) is said to have a limit ℓ as z approaches z 0 ,
written
im f ( z )
z z0
if to each positive number ε there corresponds a positive number
δ such that
f (z) whenever 0 < z z0 < δ.
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Note that: 1. By definition z approaches z 0 from any direction in the complex plane.
2. If a limit exists, then it is unique.
im f ( z ) if and only if im Re f ( z ) Re and im Im f ( z ) Im .
z z0
z z0
z z0
3.
.
Definition 4.7 A function f (z) is said to be continuous at z = z 0 if f ( z 0 ) is defined and
im f ( z ) f ( z )
0
z z0
Note that: If
im f ( z ) , then f (z) is defined in some neighborhood of z . Furthermore;
0
z z0
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 z 0 written
f ' ( z0 )
is defined by
f ' ( z0 ) =
im f ( z ) f ( z 0 )
z z0
z z0
Provided that this limit exists. In this case, we say f is differentiable at z 0 .
If we write z z z 0 , we also have
f ( z) f ( z0 )
.
z 0
z
f ' ( z 0 ) = im
Example 16 Show that f (z) = z 2 is differentiable for all complex numbers and f ' ( z ) = 2z.
Solution Let z 0 be any complex number.
2
im f ( z ) f ( z 0 ) = im z z 0
z z0
z z0 z z0
z z0
2
=
im z z 2 z .
0
0
z z0
Therefore, for any complex number z, f ' ( z ) = 2z.
4.2.4.1 Differentiation Rules
If f and g are differentiable functions, then
i) (c f ) ' ( z ) c f ' ( z )
ii) ( f g ) ' ( z ) f ' ( z ) g ' ( z )
iii) ( f g ) ' ( z ) f ' ( z ) g ( z ) f ( z ) g ' ( z )
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f '
f ' ( z) g ( z) f ( z) g ' ( z)
( z )
, provided that g (z) 0.
g 2 ( z)
g
iv)
v) ( f g ) ' ( z ) f ' ( g ( z )) g ' ( z ) .
vi) ( z n ) ' n z
n 1
for any n N.
Note that: If f (z) is differentiable at z 0 , then f (z) is continuous at z 0 .
Let f (z) = u ( x, y ) i v ( x, y ) . If f (z) is differentiable at z 0 . Then the partial derivatives of u and v at
z 0 exist.
Example 17 Show that f (z) = z is not differentiable.
Solution Let z 0 be any complex number.
Now consider
im f ( z ) f ( z 0 ) = im z = im x i y .
z z0
z z0
z 0 z z 0 x i y
i) Suppose x = 0, then
im f ( z ) f ( z 0 ) = 1.
z z0
z z0
ii) Suppose y = 0, then
im f ( z ) f ( z 0 ) = 1.
z z0
z z0
Therefore,
im f ( z ) f ( z 0 ) does not exist, and hence f (z) = z is not differentiable.
z z0
z z0
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 = z 0 in D
if f (z) is analytic in a neighborhood of z 0 , including the point z 0 .
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
f ( z)
n
ak z k
k 0
(*)
where a 0 , a1 , a 2 , . . ., a n 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.
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The quotient of two polynomials g (z) and h (z)
f ( z)
g ( z)
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 f ( z )
z
, where z x i y .
1 z
2. Determine whether f (z) is continuous at the origin or not, where
if z 0
0
a) f ( z ) Re z
z
0
z
b) f ( z ) Im
1 z
if z 0
3. Differentiate a) z i
3
2
b)
if z 0
if z 0
z2 4
z2 1
4. Show that f (z) = Re z is not differentiable at any z.
5. Show that f (z) = z
(Hint: use z z
2
is differentiable only at z = 0; hence it is nowhere analytic.
2
= ( z z ) ( z z ) ).
4.3 The Cauchy - Riemann Equations
Let = f (z) = u ( x, y ) i v ( x, y ) . 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
u x v y and u y v x
(*)
everywhere in D.
The equations in (*) are called the Cauchy - Riemann equations.
Theorem 4.1 (Cauchy Riemann Equations)
Let f (z) = u ( x, y ) i v ( x, y ) be defined and continuous in some
neighborhood of a point z 0 x0 i y0 and differentiable at z 0 itself.
Then at z 0 , 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 z 0 along any curve that passes through z 0 .
i) Suppose we approach z 0 along any curve for which x = x0 .
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im
Then f ' ( z 0 )
z z0
u ( x0 , y ) u ( x0 , y 0 )
i y y0
+
im i v ( x0 , y ) v ( x0 , y 0 )
z z0
i y y0
Therefore, f ' ( z 0 ) = i u y v y .
0
0
ii) Suppose we approach z 0 along any curve for which y = y 0 .
im
Then f ' ( z 0 )
z z0
u ( x, y 0 ) u ( x 0 , y 0 )
x x0
+
im i v ( x, y 0 ) v ( x0 , y 0 )
z z0
x x0
Thus, f ' ( z 0 ) = u x i v x .
0
0
Therefore, u x v y and u y v x .
0
0
0
0
Therefore, if f (z) is analytic in a domain D, then it satisfies the Cauchy Riemann equations.
Example 19 Let = f ( z ) z 2 3 z . Then u ( x, y ) x 2 3 x y 2 and v ( x, y ) 2 xy 3 y .
Now u x = 2x + 3 = v y and u y 2 y v x .
Therefore, f (z) satisfy the Cauchy Riemann equations.
Example 20 Let = f ( z ) z . Then u ( x, y ) x and v ( x, y ) y .
Now u x = 1, v y = 1 and u y 0 v x . Since u x v y , the Cauchy Riemann equations all not satisfied
Therefore, f ( z ) z is not analytic.
Theorem 4.2 (Cauchy - Riemann Equations)
If two real -valued continuous functions u ( x, y ) and v ( x, y ) 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 f ( z ) u ( x, y ) i v ( x, y ) 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 u ( x, y ) = x 2 y 2 x .
Solution f (z) is analytic implies the Cauchy- Riemann equations are satisfied.
Thus u x = 2 x 1 = v y and hence v ( x, y ) = 2 xy y h ( x)
dh
dh
= v x . Thus,
= 0 h (x) = k, constant.
dx
dx
Therefore, f (z) = x 2 y 2 x i 2 xy y k = z 2 z i k , in terms of z.
and u y = 2 y = 2 y
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Functions of Complex Variables
Example 22 If f (z) is analytic in D and f (z ) = k constant in D, then f (z) is constant in D.
Solution Let f ( z ) u ( x, y ) i v ( x, y ) . Then f (z ) 2 = u 2 v 2 = k 2 .
Thus, 2 u u x 2 v v x = 0 u u x v v x = 0 and 2 u u y 2 v v y = 0 u u y v v y = 0.
Now f (z) is analytic in D implies u x = v y and u y = v x .
2
u v y v vx 0
u v y uv v x 0
u2 v2 vy 0
2
u
v
v
v
0
u
vv
v
v
0
x
y
x
y
Hence
v u v y v 2 v x 0
u 2 v 2 vx 0 .
2
u v x uv v y 0
and
Thus u 2 v 2 v y 0 and u 2 v 2 v x 0 .
i) If u 2 v 2 0 , then f (z ) = 0 and hence f ( z ) 0 .
ii) If u 2 v 2 0 , then v x 0 and v y 0 .
Thus u x v y 0 and u y v x 0 and hence u ( x, y ) and v ( x, y ) are constants.
Therefore, f ( z ) u ( x, y ) i v ( x, y ) 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:
ur
1
1
v and v r u .
r
r
(Verify!)
Laplace's Equations, Harmonic Function
Theorem 4.3 (Laplace's Equation)
If f ( z ) u ( x, y ) i v ( x, y ) is analytic in a domain D, then u and v
satisfy the Lap lace's equations. i.e
2 u u xx u yy 0 and 2 v v xx v yy 0
respectively, in D and have continuous second partial derivatives in D.
2 is called the Laplacian operator and read as “nabla squared”.
Proof Since f (z) is analytic in D, u x v y and u y v x and f ' ( z ) is analytic in D.
Hence, u xx u yy v yx v xy 0 and v xx v yy u xy u yx 0 .
Therefore, 2 u 0 and 2 v 0 .
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Unit IV
Functions of Complex Variables
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 u ( x, y ) x 2 y 2 y is harmonic in the whole complex plane and find
a conjugate harmonic function v of u.
Solution Now u x 2 x and u xx 2 and u y 2 y 1 and u yy 2 .
Thus 2 u 0 and hence u ( x, y ) x 2 y 2 y is a harmonic function.
Now u x 2 x v y , since v satisfies the Cauchy-Riemann Equation.
Then v ( x, y ) 2 xy h ( x) and hence v x 2 y
dh
2 y 1 h (x) = x + c.
dx
Thus, v ( x, y ) 2 xy x c .
Therefore, f ( z ) x 2 y 2 y i 2 xy x c = z 2 i z i c .
Exercise 4.2
1. Determine whether the following functions are analytic or not.
1
b) f ( z ) e x (cos y i sin y)
4
z
c) f ( z) n z i Arg z .
a) f ( z )
2. Are the following functions harmonic? If so, find a corresponding analytic function
f ( z) u i v .
a) u ( x, y ) x 3 3 x y 2
b) u ( x, y) e x cos 2 y
c) v ( z) i n z
3. Determine a, b and c such that the given functions are harmonic and find the conjugate harmonic
functions.
a) u ( x, y ) e 2 x cos ay
b) u ( x, y ) cos bx cosh y
c) u ( x, y ) sin x sinh by
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.
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Unit IV
Functions of Complex Variables
4.3 Exponential Functions
The exponential function e z , also written exp z, is defined by:
(*)
e z e x (cos y i sin y)
z
Note that: The definition of e is a natural extension of the real exponential function e x . i.e.
i) If z = x, then e z = e x .
ii) e z is an entire function, i.e. an analytic function and e z ' = e z .
Now let z1 and z 2 be two complex numbers. Then
z1 z2
z1
z
e 2
In particular if z1 = x and z 2 i y , then
e
eZ
exe
iy
e
, e Z e x 0, arg e Z y 2 n , where n Z and hence e Z 0 z in C
and if x = 0, then from (*) we get:
e
iy
cos y i sin y , since e
iy
1
called the Euler formula .
Moreover; the polar form of a complex number z r (cos i sin ) can be written as
z r e i
4.3.1 Periodicity of e z
e
z 2 i
e z for all z.
Hence, e z is periodic with pure imaginary period 2 i . Thus, all the values that w = e z can assume are
already in the horizontal strip of width 2.
y
This infinite strip is called a fundamental region of e z .
Example 24 Find all solutions of e z 2 2 i .
3
n 2 .
Solution e z e x 2 2 x
2
1
Then e x cos y = 2 cos y =
and e x sin y = 2 sin y =
Hence, cos y =
1
and sin y =
2
1
y
2
2
3
3
Therefore, z = n 2 i ( 2 n ) , where n Z.
2
4
1
.
2
3
2 n , where n Z.
4
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Unit IV
Functions of Complex Variables
Example 25 Find all solutions of e 3 z 3 .
1
3 ( x i y)
3 x 3i y
Solution e3z e
e
e and hence e 3Z e 3x 3 x n 3 .
3
2
Then cos 3 y 1 and sin 3y = 0 if and only if y n , where n Z.
3
1
2
n i , where n Z.
Therefore, Z = n 3
3
3
Exercise 4.3
Find all values of k such that f ( z ) e x (cos ky i sin ky) is analytic.
4.4 Trigonometric and Hyperbolic Functions
4.4.1 Trigonometric Functions
Consider the Euler formulas
e
ix
cos x i sin x and e
i x
cos x i sin x .
(*)
From (*) we get:
1
1
cos x ei x e i x and sin x ei x e i x
2
2
i
This suggests the following definitions.
(**)
Definition 4.10 For any complex number z x i y
1 iz
1
iz
i z
i z
e e
e e
and sin z
2
2i
sin z
cos z
1
Furthermore; tan z
, cot z
, sec z
cos z
sin z
cos z
1
and csc z
.
sin z
cos z
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
cos z '
sin z , sin z ' cos z and tan z ' sec 2 z .
Furthermore, the Euler's formula is valid, i.e
e
iz
cos z i sin z
4.4.2 The Real and Imaginary Parts of sin z and Cos z
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Unit IV
Functions of Complex Variables
Example 26 For any complex number z x i y , show that
a) cos z cos x cosh y i sin x sinh y and sin z sin x cosh y i cos x sinh y
b) cos z 2 cos 2 x sinh 2 y and sin z 2 sin 2 x sinh 2 y
solution By definition:
1 i ( x i y)
1 ix y
i ( x i y)
i x y
e
e
e e
e
e
=
2
2
1 y
y
e
(cos x i sin x) e (cos x i sin x)
=
2
1
i
y
y
y
y
cos x (e e ) sin x (e e ) = cos x cosh y i sin x sinh y .
=
2
2
Therefore, cos z = cos x cosh y i sin x sinh y .
a) cos z =
1 i ( x i y)
1 ix y
i ( x i y)
i x y
e
e
e e
e
e
=
2i
2i
1
y
y
e
(cos x i sin x) e (cos x i sin x)
=
2i
i
1
y
y
y
y
cos x (e e ) sin x (e e ) = sin x cosh y i cos x sinh y .
=
2
2
Therefore, sin z = sin x cosh y i cos x sinh y .
sin z =
b) From the result of part a) we get:
cos z
2
cos 2 x cosh 2 y sin 2 x sinh 2 y
= cos 2 x (1 sinh 2 y ) sin 2 x sinh 2 y
= cos 2 x cos 2 x sin 2 x sinh 2 y = cos 2 x sinh 2 y .
Therefore, cos z 2 cos 2 x sinh 2 y
sin z
2
sin 2 x cosh 2 y cos 2 x sinh 2 y
= sin 2 x 1 sinh 2 y cos 2 x sinh 2 y
= sin 2 x sin 2 x cos 2 x sinh 2 y = sin 2 x sinh 2 y
Therefore, sin z 2 sin 2 x sinh 2 y
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.
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Unit IV
Functions of Complex Variables
Definition.4.11 A number z 0 is a zero of a complex valued function f (z) if f ( z 0 ) = 0.
Example 27 Find the zeros of sin z and cos z.
Solutions Let z x i y . Then cos z cos x cosh y i sin x sinh y
and sin z sin x cosh y i cos x sinh y .
Hence, cos z 0 if and only if cos x cosh y 0 and sin x sinh y 0 .
cos x cosh y 0 and sin x sinh y 0 cos x 0 and sin x 0 or sinh y 0 .
cos x 0 and sinh y 0 x
Therefore, cos z 0 if and only if z (1 2 n )
n and y 0 .
2
.
2
sin z 0 sin x cosh y i cos x sinh y 0 sin x cosh y 0 and cos x sinh y 0
sin x = 0 and sinh y = 0 x n and y = 0, where n Z.
Therefore, sin z = 0 if and only if z n , where n Z.
Example 28 Solve cos z = 2.
Solution cos z
Now e
iz
e
e
iz
e
y
1
iz
i z
iz
i z
e e
. Then cos z = 2 if and only if e e
2
i z
4e
2 3 e
cos x 2
x (2n 1) and e
y
2i z
4e
iz
1 0 e i z 2 2 3
(cos x i sin x) 2
3
3 and sin x 0 x n and e
y
2
4
y
cos x 2
3
3 x (2n 1) and y n (2 3 ) .
Therefore, cos z = 2 if and only if z = (2n 1) i n (2 3 ) , where n Z.
Exercises 4.4 For any complex numbers z, z1 and z 2 , show that:
i) cos ( z1 z 2 ) = cos z1 cos z 2 sin z1 sin z 2
ii) sin ( z1 z 2 ) = sin z1 cos z1 cos z 2 sin z 2 cos z1
iii) sin 2 z cos 2 z = 1
iv) cos 2 z sin 2 z = cos 2 z
4.5 Hyperbolic Functions
The Complex hyperbolic cosine and sine functions are defined by:
cosh z
1 z
1
(e e z ) and sinh z (e z e z )
2
2
These functions are entire with derivatives.
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Unit IV
Functions of Complex Variables
cosh z ' sinh z
and sinh z ' cosh z .
The other hyperbolic functions are defined by
sinh z
cosh z
1
1
, coth z
, sec h z
and csc h z
.
cosh z
cosh z
sinh z
sinh z
Remark: If in the definition of cosh z and sinh z replacing z by iz we get:
tanh z
cosh i z = cos z and sinh i z = i sin z .
similarly if we replace z by iz in the definition of cos z and sin z we get:
cos i z = cosh z and sinh i z = i sinh z .
Example 29 For any complex number z x i y , show that:
cosh z = cosh x cos y i sinh x sin y and sinh z = sinh x cos y i cosh x sin y
Solutions Let z x i y .
1
1 x
x
e (cos y i sin y ) e (cos y i sin y )
Then cosh z (e z e z ) =
2
2
i
1 x
x
x
e e
cos y e x e sin y
=
2
2
= cosh x cos y i sinh x sin y
Therefore, cosh z = cosh x cos y i sinh x sin y .
1
1 x
x
e (cos y i sin y ) e (cos y i sin y )
Similarly sinh z (e z e z ) =
2
2
i
1 x
x
x
e e
cos y e x e sin y
=
2
2
= sinh x cos y i cosh x sin y
Therefore, sinh z = sinh x cos y i cosh x sin y .
Exercise 4.5
1
i
2
d)
1. Compute e z in the form u i v and if z equals
b) 1 i
a) 3 i
c)
2
9
i
2
2. Find the real and the imaginary parts of
a) e
z2
z
b) e
3. Write in polar form a) 1 i
4. Find all solutions of a) e
3z
3
c) e
b)
2z
i
b) e z 2
c) n z
c) e z 3 4 i
x2
y2
is harmonic and find its conjugate.
cos
2
2
6. Show that Re cos z and Im sin z are harmonic.
5. Show that u e
xy
7. Find all solutions of the following equations
a) sin z i sinh 1
b) cos z 3 i
c) sin z cosh 3
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Unit IV
Functions of Complex Variables
8. Find all values of z for which a) sin z and b) cos z have real values.
9. Find Re tan z and Im tan z .
10. Show that sinh y cos z cosh y , sinh y sin z cosh y 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 z x i y is denoted by n z and is defined by
n z if and only if z e .
Suppose that u i v and z r e
e z e
u iv
i
= re
i
, then
e u = r and arg e = arg z
u n r and v Arg z 2 n , n Z.
Therefore, n z n z + i ( Arg z 2 n ) , where n Z.
Now let Ln z n z i Arg z . Then Ln z is called the principle value of n z and
n z Ln z 2 n i , where n Z.
Note that: 1. The complex natural logarithm n z is infinitely many valued function having the
same real part but differing in their imaginary parts by an integral multiple of 2.
2. Ln z n z i Arg z is single valued.
Example 30 Given z x i y with y = 0. Then find n z .
Solution Consider the following cases.
Case i) x 0
z x z x and Arg z 0 .
Thus, n z = n x 2 n i , where n Z.
ii) x 0
z x z x and Arg z .
Thus, n z = n x (2 n 1) i , where n Z.
Example 31 Find Ln z and n z , where z 1 i .
Solution
z 2 and Arg z
Therefore, Ln z =
4
.
1
1
n 2 i and n z =
n 2 2 n i , where n Z.
2
4
2
4
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Unit IV
Functions of Complex Variables
Properties
In general for any non-zero complex numbers z1 and z 2
z
b) n 1 = n z1 n z 2
z
a) n z1 z 2 = n z1 + n z 2
2
Note that: a) and b) need not hold true if we replace n z by Ln z .
Example 32 Let z1 = z 2 = e
i
= 1.
Then Ln z1 Ln z 2 = i , while Ln z1 z 2 Ln 1 0 2 i = Ln z1 Ln z 2
But n z1 z 2 = n 1 2 i and n z1 + n z 2 = 2 i .
Let z1 and z 2 be non-zero complex numbers. Then
Ln z1 z 2 n z1 z 2 i Arg ( z1 z 2 )
= n z1 n z 2 i Arg z1 Arg z 2 2 n , n Z.
Now consider e
e
n z
n z
where z is any non-zero complex number.
n z i ( Arg z 2 n )
= e
n z
= e
n z
e i ( Arg z 2 n ) = e
e i Arg z
z (cos Arg z i sin Arg z) z .
=
Therefore, e
, where n Z.
n z
= z.
But note that n e z need not equal to z.
n e z = n e z + i ( Arg e z 2 n ) , where n Z.
= n e x + i ( y 2 n ) = x i y 2 n i = z 2 n i , where n Z.
Therefore, n e z = z 2 n i , 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 z x i y , show that n z ' =
Solution Let n z = u i v , where n z =
where n Z.
Then u x
x
= vy
x2 y2
1
y
1
x
2
1
.
z
1
y
n x 2 y 2 and v arg z = arctan 2n ,
2
x
y
1
vx
and u y 2
x
x y2
y
2 .
2
x
y
1
x
1
Consequently from theorem 4.1 we get:
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Unit IV
Functions of Complex Variables
n z ' = u x i v x = v y i u y =
Therefore, n z ' =
x
y
x2 y2
x2 y2
i
x i y
1
= .
z
x2 y2
1
.
z
General Power
General powers of a complex number z x i y are defined by
zc e
c n z
, where c is any non-zero complex number.
Since n z is infinitely many-valued, z c will, in general, multi-valued. The particular value
zc e
c Ln z
is called the principal value of z c .
c Ln z
Remark: i) If c Z and c 0, then z c e
is single valued and is identical with the
integer power of z.
1
1
n
ii) If c = , where n N and n 1, then z c z e
n
th
n root of z has n distinct values.
1
n n z
, where z 0, is the
p
q
p
iii) If c =
the quotient of two positive integers, then z c z e
q
p
q n z
, where z 0
has only finitely many distinct values.
iv) If c is irrational or genuinely complex (i.e. c a i b , b 0), then z c is infinitely many
valued function.
Example 34 Find the principle value of
i) i i
i
Solutions i) i = e
i n i
= e
ii) (1 i)
2i
= e (2 2 n ) , where n N
i n i i ( Arg i 2 n )
Therefore, the principle value, n = 0, is e
ii) (1 i)
2i
= e
2
.
, where n N
( 2 i ) n 1 i i ( Arg (1 i ) 2 n )
n
(2 i)
=e
=e
( n 2
= 2e
4
2 n )
2 n ) i ( 4 n n 2 )
4
2 n
4
2 n
= 2e4
2 i(
2
e
( 2n ) i
2
e
i n 2
(sin n 2 i cos n 2 ) .
Therefore, the principle value is 2 e 4 (sin n 2 i cos n 2 ) .
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Unit IV
Functions of Complex Variables
Exercises
1. Compute the principle value Ln z if z equals
i) 3 4 i
2. Solve for z:
ii) 0.6 0.8 i
ii) n z (4 i )
i) n z 3 i
3. Find the principle value of
i)
ii) (2 i)
i
1 i
2
iii) ( 5)
2 4i
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
z (t ) x (t ) i y (t ) for a t b
Example 35 Find a parameterization of the line segment with endpoints
a) z 4 2 i and z 3 5 i
b) z 4 i and z 7 38 i
Solution a) x (t) = 4 t and y (t) = 2 + 3 t
Therefore, z (t ) 4 t (2 3 t ) i , where 0 t 1.
b) x (t) = 7 t and y (t) = 4 42 t
Therefore, z (t ) 7 t (42 t 4) i , where 0 t 1.
Definition 4.12 A curve C is called a smooth curve if it has a continuous and
non-zero derivative
z ' (t ) x' (t ) i y ' (t )
at each point t.
Geometrically
Z (t)
C
Δt
z(t)
z1
O
z(t + t))
z2
zn
O
z0
z (t t ) z (t )
z ' (t ) t im
0
t
The arrow on the curve indicates the sense of increasing t.
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Unit IV
Functions of Complex Variables
The Complex Line Integral
Let C be a smooth curve in the complex plane represented by
z (t ) x (t ) i y (t ) for a t b,
let f (z) be a continuous function defined at least at each point on C and let = t 0 , t1 , t 2 , ..., t n be a
partition of a, b , where a = t 0 t1 t 2 ... t n b .
Suppose to the subdivisions of there correspond a subdivision of C by points z 0 , z1 , z 2 , ..., z n , where
z k z (t k )
For each m = 1, 2, 3, ….., n, let m be a point on C between z m 1 and z m , where
m (t ) z (t ) for some t [t m 1 , t m ] .
Then consider the sum
Sn
n
f ( m ) z m , where z m z m z m 1 .
m 1
as the number of partition of increases the max t m tends to zero and hence max z m tends to zero.
In this case im S n =
n
f ( m ) z m exists and we define this limit as the line integral of f (z) over the
m 1
oriented curve C and is denoted by
c f ( z) dz
If C is a closed curve for the line integral of f (z) over C we also use the notation
f ( z ) dz
C
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.
f ( z ) dz
C
exists.
Proof: Let f ( z ) u ( x, y ) i v ( x, y ) , let m m i m and z m xm i y m .
n
Then
m 1
f ( m ) z m =
n
u ( m , m ) i v ( m , m ) xm i y m
m 1
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Unit IV
Functions of Complex Variables
n
=
u xm
m 1
n
v y m
m 1
n
+ i u y m
m 1
n
v xm
m 1
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,
f ( z ) dz = C u dx C v dy + i C u dy C v dx exists.
C
Properties of complex line integrals
1. Complex integration is a linear operation.
Let k1 and k 2 be complex constants, and f1 and f 2 complex continuous function on C, then
2
2
k m f m ( z) dz m 1 C k m f m ( z) dz
C m 1
2. Decomposing C into two portions C1 and C 2 we get:
f ( z ) dz
C
2
f ( z) dz
m 1
C
m
3. Reversing the sense of integration, we get the negative of the original value.
Z0
Z1
f ( z) dz f ( z) dz .
Z0
Z1
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 z z (t ) ,
where a t b and let f (z) be a continuous function on C.
Then
f ( z )dz
C
Example 36 Integrate f ( z )
b
a f ( z (t )) z' (t ) dt
1
on the standard unit circle oriented counter clockwise.
z
it
Solution Using Euler’s formula C: z (t ) e , where 0 t 2 , is the parameterization of the unit
it
Circle oriented counterclockwise. Furthermore; and z ' (t ) i e .
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Functions of Complex Variables
Thus,
C
Therefore,
C
dz
i
z
2
0 dt 2 i .
dz
2 i .
z
Example 37 Let f ( z ) ( z z 0 ) m where m Z and z 0 is a constant. Find
f ( z ) dz , where C is
C
a circle of radius ρ centered at z 0 and oriented counterclockwise.
Solution z z0 z z0 (cos t i sin t ) , where 0 ≤ t ≤ 2.
Then z (t ) z 0 e it , where 0 ≤ t ≤ 2 and hence z ' (t ) i e and ( z z 0 ) m m e
it
Thus,
C
f ( z ) dz = i
m 1
2
i ( m 1) t
dt =
e
m 1
0
If m = 1, then
m 1
i
2
0
i mt
.
2
cos ( m 1) t dt sin (m 1) t dt
0
2
2
= 1, and cos (m 1) t dt = 2 and sin (m 1) t dt = 0.
0
0
2
2
0
0
If m 1, then cos (m 1) t dt = 0 = sin (m 1) t dt .
Therefore,
(z z 0 )
C
m
2 i
0
if m 1
and m Z.
if m 1
dz =
Remark: A complex line integral depends not only on the endpoints of the path but in general also on
the path itself.
1 i
Example 38 Let f ( z ) Re z x . Find
a) C *
0 f ( z ) dz along
b) along C consisting of C1 and C 2 as shown below.
Solutions a) C * : z (t ) t i t , where 0 t 1
y
Then z '(t ) 1 i and f ( z (t )) t .
1
1
Hence Re z dz t (1 i) dt = (1 i ) .
2
0
*
z=1+i
C*
C
C2
C1
x
b) C1 : z (t ) t , where 0 t 1. Then z '(t ) 1 and f ( z (t )) t
and C 2 : z (t ) 1 i t , where 0 t 1. Then z '(t ) i and f ( z (t )) 1 .
1
Hence
1
1
Re z dz 0 t dt i 0 dt 2
i.
C*
This shows that complex line integral is path dependent.
Second Method: Indefinite integral
In real integration if f (x ) is continuous on a, b and if F (x ) is an antiderivative of f (x ) on a, b , then
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Unit IV
Functions of Complex Variables
b
a f ( x)dx F (b) F (a)
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 z 0 and z1 in D, the integral of f (z) from z 0 to z1 can be evaluated
by:
z
1
f ( z) dz = F ( z1 ) F ( z 0 )
z
0
1 i
0 z
Example 39 Evaluate
1 i
Solution
2
z dz =
0
0 z
2
dz .
1 3 1 i 2
= (i 1) .
z
3
3
0
1 i
Therefore,
2
dz =
2
(i 1) .
3
i
Example 40 Evaluate
cos z dz .
i
i
Solution
i
cos z dz = sin z
= 2 sin i = 2i sinh .
i
i
i
Therefore,
cos z dz = 2i sinh .
i
8 3 i
Example 41 Evaluate
z
e2
dz .
8 i
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Unit IV
Functions of Complex Variables
8 3 i
Solution
z 8 3 i
z
e2
dz =
2e 2
8i
8 i
8 3 i
Therefore,
i
3 i
4 i
2 i
= 2 e4 e 2 e 2 = 2 e 2 e
1 = 0.
z
e 2 dz = 0.
8 i
i
Example 42 Evaluate
i
Solution
i
dz
.
z
dz
i i i .
= Ln (i ) Ln ( i ) =
2
z
2
i
Therefore,
i
i
dz
=i.
z
Note that: Simply connectedness in the above theorem is essential.
dz
2 i , where C is oriented counterclockwise over the unit circle.
z
1
1
3
Although
is analytic in the annulus < z < , this domain is not simply connected.
z
2
2
Bound for the Absolute Value of Integrals
Example 43
C
Let C be a curve and f (z) be a continuous function on C. Then
f ( z ) dz
MLinequality
≤ML
C
where L is the length of C and M is a constant for which f (z ) ≤ M on C.
To show that this holds true, consider the partial sum S n . By the generalized triangle inequality
n
Sn =
Therefore,
m 1
f ( m ) z m ≤
n
m 1
f ( m ) z m ≤ M
n
m 1
z m = ML.
C f ( z ) dz ≤ M L.
Example 43 Find an upper bound for the absolute value of the integral e z dz , where C is the line
segment from the origin to 3 4 i .
Solution L = 5 and e z ≤ e 3 . Hence,
Therefore,
e
C
z
e
C
z
C
dz ≤ 5 e 3 .
dz ≤ 5 e 3 .
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Unit IV
Functions of Complex Variables
Exercises
1. What curves are represented by the following functions?
i) 1 2 i t ,
iii) t 3 t 2 i ,
v) t
i
,
t
it
0t 3
ii) 2 i 3 e
1 t 2
iv) cos t 2 i sin t ,
0 t 2
,
t
1
t5
2
2. Represent the following curves in the form z z (t ) .
i)
z 3 4i = 4
ii) y
iii) 4 x 1 + 9 y 2 = 36
2
1
1
from (1, 1) to 3,
x
3
2
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,
C f ( z ) dz 0
Example 44 For any closed path C,
C e
z
dz 0 , sin z dz 0 , cos z dz 0 and
C
C
C z
n
dz 0
where n W . Since these functions are entire.
Example 45
dz
C sec z dz 0 and C z 2 4 0 , where C is the standard unit circle.
Now sec z is not analytic at z (2 n 1)
2
, n Z. But these points are outside of the circle.
Similarly for the second integral, whose integrand is not analytic at z 2 i outside C.
Example 46
z dz 2 i , where C is the unit circle oriented counterclockwise.
C
Since f ( z ) z is not analytic and
Example 47
C
dz
z2
C
2
z dz e
it
dz
i e it dt 2 i .
0
0 , where C is the unit circle. Now
C
z2
i
2
e
it
dt = e
it
0
This result does not follow from the Cauchy’s theorem, since f ( z )
2
= 0.
0
1
is not analytic at z = 0.
z2
Note that: The condition that f (z) is analytic in D is sufficient rather than necessary for
f ( z ) dz 0 to hold.
C
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Unit IV
Functions of Complex Variables
Example 48 Evaluate
7z 6
C z 2 2 z dz , where C is the unit circle.
Solution By using partial fraction we get:
7z 6
=
3
4
+
.
z
z2
z 2z
7z 6
dz
dz
Then 2
+ 4
= 3 2 i 0 = 6 i .
dz = 3
C z 2z
C z
C z 2
4
(Since f (z) =
is not analytic at z = 2 outside of C.)
z2
7z 6
Therefore, 2
dz = 6 i .
C z 2z
Independence of Path
2
An integral of f (z) is said to be independent of path in a domain D if for every path from z1 to z 2 in D
its value depends only on z1 and z 2 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 C1 and inner boundary
curve C 2 , refer the figure below. If f (z) is analytic in any domain D * that contains D as well as its
boundary curves, then
f ( z) dz = f ( z ) dz
C1
C2
D1
C2
D2
C1
Both integrals being taken counterclockwise (or both clockwise and regardless of whether
or not the full interior of C 2 belongs to D * ).Note that: For domains of higher connectivity the idea remains
the same.
Example 49 Let C be any counterclockwise oriented simple closed path containing z 0 and m an
integer, then
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Unit IV
Functions of Complex Variables
(z z 0 )
C
dz
C
Example 50 Evaluate
m
2 i
0
if m 1
if m 1
dz =
, where C is as shown in the figure.
z 1
Solution By using partial fraction we get:
2
1
=
1 1
1
.
2 z 1 z 1
z2 1
Now let C1 and C 2 be the portion of the curve C to the
left of the origin and the right of the origin respectively.
Then
C
dz
=
z 1
C
dz
C
dz
= 0 and
z 1
=
1
2
1
Thus,
2
z2 1
Therefore,
1
C
1
dz
+
z 1 2
1
dz
C
C
z2 1
C
dz
dz
=
= 2 i .
z 1 C z 1
2
dz
= 2 i .
z 1
2
= 2 i .
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 z 0 and z1 in D, the
integral of f (z) from z 0 to z1 can be evaluated by:
z
1
f ( z) dz = F ( z1 ) F ( z 0 )
z
0
i
Example 51 Evaluate
cos z dz .
i
Solution f (z) = cos z is entire and sin z ' = cos z. Then
i
cos z dz = 2 sin i = 2i sinh .
i
i
Therefore,
cos z dz = 2i sinh .
i
2i
Example 52 Evaluate
i z
3
2
3
1 dz .
3
1 7 3 5
'
z z z3 z = z2 1 .
5
7
Solution f (z) = z 2 1 is entire and
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Unit IV
Functions of Complex Variables
2i
Thus,
3
3
1
2
z
1
dz = z 7 z 5 z 3 z
7
5
i
2i
i
=
1566
i.
35
2i
3
1566
2
z
1
dz =
i.
35
Therefore,
i
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
z 0 in D and any simple closed path C in D that encloses z 0
f ( z)
C z z0 dz = 2 i f ( z0 )
the integration being taken counterclockwise.
Example 53 Evaluate
ez
dz , where C is any closed path enclosing z 0 = 2.
Cz 2
Solution z 0 = 2 and f ( z ) e z is entire.
Hence by Cauchy’s integral formula, for any closed path enclosing z 0 = 2 we get:
ez
2
2
C z 2 dz = 2 e i , since f ( z0 ) e .
Therefore,
ez
2
C z 2 dz = 2 e i .
Example 54 Evaluate
C
z3 6
dz , where C is any simple closed path.
2z i
Solution There are two cases to consider
1
i
2
1 3
z 3
3
z 6
1
2
Then
dz . Now f ( z ) z 3 3 is entire and the region
dz =
1
2
C z i
C 2z i
2
1
i
3.
enclosed by C is a simply connected domain and f ( i )
2
16
z3 6
6 i .
Therefore,
dz =
8
C 2z i
Case I C encloses z
Case II C does not enclose z
1
i
2
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Unit IV
Functions of Complex Variables
Now g ( z )
z3 6
is analytic in the simply connected domain enclosed by C.
2z i
Therefore, by Cauchy’s integral theorem
C
z3 6
dz = 0.
2z i
Example 55 Evaluate
z2 1
z 2 1 dz , where C is a unit circle oriented counterclockwise centered at:
C
1
1
ii) z 0
iii) z 0 1 i
iv) z 0 i
2
2
Solution In i) and ii) the circle C encloses the point z 0 = 1 and does not enclose z 0 = 1.
i) z 0 = 1
z2 1
Now let f ( z )
, then f (z ) is analytic in the region enclosed by C, and f (1) = 1.
z 1
z2 1
z 2 1 dz =
Therefore,
2 i .
C
In iii) the circle C encloses the point z 0 = 1 but not z 0 = 1.
Now let f ( z )
z2 1
, then f (z ) is analytic in the region enclosed by C, and f (1) = 1.
z 1
z2 1
z 2 1 dz = 2 i .
Therefore,
C
In iv) the circle C does not enclose the points z 0 = 1 and z 0 = 1.
Hence let f ( z )
z2 1
z2 1
. Then f (z ) is analytic in the region enclosed by C.
z2 1
z 2 1 dz = 0.
Therefore,
C
Example 56 Evaluate
tan z
C z 2 1 dz , where C is the circle
Solution f (z) = tan z is not analytic at z (2 n 1)
2
z
3
.
2
, 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:
1 tan z tan z
.
z 1 2 z 1 z 1
tan z
tan z
1 tan z
dz
dz = 2 i tan 1 .
Thus, 2
dz =
2
z
1
z
1
z
1
C
C
C
tan z
Therefore, 2
dz = 2 i tan 1 .
z
1
C
tan z
2
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Unit IV
Functions of Complex Variables
Remark: (Multiply Connected Domains)
If f (z) is analytic on C1 and C 2 , and in the ring-shaped domain bounded by C1 and C 2 , and z 0 is
any point in the domain of f, then
f ( z ) dz
f ( z0 )
2 i C z z 0
1
1
C
2
f ( z)
dz
z z0
where the outer integral (over C1 ) is taken
z0
counterclockwise and the inner clockwise
C2
as shown in the figure.
C1
Exercises
1
1. Integrate
z 1
4
counterclockwise around the circle
i) z i 1
ii) z 1 1
iii) z 3 1
iv) z 3
2. Integrate the following functions counterclockwise around the unit circle z 1
cos z
i)
z
3z
sin z
ii)
2z
e
iii)
3z i
3. Integrate the following functions over the given contour C.
tan z
, where C is the boundary of the triangle with vertices 1, 1 and 2 i .
zi
Ln ( z 1)
1
ii)
, where C consists of the boundary of the triangle with vertices 1 i ,
2
2
z 1
1
1 i and 2 i .
2
sin z
i)
iii)
, where C consists of the boundary of the square with vertices
4z 2 8 z i
a) 3 and 3 i
b) 1 and i
4. Integrate the following functions counterclockwise around the unit circle z 2
i)
5. Integrate
z4
ii)
( z 3 i) 2
e 2 sin z
z2
iii)
z3
( z 1) 3
f ( z ) dz , where
C
i) f ( z) z
2
tan z , C is an ellipse with foci i
ii) f ( z) ( z 4)
3
Ln z , C is the circle z 5 3
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Unit IV
Functions of Complex Variables
ez
iii) f ( z )
2
z ( z 2 i) 2
, C consists of the boundary of
a) the square with vertices 3 and 3 i oriented counterclockwise
b) z 1 oriented clockwise.
Derivatives of Analytic Functions
Theorem 4.9 (Derivatives of Analytic Functions)
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 z 0 in D are given by the formula
f
( n)
n!
2 i
( z0 )
f ( z)
(z z
C
0)
n 1
dz , where n N
here C is any simple closed path in D that encloses z 0 and whose full
interior belongs to D; and we integrate counterclockwise around C.
cos z
( z i) 2
Example 57 Evaluate
dz , where C is any simple closed path oriented counterclockwise
C
that encloses z 0 = i .
Solution f ( z ) cos z is entire and f ' ( z ) sin z .
cos z
1
Thus, sin i =
2 i
cos z
( z i) 2
( z i) 2
dz
C
dz = 2 i sin i = 2 sinh .
C
Therefore,
cos z
( z i) 2
dz = 2 sinh .
C
Example 58 Evaluate
z4 3 z2 6
C
( z i) 3
dz , where C is any contour enclosing z 0 = i oriented
counterclockwise.
.
Solution Let f ( z ) z 4 3 z 2 6 and z 0 = i . Then f (z ) is entire with f ' ' ( z ) 12 z 2 6 and
hence f ' ' ( z ) 18 .
Thus, 18 =
Therefore,
C
2!
2 i
z4 3 z2 6
( z i) 3
C
z4 3 z2 6
( z i) 3
dz
dz = 18 i .
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Unit IV
Functions of Complex Variables
ez
dz , where C is any contour enclosing 1 but not 2 i
2
2
C ( z 1) ( z 4)
Example 59 Evaluate
oriented counterclockwise.
Solution Let
ez
and z 0 = 1. Then f (z) is analytic in the domain enclosed by C and
z2 4
f ( z)
f ' ( z) =
e z ( z 2 2 z 4)
( z 2 4) 2
ez
dz .
2
2
C ( z 1) ( z 4)
1
3
e =
Thus,
2 i
25
Therefore,
.
ez
6
( z 1) 2 ( z 2 4) dz = 25 e i .
C
Theorem 4. 10 (Morera’s Theorem)
If f (z) is continuous in a simply connected domain D and if
f ( z ) dz = 0
C
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 z 0 in D. Suppose
f (z ) ≤ M on C. Then applying the ML-inequality
f
( n)
( z0 )
n!
2 i
f ( z)
(z z
C
0)
n 1
dz ≤
n! M
.
rn
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
f (z) ≤ k for all z.
Let z 0 be any point. Then by Cauchy’s inequality
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Unit IV
Functions of Complex Variables
f '( z0 ) <
k
r
for a circle or radius any r with center z 0 , since f (z) is entire for all r.
Hence, f ' ( z 0 ) = 0 and z 0 is arbitrary implies f ' ( z ) = 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 z n called a term of the sequence,
and is written
z1 , z 2 , z 3 , . . . or { z1 , z 2 , z 3 , . . . } or simply { z n }.
A real sequence is one whose terms are real.
Convergence
A convergent sequence { z n } is one that has a limit C, written
im z = C or z → C.
n
n
n
By definition of limit this means that for every ε 0 we can find an N such that
z n C < ε for all n N.
A divergence sequence is one that does not converge.
Example 60 The sequence
i n
converges to 0.
n
i)
iv) z n with z n = 2
ii) i n is divergent.
iii)
(1 i) n is divergent.
1
2
i 1 converges to 2 i .
n
n
Theorem 4.12 (Sequences of real and imaginary parts)
A sequence z n of complex numbers z n xn i y n , where n N,
converges to C a b i if and only if the sequence of the real parts
xn converges to a and the sequence of the imaginary parts y n
converges to b.
Proof: If z n C < ε, then z n xn i y n is within the circle of radius ε about C a b i
for all n N.
Therefore, xn a < ε and y n b < ε for all n N.
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Unit IV
Functions of Complex Variables
Suppose x n converges to a and y n converges to b.
Then for all ε 0 there exists an N such that
xn a
and y n b
for all n N.
2
2
Thus, z n C = xn a i ( yn b) ≤ xn a yn b .
Therefore, z n C < ε for all n N.
Series
Given a sequence z n we may form the sequence of sums
S1 z1 , S 2 z1 z 2 , S 3 z1 z 2 z3 , . . . , S n z1 z 2 z3 ... z n , . . .
called the sequence of partial sums of the (infinite) series
zm =
m 1
z1 z 2 z3 ...
(*)
z1 , z 2 , z 3 , . . . are called terms of the series.
A convergent series is one whose sequence of partial sum S n converges. If im S n = S is called
n
the sum or value of the series and we write
S=
zm
m 1
(1)
A divergent series is one that does not converge.
If we omit the terms S n of the series
z m , there remains
m 1
Rn z n 1 z n 2 z n 3 ... .
This is called the remainder of the series (1) after z n . If (*) converges and has sum S, then
S S n Rn and hence Rn S S n .Now im S n = S implies im Rn = 0.
n
n
When S is known and we compute an approximation S n by S, then Rn is the error.
Theorem 4.13 (Real and Imaginary Parts)
A series (*) with z m xm i y m converges with sum S u i v
if and only if
x m converges to u and m1y m converges to v .
m 1
im z = 0 is a necessary condition but not sufficient.
m
m
1
1
1
0 , but
Example 61 Let z m =
. Then im
diverges, since it is a harmonic series.
m m
m
m 1 m
Note that:
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Unit IV
Functions of Complex Variables
Tests for Convergence and Divergence of Series
Theorem 4.14 (Divergence)
z m converges, then im z m = 0.
m
m 1
If a series
Theorem 4.15 (Cauchy’s Convergence Principle for Series)
A series
z n is convergent if and only if for every ε 0 (no matter
n 1
how small) we can find an N (which depends on ε, in general) such that
p
zn k
< ε for every n N and p N.
k 1
Or equivalently, S n S m < ε for every m, n N.
z m is called absolutely convergent if the series of the
Definition 4.14 A series
m 1
absolute values of the terms
If
z m converges but
m 1
Example 62 The series
m 1
z m is convergent.
m 1
z m diverges, then
1n
n 1
n
z m is called conditionally convergent.
m 1
is conditionally convergent.
Theorem 4.16 If a series
z m is absolutely convergent, then it is convergent.
m 1
Theorem 4.17 (Comparison Theorem)
If a series
z m is given and we can find a convergent series
m 1
bm with
m 1
non-negative real terms such that z m ≤ bm for m = 1, 2, 3, …, then the
given series converges, even absolutely.
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Unit IV
Functions of Complex Variables
Theorem 4.18(Geometric series)
The geometric series
1
q m converges with the sum
1 q
m0
if q < 1 and diverges if q ≥ 1.
Ratio Test
Theorem 4.19 (Ratio Test)
If a series
z n with z n ≠ 0 ( n = 1, 2, 3, . . . ) has the property that for
n 1
every n greater than some N,
zn 1
≤ q < 1 for n N
zn
(where q < 1 is fixed), this series converges absolutely. If for every n N,
zn 1
≥ 1 for n N
zn
the series diverges.
Theorem 4.20 (Ratio Test)
If a series
z m with z m ≠ 0 ( m = 1, 2, 3, . . . ) is such that
m 1
z
im m 1 = L
m zm
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
100 75 i n
n0
n!
Solution Using ratio test with z n
im
n
zn 1
zn
convergent or divergent?
100 75 i n , we get:
n!
100 75 i
im
= 0.
n
n 1
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98
Unit IV
Therefore,
Functions of Complex Variables
100 75 i n
n0
n!
Example 64 Is the series
is convergent.
i
3 n
n0 2
1
2
3 n 1
convergent or divergent?
Solution Using ratio test with z n = 3 n 1 1 2 i , we get:
1
2
zn 1
zn
Therefore,
=
2
i
3 n
n0 2
1 2 i
1
3n 4
is convergent.
1
2
3 n 1
1
2
= < 1.
8
1 2i
3 n 1
Root Test
Theorem 4.21 (Root Test)
If a series
z n is such that for every n greater than some N,
n 1
n
z n ≤ q < 1 for n N
(where q < 1 is fixed), this series converges absolutely. If for infinitely many n, n z n ≥ 1,
then the series diverges.
Caution ! n z n
≤ q < 1 implies n z n
Example 65 Consider
1
.
n 1 n
zn =
< 1, but this does not imply convergence.
1
and hence
n
n
1
< 1 for all n 1.But the series diverges.
n
Theorem 4.22 (Root Test)
If a series
z n is such that im n
n
n 1
z n = 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
Hence n
1
< 1 and
n
Example 67 Is the series
n0
1
and
n 1 n
n
1
1
2.
n 1 n
zn =
1
1
and z n = 2 .
n
n
1
1
< 1for all n 1. But
diverges, while
converges.
2
n2
n 1 n
n 1 n
(1) n
n
(
4
i
)
convergent?
2 2n 3
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Unit IV
Functions of Complex Variables
Solution z n
(1)
2
2n
and hence
Therefore,
n0
n
3
17
(4 i) n and z n
im
n
2
2n
3
. Then n z n
17
z n im
n
n
n
2
n
2
2n
3
=
17
n
2
2n
3
17
1.
4
(1) n
n
diverges.
(
4
i
)
2 2n 3
Power Series
Definition 4.15 A power series in powers of z z 0 is a series of the form
n
a n z z 0
n0
where z is a variable a 0 , a1 , a 2 , . . . are constants, called coefficients
of the series and z 0 is a constant, called the center of the series.
If z 0 = 0, then we obtain a power series in powers of z:
an z
n0
n
.
Convergence of a Power Series
A power series in powers of z z 0 may converge
i) throughout the disk with center z 0
ii) in the whole complex plane
iii) merely at the center
Consider the following examples.
Example 68 Show that the geometric series
if
z
n0
n
converges absolutely for z < 1 and diverges
z ≥ 1.
Solution Using the ratio test we get:
z n 1
im
n
Hence,
z
n0
zn
n
z
converges absolutely for z < 1 and diverges if
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z 1.
100
Unit IV
Functions of Complex Variables
If
z
n0
z = 1, then
Therefore,
z
n0
n
diverges.
converges if and only if
Example 69 Show that
n
n0
z < 1.
zn
converges on the whole complex plane.
n!
Solution Using the ratio test we get:
im
n
Therefore,
n0
zn 1
zn
z
im n 1 = 0.
n
zn
converges on the whole complex plane.
n!
Example 70 Show that
n0
n ! z n converges only at z = 0.
Solution Using the ratio test we get:
im
n
Therefore,
n0
zn 1
zn
im (n 1) z =
n
0
if z 0
.
if z 0
n ! z n converges only at z = 0.
Note that: Series such as
n0
n ! z n are useless series.
Theorem 4.23 (Convergence of a Power Series)
If a power series
n
a n z z 0
n0
i) converges at a point z = z1 ≠ z 0 , then it converges absolutely for
every z closer to z 0 than z1 .
ii) diverges at a point z = z 2 , then it diverges for every z farther away
from z 0 than z 2 .
y
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Unit IV
Functions of Complex Variables
z2
diverges
Conv.
z0
z1
x
Radius of Convergence of a Power Series
Suppose
n
n0
a n z z 0 converges for some z1 ≠ z 0 but does not converge on the whole complex plane.
n
a
z
z
n
0 is convergent
n0
Now consider the set A = z z0 : z S .
Let S = z :
A is a bounded set, hence A has a least upper bound. Let R = ℓub A. Then
a) The series
b) The series
n
a n z z 0
n0
n0
converges for all z for which z z 0 < R.
n
a n z z 0 diverges for all z for which z z 0 R.
Definition 4.16 The number R is called the radius of convergence of the series
n0
n
a n z z 0 and the circle z z 0 = R is called the circle of
convergence.
Remarks: 1. We extend R by writing
R = ∞ if the series
n0
R = 0 if the series
n
a n z z 0 converges for all z.
n
a n z z 0
n0
converges only at the center z = z 0 .
2. No general statement can be made about the convergence of the power series
n
a n z z 0
n0
on the circle of convergence itself.
Example 71 On the circle of convergence z = 1.
a)
b)
zn
zn
converges at 1 but diverges at 1.
n
1
converges every where, since
converges.
2
2
n 1 n
n 1 n
n 1
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Unit IV
Functions of Complex Variables
c)
z n diverges every where.
n 1
Theorem 4.24 (Radius of Convergence R)
Given the power series
n0
n
a n z z 0 (with radius of convergence R)
and suppose
an 1
im
= L*
n an
(Cauchy-Hadamard formula)
Then,
a) if L* = 0, then R = ∞.
c) if L* ≠ 0 and L* , then R =
Remark: If L im
an 1
n an
b) if L* = ∞, then R = 0.
1
.
L
≠ 0 and L* , then from theorem 4.24 it follows that
an
R = im
n an 1
Example 72 Determine the radius of convergence of the power series
Solution Now a n
( 2 n) !
( n !) 2
n
z 3 i
2
( 2 n) !
( n !)
.
. Then
an
R = im
n0
n an 1
(2 n) ! (( n 1) !) 2
2
n (n !) (2 (n 1)) !
= im
1
1
n 1
n = 1.
= im
= im
n 2 (2n 1) n 4 2 4
n
1
1
Therefore, the series converges in the open disk z 3 i < , of radius
and center 3 i .
4
4
Functions Given by Power Series
Consider the power series in z. i.e.
an z n
n0
(1)
Note that: For a power series in powers of z z 0 with any center z 0 we can always set
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Unit IV
Functions of Complex Variables
z z 0 = 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
f ( z)
a n z n , where
n0
z < 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 z < 1,
zn
n0
1
.
1 z
Therefore,
1
z n represents the function f ( z )
1 z
n0
for z < 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 im f ( z ) a 0 . i.e. we want to show that for all ε 0 there exists a
z 0
0 such that
Now
an z n
n0
n 1
an r
z < implies f ( z ) a0 < ε .
converges absolutely for z ≤ r < R, hence
n 1
Let S =
n 1
an r
f ( z ) a0 =
=
1
r
n 1
n 1
a n r n converges.
. Then for 0 < z ≤ r
an z
n 1
n
≤ z
Thus, z S < ε if and only if z <
n 1
an
z
n 1
≤ z
n 1
an r
n 1
= z S <ε.
. Now choose < min. r , .
S
S
Therefore, the theorem holds.
Theorem 4.26 (Identity Theorem for Power Series)
Suppose that the power series
n0
a n z n and
bn z n
n0
both converge for z < R, where R is positive and have the same
sum for all z < R , then these series are identical. i.e a n = bn for
all non-negative integer n.
Proof: (Use induction on n and continuity at z = 0)
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Unit IV
Functions of Complex Variables
a n ( z z 0 ) n for all
n0
Remark: If f (z) =
z z0 < 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
n0
a n z n and
a m z m be two power series with radius of convergence R1 and R2 respectively.
m0
Then
a n z n a m z m converges for a radius of convergence at least equal to the minimum of R1
n0
m0
and R2 .
Term wise Multiplication
Let f ( z )
k 0
a k z k and g ( z )
a m z m . Then
m0
a z k am z m =
k
k 0
m0
n0
n
n
a b
z
k
n
k
k 0
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
S ( z) f ( z) g ( z)
Term wise Differentiation and Integration
Given the power series
an z n
n0
The power series
n an z
n 1
(1)
n 1
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
n2
n n
z .
2
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Unit IV
Functions of Complex Variables
n
n!
1
Solution a n = =
= n (n 1) .
2 ( n 2) ! 2 ! 2
n2
Let f ( z )
Then g ' ( z )
1
n (n 1) z n and g ( z )
2
n 1
nz
n 1
and g ' ' ( z )
Therefore, the radius of convergence R of
n2
n 1
zn .
n0
n (n 1) z
n2
z2
f ( z) .
2
. Thus, g ' ' ( z )
n n
z is 1.
2
Theorem 4.28 (Term wise integration of a power series)
n0
The power series
an n 1
z
, obtained by integrating the
n 1
series
an z n
n0
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 z 0 , then f (z) can be represented by
n
f
m0
f ( z)
( m)
( z0 )
m!
( z z 0 ) m Rn ( z )
(1)
called the Taylor’s formula, Rn (z ) is called the Taylor remainder.
If we let n approach infinity, we obtain the power series
f ( z)
n0
(n)
f
( z0 )
n!
( z z0 ) n
(2)
The series in (2) is called the Taylor series of f (z) with center z = z 0 . If z 0 = 0, then we get the Maclaurin
series
f ( z)
n0
f
( n)
(0) n
z
n!
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Unit IV
Functions of Complex Variables
Theorem 4.30 (Taylor’s Theorem)
Let f (z) be analytic in a domain D and let z = z 0 be any point in D. Then there
exists precisely one power series with center z 0 that represents f (z). This series
is of the form
f ( z)
an ( z z0 )
n0
n
, where an
f
( n)
( z0 )
n!
(3)
This representation is valid in the largest open disk with center z 0 in which f (z)
is analytic.
Remark: In (3) a n can be given by
an
1
2 i
C
f ( z)
( z z )n 1
dz
f ( n) ( z 0 )
n!
0
where we integrate counterclockwise around a simple closed path in D that encloses z 0 .
In (1) Rn (z ) is given by
Rn (z ) =
( z z )n 1
0
2 i
C
f ( z*)
( z * z )n 1( z * z )
dz *
0
where z* is on C and z is inside C.
Definition 4.17 Singular Points
z = z 0 is called a singular point of an analytic function f (z) if f (z) is not
differentiable at z = z 0 , but every disk with center z 0 contains points at
which f (z) is differentiable.
Example 75 Let f (z) = tan z.
Then f (z) is not analytic at z = ( 2n 1)
Therefore, f has singularities at z = ( 2n 1)
2
2
, where n Z.
, where n Z.
Important special Taylor Series
Example 76 Geometric series
Let f ( z )
1
. Find the Maclaurin series of f (z).
1 z
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Unit IV
Functions of Complex Variables
Solution f
( n)
( z)
n!
(1 z )
and f
n 1
z n , where
n0
Therefore, f ( z )
( n)
(0) n!. Thus a n
f
( n)
(0)
= 1.
n!
z 1and f (z) is singular at z = 1; this point lies on
the circle of convergence.
Example 77 Exponential Function
Let f ( z ) e z . Find the Maclaurin series of e z .
Solution f
( n)
( z) e z and f
Therefore, e z =
( n)
(0) 1 . Thus a n
1
.
n!
zn
.
n!
n0
(*)
If we set z i y in (*) we obtain:
e
Thus, e
iy
iy
n0
Therefore, e
iy
n0
1
in yn
and recall that i n
n!
i
2n 1
(1) n y 2n
i
( 2n) !
if n is even
if n is odd
n0
(1) n y
(2n 1) !
cos y i sin y .
= cos y i sin y .
(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:
(i z ) n
( i z ) n (i z ) n
z
1 z
cos z (e e )
1 ( i ) n
=
2
n! n 0 n!
n 0 n!
Therefore, cos z =
n0
Similarly, sin z =
n0
=
n0
2n
( 1) n z
( 2n) !
.
2n
( 1) n z
( 2n) !
.
2 n 1
( 1) n z
(2n 1) !
, cosh z =
n0
2n
z
and sinh z =
( 2n) !
n0
2 n 1
z
.
( 2n 1) !
Example 79 Logarithm
The Maclaurin series of
Ln (1 z )
n0
(1)
n 1
zn
n
( z 1)
Replacing z by z and multiplying both sides by 1 we get:
zn
1
( z 1)
Ln (1 z ) Ln
1 z n 1 n
and adding these yields:
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Unit IV
Functions of Complex Variables
1 z
2
1 z
Ln
2n 1
n 1
z
2n 1
( z 1)
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.
1
x2
Example 80 Let f ( z ) e
. 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 f ( z )
1 z2
for z < 1.
1
=
1 z
Solution Recall that:
1
zn
n0
for z < 1.
Then substituting z 2 instead of z we get:
1
( z 2 )
=
2
1 ( z ) n 0
Therefore,
1
1 z
=
2
(1)
n0
n
n
.
z 2n .
(*)
Example 82 Integration
Determine the Maclurian series of f ( z ) tan
Solution f ' ( z )
tan
1
Therefore, tan
1
1 z2
1
n
n0
(1)
2n 1
for z < 1.
z
2n 1
z=
z for z < 1.
. Integrating (*) term by term and using f (0) = 0 we get:
z=
1
n0
n
(1)
2n 1
for z < 1.
z
2n 1
This value represents the principal value of u i v = tan
1
z defined as the value for which u
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109
.
Unit IV
Functions of Complex Variables
Example 83 Developing by using the trigonometric series
1
in powers of z a, where c ab ≠ 0 and b ≠ 0.
c bz
1
1
1
Solution Now
=
=
c bz
c b ( z a) ab
c ab b ( z a)
Develop
1
1
=
=
c ab 1 b ( z a)
c ab
1
bn
Therefore,
=
( z a) n for z
c bz n 0 c ab n 1
1
c ab
a <
n
b ( z a)
.
n 0 c ab
c
a .
b
Example 84 Binomial Series, reduction by partial fraction
Determine the Taylor series of
f ( z)
2z 2 9z 5
z 3 z 2 8z 12
i) z 0 = 0
Recall that:
with center
ii) z 0 = 1
s
x n where s is any number and
n0 n
s
n
s ( s 1) ( s 2) ... ( s (n 1))
=
for n ≥ 0 and
n!
s
= 1
0
called the binomial coefficient, is called the Binomial series.
Solution Using partial fraction reduction
f ( z)
1
( z 2)
2
1
z3
Now by the Binomial series
1
m
=
m ( z 2)
( z 2)
Thus,
i) z 0 = 0
1
m n
z .
n
n0
m n
z converges for z < 1.
n
n0
2
z n for z < 2.
=
2
( z 2)
n0 n
1
1 1
1
=
=
z
z3
3
3
1
3
n0
z
3
n
for z < 3.
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Unit IV
Functions of Complex Variables
2 n 1
z
3
n0 n
Therefore,
n0
n
z for z < 2.
3
ii) Similarly, f (z) can be written as:
1
1
1
f ( z)
.
1
1
9
2
(1 ( z 1))
1 ( z 1)
3
2
1
Thus, f ( z )
9
2 z 1
n 0 n 3
n
n
z 1
for z < 2.
n0 3
Example 85 Use of Differential Equations
Determine the Maclaurin series of f ( z ) tan z
Solution Now f ' ( z ) sec 2 z and f (0) = 0.
Thus, f ' ( z ) 1 f 2 ( z ) and f ' (0) 0 .
Then f ' ' ( z ) 2 f ( z ) f ' ( z ) ; f ' ' (0) 0 ,
f
f
' ' ' ( z ) 2 f '2 ( z ) 2 f ' ' ( z ) ; f ' ' ' (0) 2 ,
( 4)
f
( z) 6 f ' ( z) f
( 5)
( z) 6 f
...
Therefore, tan z = z
' ' ( z) 2 f ( z) f ' ' ' ( z) ; f
( 4)
' '2 8 f ' ( z ) f ' ' ' ( z ) 2 f ( z ) f
(0) 0 ,
( 4)
( z) ;
f
(5)
(0) 16 ,
1 3
2 5
17 7
z
z
z ... for z < .
3
15
315
2
Example 86 Undetermined Coefficients
Determine the Maclaurin series of f ( z ) tan z by using those of cos z and sin z .
Solution Since tan z is odd the desired expansion will be
tan z = a1 z a2 z 3 a3 z 5 ...
Now using sin z = tan z cos z , we obtain
z
z3
3!
Since tan z is analytic except at z (2n 1)
for z <
.
2
z2
z4
... = a1 z a2 z 3 a3 z 5 ... 1
...
5!
2!
4!
z5
2
where n Z, its Maclaurin series converges
Now applying Cauchy’s product and comparing coefficients we get:
a1 = 1,
a
a
a
1
1
1 3 a5 , etc.
2 a3 ,
5!
4! 2!
3!
2!
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Unit IV
Functions of Complex Variables
1
2
Hence, a1 = 1, a3 , a 5
, etc.
3
15
Therefore, tan z = z
1 3
2 5
17 7
z
z
z ... for z < .
3
15
315
2
Exercises
1. Determine the center and radius of convergence of
i)
n0
iv)
( z 4 i) n
n!
n
n 1 n
n k
vii)
n0 n
1
z
n 1
n n (3 i ) n
( z i) 2
ii)
nk
3n
n 1 n ( n 1)
v)
viii)
iii)
( 1) n z 2n
n0
2 2 n ( n !) 2
zn
vi)
( 2n) !
z
2
n 0 ( n 1) ( n !)
n 1
3 n n ( n 1)
7
n
z 2n
n 1
2. Determine the Taylor series of the given functions with the given point as center and determine the
radius of convergence.
ii) cos z;
i) e z ; 0
v)
1
1
;
z
iv) sin z;
vii) sin 2 z; 0
viii) cos ( z
2
vi) z 4 z 2 1; 1
i
iii) e 2 z ; 2i
2
2
);
2
3. Determine the Maclaurin series of the following functions and determine the radius of convergence.
i)
v)
1
ii)
1 z4
ez
4
1
z3
4 3z
iii) sin 2 z 2
(1 z )2
z
2
2
vi) e z e t dt
vii)
0
iv)
1
( z 3 4i ) 2
2 z 2 15 z 34
( z 4) 2 ( z 2)
4. Determine the Taylor series of the given functions with the given point as center and determine the
radius of convergence.
i)
1
;1
z
ii)
1
;1 i
z
iii)
iv) z 5 z 3 z ; i
v) ( z 1) 3 ; 1 i
vii) e z ; i
viii) cosh z ;
2
i
vi)
1
; 2i
( z i )2
1 z sin ( z 1)
( z 1)3
1
ix) sin z ;
2
;
1
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 z z 0 and being convergent in some annulus
(bounded by two circles with center at z 0 ) in which f (z) is analytic.
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Functions of Complex Variables
Theorem 4.32 (Laurent’s series)
If f (z) is analytic on two concentric circles C1 and C 2 with center z 0 and in the
annulus between them, then f (z) can be represented by the Laurent series
f ( z)
an ( z z0 ) n +
n0
b
n
n 1 (z z
(1)
n
0)
The coefficients of this Laurent series are given by the integrals
an
f ( z*) dz *
1
2 i C ( z * z
0)
n 1
and bn
1
( z * z0 )
2 i C
n 1
f ( z*) dz *
(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 C 2 and decreasing the
inner circle C1 until each of the two circles reaches a point where f (z) is singular.
In the important special case that z 0 is the only singular point of f (z) inside C 2 , this circle can be shrunk to
the point z 0 giving convergence in a disk except at the center.
Remark: 1. Instead of (1) and (2) we may write, denoting bn by a n
f ( z)
a n ( z z 0 ) n where a n
n
1
f ( z ) dz
2 i C ( z z
0)
n 1
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 z
Solution sin z
Hence z
Therefore, z
5
n0
5
sin z with center 0.
2n 1
(1) n z
(2n 1) !
sin z =
sin z =
5
n0
n0
for all z.
2n 4
( 1) n z
(2n 1) !
for all non-zero complex number z.
2n 4
( 1) n z
(2n 1) !
for z 0.
Here the annulus of convergence is the whole complex plane without the origin.
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Unit IV
Functions of Complex Variables
Example 88 Substitution
1
2 z
Find the Laurent series of z e with center 0.
Solution e z
n0
1
2 z
Therefore, z e =
z
n0
for z 0.
n!
Example 89 Develop f ( z )
1
in
1 z
i) non-negative powers of z
ii) negative powers of z
1
is singular at z = 1.
1 z
1
=
1 z
z
n0
1
Therefore,
=
1 z
ii)
z
for z 0.
n!
2n
n0
Solutions i) f ( z )
n
1
zn
for all z. Hence e z
n!
1
=
1 z
n
for z < 1.
z
n0
1
1
z (1 )
z
1
Therefore,
=
1 z
=
n
for z < 1.
1
z
z
n0
n0
1
=
n0
1
z
n 1
for z 1.
for z 1.
n 1
z
n
Example 90 Laurent series in different concentric annuli
Find all the Laurent series of
Solution f ( z )
1
z z4
3
1
z z4
3
with center 0.
is singular at z = 0 and z = 1, thus we consider two Laurent series for f (z).
I. On the annulus 0 < z < 1.
1 1
1
=
z
= 3
z3 z4
z 1 z n 0
Therefore,
1
z z
3
4
=
z
n0
n3
n3
.
for 0 < z < 1.
II. On the annulus z 1.
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Unit IV
Functions of Complex Variables
1
= 1 for z 1.
n 4
1 1
z3 z4
n0 z
z
1
1
Therefore, 3
=
for z 1.
n 4
4
z z
n0 z
1
1
= 4
z
Example 91 Find all the Taylor and Laurent series of f ( z )
2z 3
z 3z 2
2
with center 0.
Solution In terms of partial fractions
f ( z)
1
1
.
1 z 2 z
1
1
1
1
n
z
for
<
1
and
=
=
for z 1.
z
n
1
1
1 z
z
n0
n
0
z
1
z
n
n
1
1
1 1
z
2
Similarly,
= n 1 for z < 2 and
=
= n 1 for z 2.
2
2z
2z
z
n0 2
n0 z
1
z
1 n
Therefore, f (z ) = 1 n 1 z for z < 1,
n0
2
1
Now
=
1 z
zn
1
f (z ) = n 1 n 1 for 1 < z < 2,
n0 2
2
1
and f (z ) = 1 2 n n 1 for z 2.
n0
z
1
1
1
Example 92 Find the Laurent series of f ( z )
that converges in the annulus < z 1 < ,
2
4
2
1 z
and determine the precise region of convergence.
Solution f (z) is not analytic at z 1 .
Taylor’s series of f (z) at z = 0 is
1
z
=
1 z2
n0
But the annulus
powers of z 1.
2n
for z < 1.
1
1
< z 1 <
contains z = 1 as its center. Hence we must develop a series in
4
2
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Unit IV
Functions of Complex Variables
1
1
1
1
2
Now
=
( z 1) ( z 1)
z
1
1 z2
1 1 ( z 1) n
n
=
=
(
1
)
n
z 1 2 n 0 2
Therefore,
1
=
1 z2
n0
( z 1)
2
n 1
n 1
(1)
n 1
1
z 1
1
2
n0
n 1
( z 1)
2
n 1
(1)
n 1
for z 1 < 2.
for 0 < z 1 < 2.
Note that: If f (z) in Laurent’s theorem is analytic inside the inner circle C 2 the coefficients bn 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 = z 0 if f (z) is not analytic (perhaps not even defined) at z = z 0 , but every
neighborhood of z = z 0 contains points at which f (z) is analytic.
The two types of singularities
i) We call z = z 0 an isolated singularity of f (z) if z = z 0 has neighborhood without singularity of f (z).
Example 93 f (z) = tan z has an isolated singularities at z (2n 1)
2
, where n Z.
ii) Non-isolated singularity is a singularity z = z 0 for which every neighborhood of z 0 contains another
singularity of f (z).
Example 94 f (z) = tan
1
has no isolated singularity at z = 0.
z
Now consider isolated singularities only.
Suppose f (z) has an no isolated singularity at z = z 0 , then there is neighborhood of z 0 say
0 < z z0 < R for some R in which f (z) is analytic except at z 0 . Hence in the region
0 < z z0 < R f (z) has the Laurent series
f ( z)
The series
an ( z z0 ) n +
n0
bn ( z z 0 )
n 1
n
an ( z z 0 ) n is analytic at z = z 0 and the series
n0
(1)
bn ( z z 0 )
n 1
n
is called the
principal part of (1).
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Unit IV
Functions of Complex Variables
If it has infinitely many terms, then it is of the form
b1
b
b2
b
m
3
+
+...+
, where bm ≠ 0.
2
3
z z0 ( z z0 )
( z z0 ) m
( z z0 )
The singularity of f (z) at z = z 0 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 = z 0 .
Example 95 Laurent series of f (z) =
1
1
z z4
3
at z = 0 is
n3
z
=
for 0 < z < 1.
z3 z4 n 0
Therefore, f (z) has a pole of order three at z = 0.
Example 96 Poles
1
3
+
has a simple pole at z = 0 and a pole of
5
z ( z 2)
( z 2) 2
The function f (z) =
fifth order at z = 2.
Example 97 essential singularities
1
1
Let f ( z ) e z . Then the Laurent series of e z is:
1
e z=
n0
n
z
.
n!
Therefore, f (z) has an isolated essential singularity at z = 0.
Example 98 essential singularities
1
1
. Then the Laurent series of sin is:
z
z
n
(1)
1
sin =
.
z n 0 (2n 1) ! z 2n 1
Let f ( z ) sin
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 = z 0 , then
im f ( z )
z z0
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Unit IV
Functions of Complex Variables
Theorem 4.34 (Picard’s Theorem)
If f (z) is analytic and has an isolated essential singularity at a point z 0 ,
then it takes on every value, with at most one exceptional value in an
arbitrary small neighborhood of z 0 .
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 = z 0 in D if
f ( z 0 ) = 0. We also say that this zero is of order n if not only f but also the derivatives f ' , f ' ' ,
f
' ' ' , . . ., f
( n 1)
are all zero at z = z 0 but f
(n )
( z ) ≠ 0.
0
A zero of first order is called a simple zero; for it f ( z 0 ) = 0 but f ' ( z ) 0 . For a second order
0
zero f ( z 0 ) = 0 and f ' ( z ) 0 but f ' ' ( z ) 0 and so on.
0
0
Example 99 Let f ( z ) 1 cos z and g ( z ) (1 cos z ) 2 .
Now f ( z ) 0 if and only if z 2n , where n Z
and f ' ( z ) sin z , f ' ' ( z ) cos z but f ' (2n ) 0 while f ' ' (2n ) 1 .
Therefore, f has second order zeros at z 2n , where n Z.
Similarly, g ( z ) 0 if and only if z 2n , where n Z
and g ' (2n ) g ' ' (2n ) g ' ' ' (2n ) 0 while g
( 4)
(2n ) 6 0 .
Therefore, g has fourth order zeros at z 2n , 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 = z 0 and have a zero of n th order z = z 0 .
Then
1
has a pole of n th order z = z 0 .
f ( z)
The same holds for
h ( z)
if h (z) is analytic at z = z 0 and h ( z 0 ) 0 .
f ( z)
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Unit IV
Functions of Complex Variables
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 b1 to evaluate complex integrals in a very
elegant and simple fashion. b1 will be called the residue of f (z) at z = z 0 .
Residue
Now we need to evaluate complex integrals of the form
f ( z ) dz
C
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 = z 0 inside C, but is otherwise analytic on C and inside C, then f (z) has
a Laurent series
f ( z)
b
n
an ( z z0 ) n +
n0
n 1 (z z
0)
n
that converges for all points near z = z 0 , (except at z = z 0 itself) in some domain of the form
0 z z0 R . The coefficient b1 of ( z z 0 )
b1 =
1
2 i
1
is given by:
f ( z) dz
C
and hence
C f ( z ) dz = 2 i b1 .
Here we integrate counterclockwise around the simple closed path C that contains z = z 0 in its interior. The
coefficient b1 is called the residue of f (z) at z = z 0 and we denote it by
b1 = Re s f ( z )
z z0
Example 100 (Evaluation of an integral by means of a residue)
Evaluate
z
C
Solution sin z =
n0
Hence, z
4
4
sin z dz , where C is a unit circle oriented counterclockwise.
n
2n 1
( 1) z
(2n 1) !
sin z =
n0
for any z.
n
2n 3
(1) z
(2n 1) !
for z 0.
Thus, this series shows that f (z) has a pole of third order at z = 0 and the residue
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Unit IV
Functions of Complex Variables
b1 =
Therefore,
4
z
C
sin z dz =
Example 101 Integrate f ( z )
Solution f ( z )
1
1
.
3!
6
1
3
i.
1
z z
3
clockwise around the circle C: z
4
1
.
2
has singularities at z = 0 and z = 1. But z = 1 lies outside of C.
z z4
3
Hence we need to the residue of f (z) at 0.
Thus, the Laurent series that converges for 0 < z < 1 is given by:
f ( z)
1 1
z 3 1
1
=
z z 3
n0
zn =
z
n0
n3
for 0 < z < 1.
Hence, b1 = 1 and clockwise integration yields
dz
3 4
Cz z
Therefore,
= 2 i Re s f ( z ) 2 i .
z 0
dz
C z 3 z 4 = 2 i .
Two Formulas for Residue at Simple Poles
Let f (z) have a simple pole at z = z 0 . The corresponding Laurent series (with m = 1) is
f ( z)
b1
z z0
an ( z z0 ) n
n0
0 < z z0 < R
Here, b1 ≠ 0. Multiplying both sides by z z 0 we have
( z z 0 ) f ( z ) b1
Now taking
an ( z z0 )
n0
n 1
im on both sides of this equation we get:
z z0
Re s f ( z ) b im ( z z ) f ( z )
1
0
z z0
z z0
Example 102 (Residue at a simple pole)
f ( z)
9z i
z ( z 1)
2
. Then f ( z )
9z i
.
z ( z i) ( z i)
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Unit IV
Functions of Complex Variables
9z i
5i ,
z i
z i z ( z i)
9z i
and Re s f ( z ) im
4i .
z i
z i z ( z i)
Hence, Re s f ( z ) im
9z i
Re s f ( z ) im
i
z 0
z i ( z i) ( z i)
Another Simpler Method for the Residue at a Simple Pole
Let f ( z )
p ( z)
where p (z) and q (z) are analytic with p ( z 0 ) ≠ 0 and q (z) has a simple zero at
q ( z)
z = z 0 . Hence f (z) has a simple pole at z = z 0 . By definition of a simple zero, q (z) has a Taylor series of
the form
q ( z)
n 1
( z z0 )
n
n!
q
(n)
( z0 )
Thus, substitution yields
( n)
q
(z )
Re s f ( z ) im ( z z ) p ( z ) = im
0
n
(
z
z
)
0
0
n!
z z 0 n 1
z z0
z z0
q ( z)
p ( z)
p ( z0 )
n 1
=
= im ( z z 0 )
( n)
z z0
q
( z0 )
q ' ( z0 )
n
!
n
1
Re s f ( z ) p ( z 0 ) .
Therefore, b1
z z0
q ' ( z0 )
Example 103 Let f ( z )
9z i
z ( z 2 1)
1
( z z0 ) p ( z)
. Then f (z ) has simple pole at z = 0, z = i and z = i.
Now let p (z) = 9z + i and q ( z ) z 3 z .
Then p (0) = i ≠ 0 while q (0) = 0 and q ' (0) = 1.
Therefore, b1 Re s f ( z ) i .
z 0
Similarly, p (i) = 10 i ≠ 0 while q (i) = 0 and q ' (i ) = 2.
Therefore, b1 Re s f ( z ) 5 i .
z i
and p ( i) = 8 i ≠ 0 while q ( i) = 0 and q ' ( i ) = 2.
Therefore, b1 Re s f ( z ) 4 i .
z i
Example 104 Find all poles and the corresponding residues of the function
f ( z)
cosh z
z4 1
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Unit IV
Functions of Complex Variables
Solution Let p ( z ) cosh z and q ( z ) z 4 1 . p ( z ) cosh z is entire and q ( z ) z 4 1
has simple zeros at i, i, 1 and 1 . Hence f (z) has simple poles at these points.
Since q ' ( z ) 4 z 3 , the residues are equal to the values of
cosh
Therefore, , Re s
z i 4 z3
z i Re s cosh
,
4 z i 4 z3
cosh z
4 z3
at these points.
cosh
z
i
, Re s
4 z 1 4 z 3
z
cosh
4
cosh z cosh
and Re s
.
z 1 4 z 3
4
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 = z 0 . Then by definition of
such a pole, the Laurent series of f (z) at z = z 0 is:
f ( z)
b
n
an ( z z0 ) n +
n0
n 1 (z z
0)
Multiplying both sides by ( z z 0 ) m we get:
( z z0 ) m f ( z)
an ( z z0 )
n0
nm
n
where bm 0 .
+
bn ( z z 0 )
n 1
mn
(*)
Let g ( z ) ( z z 0 ) m f ( z ) . Then the residue b1 of f (z) at z = z 0 is the coefficient of the power series
( z z0 )
m 1
Thus, b1 =
Therefore,
in the Taylor series of g (z).
1
( m 1)
g
( z0 ) .
(m 1) !
( m 1)
1
Re s f ( z )
m
im d
( m 1) ( ( z z 0 ) f ( z ) )
z z0
(m 1) ! z z 0
dz
In particular, for a second order pole m = 2
1
Re s f ( z )
im
z z0
(m 1) ! z z 0
Example 105 The function f ( z )
( z z0 ) 2 f ( z)'.
50 z
( z 4) ( z 1) 2
has a pole of second order at z = 1 and a simple
pole at z = 4.
( z 1) 2 f ( z )' =
z 1
Hence, Re s f ( z ) im
z 1
im d 50 z
z 1 dz z 4
d 200
= im
= 8
2
z 1 dz
z 4
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Unit IV
Functions of Complex Variables
and Re s f ( z )
z4
im ( z 4) f ( z ) ' = im d 50 z 8 .
z4
z 4 dz z 12
Example 106 Residue from partial fraction
50 z
Let f ( z )
.
( z 4) ( z 1) 2
8
8
10
Then f ( z )
z 4 z 1 ( z 1) 2
Therefore, Re s f ( z ) = 8, and Re s f ( z ) = 8 and f (z) has a simple pole at z = 4 and a second order
z 1
z4
pole at z = 1.
Example 107 Integration around a second order pole
Evaluate
50 z
C ( z 4) ( z 1) 2 dz where C is any closed path such that z + 1 is inside C and
z = 4 is outside C.
Solution
50 z
C ( z 4) ( z 1) 2 dz = 2 i
Therefore,
50 z
C ( z 4) ( z 1) 2 dz =
Re s
z 1
50 z
16 i
2
(
z
4
)
(
z
1
)
16 i .
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 z1 , z 2 , z 3 , . . . , z k inside C. Then
f ( z ) dz = 2 i
C
k
n 1
Re s f ( z )
z zn
(1)
the integration being taken counterclockwise around the path C.
Example 108 Evaluate
4 3z
2 dz , where C is any closed path oriented counterclockwise such that
C z z
i) 0 and 1 are inside C.
iii) 1 is inside C and 0 is outside C.
ii) 0 is inside C and 1 is outside C.
iv) 0 and 1 are outside C.
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Unit IV
Functions of Complex Variables
Solutions The integrand has a simple pole at 0 and 1, with residues:
Re s 4 3 z = im 4 3z = 4 and Re s 4 3 z = im 4 3 z = 1.
z 1 z
z 0 z2 z
z 1 z 2 z
z 0 z 1
Therefore, i) 6 i
iii) 2 i
ii) 8 i
iv) 0.
Example 109 (Poles and Essential Singularities)
Evaluate
C
ze z
z
z 4 16 z e
dz , where C is the ellipse 9 x 2 y 2 9 oriented
Counterclockwise.
Solutions The first term of the integrand has a simple pole at 2 i and at 2 .
Now 2 i is inside C with residues
z
ze z
1
1
Re s z e
Re
s
and
while 2 lie outside C.
4
=
4
=
z 2 i z 16
z 2 i z 16
16
16
2
The second term of the integrand has an essential singularity at 0, with residue
as obtained
2
from
ze
Therefore,
C
z
n
n0
z n n!
= z
ze z
z
z
e
z 4 16
= z
2 1
...
2 z
2
dz = 1 2 i =
2
8
2 1
i .
4
Example 110 (Confirmation of an earlier result)
Evaluate
dz
C (z z
0)
m
, where m N and C is any simple closed path oriented
Counterclockwise enclosing z 0 .
Solutions
1
is its own Laurent series with center z = z 0 consisting of this one term principal
( z z0 ) m
part and
Therefore,
1
zz
0
dz
=
( z z0 ) m
Re s
z z0
C
= 1, m = 1 and Re s
z z0
2 i
0
1
z z0 m
= 0, m = 2, 3, 4, . . .
if m 1
.
if m 2, 3, 4, ...
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
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Unit IV
Functions of Complex Variables
2
0 F (cos , sin ) d
I
(1)
where F (cos , sin ) is a real rational function of cos and sin and is finite on the interval of
integration.
Setting z e
i
cos =
, we get:
1
1
1
1
( z ) and sin =
(z ) .
2
z
2i
z
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 z = 1 in counterclockwise. Since z e
d
i
, we have
dz
and the integral takes the form
iz
dz
I f ( z)
iz
C
the integration being taken counterclockwise around the unit circle.
2
d
0
2 .
2 cos
1
1
dz
Solution We use cos = ( z ) and d
. Then the integral becomes
2
z
iz
dz
iz
dz
2
=
C
1
1
i C z 2 1 z 2 1
2 z
2
z
Example 111 Show that
2 1 and the other at 2 1 .
Thus, the integrand has two simple poles, one at
outside the unit circle z = 1 and
Re s
z 2 1 z
2
Therefore,
0
d
2 cos
2 1 lies
2 1 lies inside C where the residue is:
1
2 1 z
1
= .
2
2 1
2 .
Improper Integrals of Rational Functions
We consider the real integrals of the type
f ( x) dx
(1)
The improper integral has the meaning
0
f ( x) dx =
b
im
f ( x) dx im f ( x) dx
a a
b 0
Prepared by Tekleyohannes Negussie
(2)
125
Unit IV
Functions of Complex Variables
If both limits exist, we may couple the two independent passages to and and write
R
f ( x) dx =
im
f ( x) dx
R R
(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
b
R
2
2
im
im R R 0 , but im
x dx 0 .
x
dx
b 0
R R
R 2
2
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
f ( x) dx = 2 i
Re s f ( z) =
S
f ( z ) dz
R
R f ( x) dx
where S is as shown in the figure.
y
R
R
x
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:
R
f ( x ) dx = 2 i
R
Re s f ( z) f ( z) dz
S
Now as R tends to infinity the value of the integral over S approaches to zero and hence
f ( x) dx = 2 i
Re s f ( z)
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 f ( z )
0
dx
1 x
4
.
2 2
i
1
3
has four simple poles at the points z1 e 4 , z 2 e 4
4
1 z
i
i
3
, z3 e 4
i
and z 4 e 4 .
Now z1 and z 2 lie in the upper half-plane while z 3 and z 4 lie in the upper half-plane, and
Prepared by Tekleyohannes Negussie
126
Unit IV
Functions of Complex Variables
1
1 i
1
Re s f ( z ) = im
= e 4 = e
3
z z1 4 z
z z1
4
4
3
1
1 4 i 1
Re s f ( z ) = im
e
e
=
=
z z2 4 z 3
z z2
4
4
9
and
2
Thus,
Re s f ( z ) = 1 e
4
k 1 z zk
Therefore,
Since f ( x)
dx
1 x
4
1
sin
4
2
i
4
1
e
4
+
i
4
=
1
4
i
4
i
4
e
.
i
4
e
i
4
i
= sin
2
4
.
is an even function, we get:
1 x4
dx
0
1 x
4
.
2 2
Example 114 Another improper integral
Show that
Solution f ( z )
x2 1
x 5x 4
4
2
z2 1
z 4 5z 2 4
dx
6
.
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:
2
Re s f ( z ) = im z 1 = 5 i
z 2i
z 2 i 4 z 3 10 z
12
z2 1
1
i.
and Re s f ( z ) = im 3
=
z i
z i 4 z 10 z
3
2
Re s f ( z ) = 1 i .
Thus,
12
k 1 z zk
Therefore,
x2 1
x 4 5x 2 4
dx
6
.
Prepared by Tekleyohannes Negussie
127
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