Department Of Mathematics

UNIT – II
COMPLEX ANALYSIS
Topic Learning Outcomes:
At the end of this unit, student will be able to

1. Understand the concept of analytic functions, C-R equations, and harmonic

functions.
2. Construct analytic functions using Milne-Thomson method.
3. Determine harmonic conjugates, orthogonal trajectories of family of curves,
potential functions and stream functions in fluid flow.
4. Develop Taylor’s and Laurent’s series of a complex function in the given region.
5. Identify the singularities and poles of a complex function.
6. Evaluate the integral of a complex function over a simple closed curve using
Cauchy’s theorem and Residue theorem.

Complex analysis is the theory of functions of a complex variable. It is the branch of

mathematical analysis that investigates functions of complex variable. The term complex
number came from Carl Friedrich Gauss in 1831 from the idea of geometric representation
of complex numbers. Gauss mentions the theorem that was known later as Cauchy’s
theorem. Augustin-Luis Cauchy (1789-1857) was a revolutionary in Mathematics and a
highly original founder of modern complex function theory who discovered and
rediscovered countless amazing results in the area of complex analysis along with
constructing the set of complex numbers in 1847.Some of these results that will be
emphasized are Cauchy’s Integral Theorem and Residue Theorem. The residue calculus is
an important tool in evaluating definite integrals, summing series, and discovering integral
expressions for the roots of equations and the solutions of differential equations. It is also
useful calculating real definite integrals, integrals involving sines and cosines, improper
integrals.
Complex analysis is useful in many branches of Mathematics, including algebraic
geometry, number theory and analytic combinatory. It is also useful in physics including
hydro dynamics, thermodynamics and particularly quantum mechanics. Complex analysis
has many applications in engineering fields such as nuclear, aerospace, mechanical and
electrical engineering. In control theory knowing the location of poles and zeros gives the
conclusion about the stability of the system.
Complex equations and their graphs are used to visualize electrical & fluid flow in the real
world. Murray R Spiegel described complex analysis as “one of the beautiful as well as
useful branches of Mathematics”.

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Basic Concepts:

• If 𝑥, 𝑦 are real numbers & 𝑖 = √−1 then 𝑧 = 𝑥 + 𝑖𝑦 is a complex number whose

real part is 𝑥 & imaginary part is 𝑦.
• Two complex numbers 𝑧1 = 𝑥1 + 𝑖𝑦1 & 𝑧2 = 𝑥2 + 𝑖𝑦2 are equal if and only if
𝑥1 = 𝑥2 , 𝑦1 = 𝑦2 .
• 𝑧 = 𝑥 + 𝑖𝑦 is equal to zero if 𝑥 = 0 & 𝑦 = 0.
• If 𝛼 is a real number & 𝑧 = 𝑥 + 𝑖𝑦 is complex number then 𝛼𝑧 = 𝛼𝑥 + 𝑖𝛼𝑦 is a
complex number.
• If 𝑧1 = 𝑥1 + 𝑖𝑦1 & 𝑧2 = 𝑥2 + 𝑖𝑦2 , then
o 𝑧1 ± 𝑧2 = (𝑥1 ± 𝑥2 ) + 𝑖(𝑦1 ± 𝑦2 )
o 𝑧1 ∙ 𝑧2 = (𝑥1 𝑥2 − 𝑦1 𝑦2 ) + 𝑖(𝑥1 𝑦2 + 𝑥2 𝑦1 )
𝑧1 (𝑥1 𝑥2 +𝑦1 𝑦2 )+𝑖(𝑥2 𝑦1 −𝑥1 𝑦2 )
o =
𝑧2 𝑥 2 +𝑦 2
1 𝑥−𝑖𝑦
o = 𝑥 2 +𝑦 2
𝑧
• If 𝑧 = 𝑥 + 𝑖𝑦 then 𝑧̅ = 𝑥 − 𝑖𝑦 is called complex conjugate of 𝑧.
• If 𝑧 = 𝑥 + 𝑖𝑦 then |𝑧| = √𝑥 2 + 𝑦 2 is non-negative real number.
• 𝑧 = 𝑥 + 𝑖𝑦 is represented by a point 𝑃(𝑥, 𝑦) in the 𝑋𝑌 plane, 𝑥 − axis is real axis,
𝑦 − axis is imaginary axis, plane is complex plane.
• 𝑧 = 𝑟𝑒 𝑖𝜃 is the polar form of complex number 𝑧 where 𝑟 = √𝑥 2 + 𝑦 2 and 𝜃 =
𝑦
𝑡𝑎𝑛−1 ( ⁄𝑥).
• |𝑧1 − 𝑧2 | = √(𝑥1 − 𝑥2 )2 + (𝑦1 − 𝑦2 )2 represents the distance between the points
𝑧1 &𝑧2 in complex plane.
• |𝑧 − 𝑧0 | = 𝑅 represents complex equation of circle with centre 𝑧0 & radius 𝑅.
• |𝑧 − 𝑧0 | < 𝑅 represents the region with in, but not on, a circle of
radius R centred at the point 𝑧0 , the point 𝑧0 is said to be interior point.
• |𝑧 − 𝑧0 | ≤ 𝑅 represents the region with in, and on, a circle of
radius R centred at the point 𝑧0 .
• |𝑧 − 𝑧0 | > 𝑅 represents the region outside the circle with centre 𝑧0 & radius 𝑅.

Neighbourhood of a point:
A neighbourhood of a point 𝑧0 in the complex plane is the set of all points 𝑧 such that
|𝑧 − 𝑧0 | < 𝛿, where 𝛿 is a small positive real number. Geometrically, a neighbourhood of a
point 𝑧0 is the set of all points inside a circle having 𝑧0 as the centre & 𝛿 as the radius.

Function of a Complex Variable:

Let S be a set of complex numbers. If to each complex number 𝑧 in S there corresponds a
unique complex number 𝑤 according to some rule, then 𝑤 = 𝑓(𝑧) is called a complex
function of 𝑧 defined on S.

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In Cartesian form w = f(z) = u(x, y) + i v(x, y)
In polar form w = f(z) = u(r, θ) + i v(r, θ)

Example:
f(z) = z 2
f(z) = (x + iy)2
= (x 2 − y 2 ) + i(2xy)
= u + iv
∴ u = x − y 2 , v = 2xy
2
2
In polar form, f(z) = (reiθ ) = r 2 e2iθ = r 2 (cos2θ + isin2θ)

∴ 𝑢 = 𝑟 2 𝑐𝑜𝑠2𝜃, 𝑣 = 𝑟 2 𝑠𝑖𝑛2𝜃.

Analytic function:
A function 𝑓(𝑧) is said to be analytic at a point 𝑧0 , if f(z) is differentiable not only at 𝑧0 but
at every point in some neighborhood of 𝑧0 .

• A function 𝑓(𝑧) is analytic in a domain if it is analytic at every point of the domain.

• An analytic function is also known as holomorphic, regular and monogenic

function.

Entire Function:
A function which is analytic everywhere (for all 𝑧 in the complex plane) is known as entire
function.

Example: Polynomials, rational functions, 𝑒 𝑧 , 𝑠𝑖𝑛𝑧, 𝑐𝑜𝑠𝑧 are entire.

A fundamental result of complex analysis is the C-R equations which gives the conditions,
a function must satisfy in order for a complex generalization of the derivative, the so-called
complex derivative, to exist. When the complex derivative is defined ‘everywhere’, the
function is said to be analytic.

Cauchy Riemann (C-R) Equations in Cartesian form:

If 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖 𝑣(𝑥, 𝑦) is differentiable at 𝑧 = 𝑥 + 𝑖𝑦 then at this point the first
order partial derivatives of 𝑢 and 𝑣 exist and satisfy the equations

u = v & u =−v
x y y x
are called the Cauchy- Riemann equations.

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Note: These conditions are only necessary for f (z) to be analytic. The sufficient
conditions for 𝑓(𝑧) to be analytic are that 𝑢, 𝑣, ux, uy, vx&vy are continuous.

Example: (i) 𝑓(𝑧) = 𝑒 𝑧 = 𝑒 𝑥 (𝑐𝑜𝑠𝑦 + 𝑖𝑠𝑖𝑛𝑦)

𝑢(𝑥, 𝑦) = 𝑒 𝑥 𝑐𝑜𝑠𝑦, 𝑣(𝑥, 𝑦) = 𝑒 𝑥 𝑠𝑖𝑛𝑦
𝑢𝑥 = 𝑒 𝑥 𝑐𝑜𝑠𝑦, 𝑢𝑦 = −𝑒 𝑥 𝑠𝑖𝑛𝑦
𝑣𝑥 = 𝑒 𝑥 𝑠𝑖𝑛𝑦, 𝑣𝑦 = 𝑒 𝑥 𝑐𝑜𝑠𝑦
Partial derivatives 𝑢𝑥 , 𝑣𝑥 , 𝑢𝑦 , 𝑣𝑦 are continuous & C-R equations 𝑢𝑥 = 𝑣𝑦 & 𝑢𝑦 = −𝑣𝑥 are
satisfied for all 𝑥 & 𝑦.
∴ 𝑓(𝑧) = 𝑒 𝑧 is analytic everywhere.

Harmonic Functions:
A function ∅ is said to be a Harmonic function if it satisfies Laplace equation ∇2 ∅ = 0.
𝜕2∅ 𝜕2∅
In Cartesian form ∅(𝑥, 𝑦) is Harmonic if + 𝜕𝑦 2 = 0.
𝜕𝑥 2

Consequences of Analytic function:

1. The real and imaginary part of an analytic function are harmonic. If 𝑓(𝑧) = 𝑢 + 𝑖𝑣 is an
analytic function, then 𝑢 & 𝑣 are harmonic conjugate function to each other.
Note: The converse of the above result is not true i.e., we can give examples of
functions like 𝑢 = 𝑒 𝑥 𝑐𝑜𝑠𝑦 & 𝑣 = 2𝑥𝑦, satisfying Laplace equation but not
satisfying C-R equations.
Example: Show that 𝑢 = 𝑥 3 − 3𝑥𝑦 2 + 3𝑥 2 − 3𝑦 2 + 1 is harmonic

𝑢 = 𝑥 3 − 3𝑥𝑦 2 + 3𝑥 2 − 3𝑦 2 + 1

𝑢𝑥 = 3𝑥 2 − 3𝑦 2 + 6𝑥 ; 𝑢𝑥𝑥 = 6𝑥 + 6

𝑢𝑦 = −6𝑥𝑦 − 6𝑦 ; 𝑢𝑦𝑦 = −6𝑥 − 6

∴ 𝑢𝑥𝑥 + 𝑢𝑦𝑦 = 6𝑥 + 6 − 6𝑥 − 6 = 0

Hence 𝑢 is harmonic.

2. If 𝑓(𝑧) = 𝑢 + 𝑖𝑣 is analytic function, then the family of curves 𝑢(𝑥, 𝑦) =

𝑢0 & 𝑣(𝑥, 𝑦) = 𝑣0 , 𝑢0 & 𝑣0 being constants, intersect each other orthogonally.
Note: Converse of the above result is not true.

Problems:
1. Show that analytic function with constant real part is constant.
Proof: Given 𝑓(𝑧) = 𝑢 + 𝑖𝑣 is analytic

Given u is constant = k

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𝜕𝑢 𝜕𝑢
⇒ =0 =0
𝜕𝑥 𝜕𝑦

𝜕𝑣 𝜕𝑣
Since f is an analytic function =0 =0
𝜕𝑥 𝜕𝑦
⇒ 𝑣 is constant
𝑓(𝑧) = 𝑢 + 𝑖𝑣
𝜕𝑢 𝜕𝑣
𝑓 ′ (𝑧) = +𝑖 = 0 + 𝑖(0) = 0
𝜕𝑥 𝜕𝑥
⇒ 𝑓(𝑧) is constant.

2. Prove that analytic function with constant modulus is constant.

Proof: 𝑓(𝑧) = 𝑢 + 𝑖𝑣

|𝑓(𝑧)| = √𝑢2 + 𝑣 2 = 𝑐 (given)

|𝑓(𝑧)|2 = 𝑢2 + 𝑣 2 = 𝑐 2
differentiating w. r. t 𝑥 & 𝑦, we get
𝜕𝑢 𝜕𝑣
2𝑢 + 2𝑣 =0
𝜕𝑥 𝜕𝑥
𝑢𝑢𝑥 + 𝑢𝑢𝑦 = 0 … … (𝑖)
And
𝜕𝑢 𝜕𝑣
𝑢 +𝑣 =0
𝜕𝑦 𝜕𝑦
Using C-R equations
𝜕𝑣 𝜕𝑢
−𝑢 +𝑣 =0 … … . (𝑖𝑖)
𝜕𝑥 𝜕𝑥
Squaring and adding (i) and (ii)
2 2)
𝜕𝑢 2 𝜕𝑣 2
⇒ (𝑢 + 𝑣 [( ) + ( ) ]=0
𝜕𝑥 𝜕𝑥
⇒ |𝑓 ′ (𝑧)|2 = 0
⇒ 𝑓(𝑧) = constant
3. If 𝑓(𝑧) is analytic function , show that
𝜕2 𝜕2
[ 2 + 2 ] |𝑓(𝑧)|2 = 4|𝑓 ′ (𝑧)|2
𝜕𝑥 𝜕𝑦

Proof: let 𝑓(𝑧) = 𝑢 + 𝑖𝑣 be analytic.

∴ |𝑓(𝑧)| = √𝑢2 + 𝑣 2 or |𝑓(𝑧)|2 = 𝑢2 + 𝑣 2

differentiate w. r. t. 𝑥 partially twice we get
𝜕2
|𝑓(𝑧)|2 = 2[𝑢 𝑢𝑥𝑥 + 𝑢𝑥2 + 𝑣 𝑣𝑥𝑥 + 𝑣𝑥2 ]----------------(i)
𝜕𝑥 2

Similarly, we can also get

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𝜕2
|𝑓(𝑧)|2 = 2[𝑢 𝑢𝑦𝑦 + 𝑢𝑦2 + 𝑣 𝑣𝑦𝑦 + 𝑣𝑦2 ]----------------(ii)
𝜕𝑦 2

Adding (𝑖) and (𝑖𝑖) we have,

𝜕2 𝜕2
[ + ] |𝑓(𝑧)|2 = 2[𝑢(𝑢𝑥𝑥 + 𝑢𝑦𝑦 ) + 𝑣(𝑣𝑥𝑥 + 𝑣𝑦𝑦 ) + 𝑢𝑥2 + 𝑣𝑥2 + 𝑢𝑦2 + 𝑣𝑦2 ]
𝜕𝑥 2 𝜕𝑦 2

Since 𝑓(𝑧) = 𝑢 + 𝑖𝑣 is analytic, 𝑢 and 𝑣 are harmonic. hence

𝑢𝑥𝑥 + 𝑢𝑦𝑦 = 0, 𝑣𝑥𝑥 + 𝑣𝑦𝑦 = 0

Further we also have C-R equations 𝑣𝑦 = 𝑢𝑥 , 𝑢𝑦 = −𝑣𝑥

using these results in the R.H.S, we have

𝜕2 𝜕2
[ 2 + 2 ] |𝑓(𝑧)|2 = 2[𝑢. 0 + 𝑣. 0 + 𝑢𝑥2 + 𝑣𝑥2 + (−𝑣𝑥 )2 + (𝑢𝑥 )2 ]
𝜕𝑥 𝜕𝑦

𝜕2 𝜕2
𝑖. 𝑒. , [ 2 + 2 ] |𝑓(𝑧)|2 = 2[2𝑢𝑥2 + 2𝑣𝑥2 ] = 4[𝑢𝑥2 + 𝑣𝑥2 ] − − − (iii)
𝜕𝑥 𝜕𝑦

But 𝑓 ′ (𝑧) = 𝑢𝑥 + 𝑖𝑣𝑥

∴ |𝑓 ′ (𝑧)| = √𝑢𝑥2 + 𝑣𝑥2 𝑜𝑟 |𝑓 ′ (𝑧)|2 = 𝑢𝑥2 + 𝑣𝑥2

𝜕2 𝜕2
Using this in R.H.S of (iii) we have [𝜕𝑥 2 + 𝜕𝑦 2] |𝑓(𝑧)|2 = 4|𝑓 ′ (𝑧)|2

4. If 𝑓(𝑧) is a regular function of 𝑧 show that

2 2
𝜕 𝜕
{ |𝑓(𝑧)|} + { |𝑓(𝑧)|} = |𝑓 ′ (𝑧)|2
𝜕𝑥 𝜕𝑥

Proof: let 𝑓(𝑧) = 𝑢 + 𝑖𝑣 be regular (analytic)function.

|𝑓(𝑧)| = √𝑢2 + 𝑣 2
𝜕 1 𝜕𝑢 𝜕𝑣
|𝑓(𝑧)| = [2𝑢 + 2𝑣 ]
𝜕𝑥 2
2√𝑢 + 𝑣 2 𝜕𝑥 𝜕𝑥
2
𝜕 1
{ |𝑓(𝑧)|} = (𝑢2 𝑢𝑥2 + 𝑣 2 𝑣𝑥2 + 2𝑢𝑣 𝑢𝑥 𝑣𝑥 )
𝜕𝑥 2
√𝑢 + 𝑣 2

𝜕 2 1
Similarly {𝜕𝑦 |𝑓(𝑧)|} = √𝑢2 (𝑢2 𝑢𝑦2 + 𝑣 2 𝑣𝑦2 + 2𝑢𝑣 𝑢𝑦 𝑣𝑦 )
+𝑣 2

𝜕 2 𝜕 2 1
Adding {𝜕𝑥 |𝑓(𝑧)|} + {𝜕𝑥 |𝑓(𝑧)|} = 𝑢2 +𝑣2[(𝑢2 𝑢𝑥2 + 𝑣 2 𝑣𝑥2 + 2𝑢𝑣 𝑢𝑥 𝑣𝑥 ) + (𝑢2 𝑢𝑦2 +
𝑣 2 𝑣𝑦2 + 2𝑢𝑣 𝑢𝑦 𝑣𝑦 )]

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Since 𝑓(𝑧) = 𝑢 + 𝑖𝑣 is analytic we have C-R equations 𝑣𝑦 = 𝑢𝑥 , 𝑢𝑦 = −𝑣𝑥 ,using these in
the second bracket of the R.H.S we have,
2 2
𝜕 𝜕
{ |𝑓(𝑧)|} + { |𝑓(𝑧)|}
𝜕𝑥 𝜕𝑥
1
= 2 [(𝑢2 𝑢𝑥2 + 𝑣 2 𝑣𝑥2 + 2𝑢𝑣 𝑢𝑥 𝑣𝑥 ) + (𝑢2 𝑣𝑥2 + 𝑣 2 𝑢𝑥2 − 2𝑢𝑣 𝑢𝑥 𝑣𝑥 )]
𝑢 + 𝑣2
1
= 2 2
[𝑢2 (𝑢𝑥2 + 𝑣𝑥2 ) + 𝑣 2 (𝑢𝑥2 + 𝑣𝑥2 )]
𝑢 +𝑣
1
= 2 (𝑢2 + 𝑣𝑥2 )(𝑢2 + 𝑣 2 )
𝑢 + 𝑣2 𝑥
= 𝑢𝑥2 + 𝑣𝑥2
= |𝑓′(𝑧)|2

Construction of Analytic Function Using Milne-Thomson Method

Working procedure:
• Given 𝑢 or 𝑣 as functions 𝑥, 𝑦 we find 𝑢𝑥 , 𝑢𝑦 or 𝑣𝑥 , 𝑣𝑦 and consider 𝑓′(𝑧) = 𝑢𝑥 +
𝑖𝑣𝑥
• Given 𝑢, we use 𝐶 − 𝑅 equation 𝑣𝑥 = −𝑢𝑦 or given 𝑣 we use 𝑢𝑥 = 𝑣𝑦 so that
• 𝑓′(𝑧) = 𝑢𝑥 − 𝑖𝑢𝑦 or 𝑓′(𝑧) = 𝑣𝑦 + 𝑖𝑣𝑥
• We substitute the expression for the partial derivatives in R.H.S and then put 𝑥 = 𝑧,
𝑦 = 0 to obtain 𝑓′(𝑧) as a function of 𝑧
• Integrating w.r.t 𝑧, we get 𝑓(𝑧)

Applications:
Complex potential, steam and potential functions:
In a fluid flow, the analytic function 𝑤(𝑧) = ∅(𝑥, 𝑦) + 𝑖𝜓(𝑥, 𝑦) is known as
complex potential, ∅(𝑥, 𝑦)is known as velocity potential, 𝜓(𝑥, 𝑦) is known as
stream function. The velocity potential function and stream function are harmonic
functions that satisfy Laplace equation. The equipotential lines ∅(𝑥, 𝑦) = 𝑐, the
stream lines 𝜓(𝑥, 𝑦) = 𝑑 cut orthogonally.
In the study of electrostatics and gravitational fields, the curves ∅(𝑥, 𝑦) = 𝑐and
𝛹(𝑥, 𝑦) = 𝑑 are called equipotential lines and lines of force respectively.
In heat flow problems, the curves ∅(𝑥, 𝑦) = 𝑐 and 𝜓(𝑥, 𝑦) = 𝑑 are known as
isothermals and heat flow lines respectively.

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Problems:

1. If 𝑤(𝑧) = 𝑢 + 𝑖𝑣 is the complex potential function. Construct analytic function 𝑤(𝑧)

whose potential function is 𝑢 = 𝑙𝑜𝑔√𝑥 2 + 𝑦 2. Also find the flux function 𝑣
1
Solution: 𝑢 = 𝑙𝑜𝑔√𝑥 2 + 𝑦 2 = 𝑙𝑜𝑔(𝑥 2 + 𝑦 2 )1⁄2 = 2 𝑙𝑜𝑔(𝑥 2 + 𝑦 2 )

1 1 𝑥
∴ 𝑢𝑥 = . 2 . 2𝑥 =
2 𝑥 + 𝑦2 𝑥2 + 𝑦2

1 1 𝑦
𝑢𝑦 = . 2 2
. 2𝑦 = 2
2 𝑥 +𝑦 𝑥 + 𝑦2

Consider 𝑓′(𝑧) = 𝑢𝑥 + 𝑖𝑣𝑥 but 𝑣𝑥 = −𝑢𝑦 (𝐶 − 𝑅 equation)

𝑥 𝑦
∴ 𝑓 ′ (𝑧) = 𝑢𝑥 − 𝑖𝑢𝑦 = −𝑖 2 … . . (1)
𝑥2 +𝑦 2 𝑥 + 𝑦2

Putting 𝑥 = 𝑧 and 𝑦 = 0,

𝑧 1 1
𝑓 ′ (𝑧) = − 𝑖0 = ∴ 𝑓(𝑧) = 𝑢𝑥 + 𝑖𝑢𝑦 = ∫ 𝑑𝑧 + 𝑐
𝑧2 +0 𝑧 𝑧

i.e., 𝑤(𝑧) = 𝑙𝑜𝑔𝑧 + 𝑐, where 𝑐 is complex constant.

𝑦
Separating real and imaginary parts, 𝑣 = 𝑡𝑎𝑛−1 .
𝑥

2. Find the analytic function 𝑓(𝑧) whose imaginary part is 𝑒 𝑥 (𝑥𝑠𝑖𝑛 𝑦 + 𝑦 cos 𝑦)

Solution: Let 𝑣 = 𝑒 𝑥 (𝑥𝑠𝑖𝑛 𝑦 + 𝑦 cos 𝑦)

∴ 𝑣𝑥 = 𝑒 𝑥 (𝑠𝑖𝑛 𝑦) + (𝑥𝑠𝑖𝑛𝑦 + 𝑦 cos 𝑦)𝑒 𝑥 [by product rule]

i.e. 𝑣𝑥 = 𝑒 𝑥 (𝑠𝑖𝑛 𝑦 + 𝑥𝑠𝑖𝑛𝑦 + 𝑦 cos 𝑦) … … . (𝑖)

also 𝑣𝑦 = 𝑒 𝑥 (𝑥 𝑐𝑜𝑠 𝑦 − 𝑦 𝑠𝑖𝑛𝑦 + cos 𝑦) … … . (𝑖𝑖)

Consider 𝑓′(𝑧) = 𝑢𝑥 + 𝑖𝑣𝑥 but 𝑢𝑥 = 𝑣𝑦 (𝐶 − 𝑅 equation)

i.e., 𝑓 ′ (𝑧) = 𝑣𝑦 + 𝑖𝑣𝑥 = 𝑒 𝑥 (𝑥 𝑐𝑜𝑠 𝑦 − 𝑦 𝑠𝑖𝑛𝑦 + cos 𝑦) +

𝑖𝑒 𝑥 (𝑠𝑖𝑛 𝑦 + 𝑥𝑠𝑖𝑛𝑦 + 𝑦 cos 𝑦)

Putting 𝑥 = 𝑧 and 𝑦 = 0,

𝑓 ′ (𝑧) = 𝑒 𝑧 (𝑧 + 1) (since 𝑠𝑖𝑛 0 = 0, 𝑐𝑜𝑠 0 = 1)

∴ 𝑓(𝑧) = ∫(𝑧 + 1)𝑒 𝑧 𝑑𝑧 + 𝑐

Integrating by parts,
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𝑓(𝑧) = (𝑧 + 1)𝑒 𝑧 − ∫ 𝑒 𝑧 . 1𝑑𝑧 + 𝑐 = (𝑧 + 1)𝑒 𝑧 − 𝑒 𝑧 + 𝑐

i.e. 𝑓(𝑧) = 𝑧𝑒 𝑧 + 𝑐.
𝑥 4 −𝑦 4 −2𝑥
3. Find the analytic function whose real part is hence determine 𝑣.
𝑥 2 +𝑦 2

𝑥 4 −𝑦 4 −2𝑥
Solution: Given 𝑢= 𝑥 2 +𝑦 2

(𝑥 2 + 𝑦 2 )(4𝑥 3 − 2) − (𝑥 4 − 𝑦 4 − 2𝑥)2𝑥
∴ 𝑢𝑥 =
(𝑥 2 + 𝑦 2 )2

(𝑥 2 + 𝑦 2 )(−4𝑦 3 ) − (𝑥 4 − 𝑦 4 − 2𝑥)2𝑦
𝑢𝑦 =
(𝑥 2 + 𝑦 2 )2

Consider 𝑓′(𝑧) = 𝑢𝑥 + 𝑖𝑣𝑥 but 𝑣𝑥 = −𝑢𝑦 (𝐶 − 𝑅 equation)

∴ 𝑓 ′ (𝑧) = 𝑢𝑥 − 𝑖𝑢𝑦

Putting 𝑥 = 𝑧 and 𝑦 = 0,

𝑓 ′ (𝑧) = [𝑢𝑥 ](𝑧,0) − 𝑖[𝑢𝑦 ](𝑧,0)

𝑧 2 (4𝑧 3 −2)−(𝑧 4 −2𝑧)2𝑧

i.e., 𝑓 ′ (𝑧) = (𝑧 2 )2
− 𝑖(0)

4𝑧 5 − 2𝑧 2 − 2𝑧 5 + 4𝑧 2 2𝑧 5 + 2𝑧 2
= =
𝑧4 𝑧4
𝑧5 𝑧2 2
i.e., 𝑓 ′ (𝑧) = 2 𝑧 4 + 2 𝑧 4 = 2𝑧 + 𝑧 2

2 2
∴ 𝑓(𝑧) = ∫ (2𝑧 + 2
) 𝑑𝑧 + 𝑐 = 𝑧 2 − + 𝑐
𝑧 𝑧
2
Thus 𝑓(𝑧) = 𝑧 2 − 𝑧 + 𝑐

To find 𝑣, separate the R.H.S of 𝑓(𝑧) into real and imaginary parts.
2
i.e., 𝑢 + 𝑖𝑣 = (𝑥 + 𝑖𝑦)2 − 𝑥+𝑖𝑦 + 𝑐

2(𝑥 − 𝑖𝑦)
= (𝑥 2 + 𝑖 2 𝑦 2 + 2𝑥𝑖𝑦) − +𝑐
(𝑥 + 𝑖𝑦)(𝑥 − 𝑖𝑦)

2(𝑥 − 𝑖𝑦)
= (𝑥 2 − 𝑦 2 ) + 2𝑥𝑖𝑦 − +𝑐
𝑥2 + 𝑦2

2𝑥 2𝑦
= [𝑥 2 − 𝑦 2 − ] + 𝑖 [2𝑥𝑦 + 2 ]+𝑐
𝑥2 +𝑦 2 𝑥 + 𝑦2

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𝑥 4 − 𝑦 4 − 2𝑥 2𝑥 3 𝑦 + 2𝑥𝑦 3 + 2𝑥
𝑢 + 𝑖𝑣 = [ ]+𝑖[ ]+𝑐
𝑥2 + 𝑦2 𝑥2 + 𝑦2

Equating real and imaginary parts we observe that the real part 𝑢 is same as in the given
2𝑥 3 𝑦+2𝑥𝑦 3 +2𝑥
problem and the required 𝑣 = + 𝑐.
𝑥 2 +𝑦 2

3. Determine the analytic function 𝑓(𝑧) = 𝑢 + 𝑖𝑣 Given that

𝑢 = 𝑒 2𝑥 (𝑥𝑐𝑜𝑠2𝑦 − 𝑦𝑠𝑖𝑛2𝑦).

Solution: 𝑢 = 𝑒 2𝑥 (𝑥𝑐𝑜𝑠2𝑦 − 𝑦𝑠𝑖𝑛2𝑦)

𝑢𝑥 = 𝑒 2𝑥 . 𝑐𝑜𝑠2𝑦 + 2𝑒 2𝑥 (𝑥𝑐𝑜𝑠2𝑦 − 𝑦𝑠𝑖𝑛2𝑦)

= 𝑒 2𝑥 (𝑐𝑜𝑠2𝑦 + 2𝑥𝑐𝑜𝑠2𝑦 − 2𝑦𝑠𝑖𝑛2𝑦)

𝑢𝑦 = 𝑒 2𝑥 (−2𝑥𝑠𝑖𝑛2𝑦 − 2𝑦𝑐𝑜𝑠2𝑦 − 𝑠𝑖𝑛2𝑦)

= −𝑒 2𝑥 (2𝑥𝑠𝑖𝑛2𝑦 + 2𝑦𝑐𝑜𝑠2𝑦 + 𝑠𝑖𝑛2𝑦)

Consider 𝑓′(𝑧) = 𝑢𝑥 + 𝑖𝑣𝑥 = 𝑢𝑥 − 𝑖𝑢𝑦 (𝑏𝑦 𝐶 − 𝑅 equation)

Putting 𝑥 = 𝑧 and 𝑦 = 0,

𝑓 ′ (𝑧) = [𝑢𝑥 ](𝑧,0) − 𝑖[𝑢𝑦 ](𝑧,0)

i.e., 𝑓 ′ (𝑧) = 𝑒 2𝑧 (1 + 2𝑧)

∴ 𝑓(𝑧) = ∫(1 + 2𝑧)𝑒 2𝑧 𝑑𝑧

𝑒 2𝑧 𝑒 2𝑧 𝑒 2𝑧 𝑒 2𝑧
∴ 𝑓(𝑧) = (1 + 2𝑧) −2 = + 𝑧𝑒 2𝑧 − +𝑐
2 4 2 2
Thus 𝑓(𝑧) = 𝑧𝑒 2𝑧 + 𝑐

Also 𝑓(𝑧) = 𝑢 + 𝑖𝑣 = (𝑥 + 𝑖𝑦)𝑒 2(𝑥+𝑖𝑦) + 𝑐

= 𝑒 2𝑥 (𝑥 + 𝑖𝑦)(𝑐𝑜𝑠2𝑦 + 𝑖 𝑠𝑖𝑛2𝑦) + 𝑐

i.e., 𝑓(𝑧) = 𝑒 2𝑥 (𝑥𝑐𝑜𝑠2𝑦 − 𝑦 𝑠𝑖𝑛2𝑦) + 𝑒 2𝑥 (𝑥 𝑠𝑖𝑛2𝑦 + 𝑦 𝑐𝑜𝑠2𝑦) + 𝑐

𝑠𝑖𝑛2𝑥
4. Find the analytic function 𝑓(𝑧) whose real part is and hence find the
cosh 2𝑦−cos 2𝑥
imaginary part.
Solution: Given
𝑠𝑖𝑛2𝑥
𝑢=
cosh 2𝑦 − cos 2𝑥

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(cosh 2𝑦 − cos 2𝑥)(2 cos 2𝑥) − (sin 2𝑥)(2 sin 2𝑥)

∴ 𝑢𝑥 =
(cos h 2𝑦 − cos 2𝑥)2

− sin 2𝑥 (2 sin h 2𝑦)

𝑢𝑦 =
(cos h 2𝑦 − cos 2𝑥)2

Consider 𝑓′(𝑧) = 𝑢𝑥 + 𝑖𝑣𝑥 = 𝑢𝑥 − 𝑖𝑢𝑦 (𝑏𝑦 𝐶 − 𝑅 equation)

Putting 𝑥 = 𝑧 and 𝑦 = 0,

𝑓 ′ (𝑧) = [𝑢𝑥 ](𝑧,0) − 𝑖[𝑢𝑦 ](𝑧,0)

i.e.,

(1 − 𝑐𝑜𝑠2𝑧)(2𝑐𝑜𝑠2𝑧) − 2𝑠𝑖𝑛2 2𝑧
𝑓 ′ (𝑧) =
(1 − cos 2𝑧)2

2 cos 2𝑧 − 2 (2 cos 2 2𝑧 + sin2 2𝑧)

=
(1 − 𝑐𝑜𝑠2𝑧)2

−2(1 − cos 2𝑧) −2 −2

= = =
(1 − 𝑐𝑜𝑠2𝑧)2 (1 − 𝑐𝑜𝑠2𝑧) 2𝑠𝑖𝑛2 𝑧

Thus 𝑓 ′ (𝑧) = −𝑐𝑜𝑠𝑒𝑐 2 𝑧 ⇒ 𝑓(𝑧) = 𝑐𝑜𝑡𝑧 + 𝑐

Separate 𝑐𝑜𝑡𝑧 = cot (𝑥 + 𝑖𝑦) into real and imaginary parts to determine the imaginary part
𝑣

Consider 𝑓(𝑧) = 𝑐𝑜𝑡𝑧 + 𝑐

cos(𝑥+𝑖𝑦)
i.e., 𝑢 + 𝑖𝑣 = cot(𝑥 + 𝑖𝑦) = +𝑐
sin(𝑥+𝑖𝑦)

cos(𝑥 + 𝑖𝑦) sin (𝑥 + 𝑖𝑦)

= +𝑐
sin(𝑥 + 𝑖𝑦) sin (𝑥 + 𝑖𝑦)

1
[sin(𝑥 − 𝑖𝑦 + 𝑥 + 𝑖𝑦) + sin (𝑥 − 𝑖𝑦 − 𝑥 − 𝑖𝑦)]
= 2 +𝑐
1
[cos(𝑥 + 𝑖𝑦 − 𝑥 + 𝑖𝑦) − cos (𝑥 + 𝑖𝑦 + 𝑥 − 𝑖𝑦)]
2
sin 2𝑥 + sin (−2𝑖𝑦) sin 2𝑥 − 𝑖 sin ℎ2𝑦
= = +𝑐
cos(2𝑖𝑦) − cos 2𝑥 cos ℎ2𝑦 − cos 2𝑥
sin 2𝑥 −𝑠𝑖𝑛ℎ2𝑦
Thus 𝑢 + 𝑖𝑣 = [cos ℎ2𝑦−𝑐𝑜𝑠2𝑥] + 𝑖 [𝑐𝑜𝑠ℎ2𝑦−𝑐𝑜𝑠2𝑥] + 𝑐

(It may be observed that the real part 𝑢 is the given problem)

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−𝑠𝑖𝑛ℎ2𝑦
The required imaginary part 𝑣 = 𝑐𝑜𝑠ℎ2𝑦−𝑐𝑜𝑠2𝑥 + 𝑐.

5. If 𝑓(𝑧) = ɸ + 𝑖𝜓 represents complex potential of a two dimensional fluid flow

where ɸ is velocity potential, 𝜓 is stream function. the velocity potential ɸ for given
𝑥
stream function 𝜓 = (𝑥 2 − 𝑦 2 ) + 𝑥 2 +𝑦 2 .

(𝑥 2 +𝑦 2 )1−𝑥.2𝑥 𝑦 2 −𝑥 2
Solution: 𝜓𝑥 = 2𝑥 + (𝑥 2 +𝑦 2 )2
= 2𝑥 + (𝑥 2+𝑦 2)2

(𝑥 2 + 𝑦 2 )0 − 𝑥. 2𝑦 2𝑥𝑦
𝜓𝑦 = −2𝑦 + 2 2 2
= −2𝑦 − 2
(𝑥 + 𝑦 ) (𝑥 + 𝑦 2 )2

Consider 𝑓 ′ (𝑧) = ɸ𝑥 + 𝑖𝜓𝑥 but ɸ𝑥 = 𝜓𝑦 (𝑏𝑦 𝐶 − 𝑅 equation)

Putting 𝑥 = 𝑧 and 𝑦 = 0 we have,

𝑓 ′ (𝑧) = [𝜓𝑦 ](𝑧,0) − 𝑖[𝜓𝑥 ](𝑧,0)

−𝑧 2 1
i.e. 𝑓 ′ (𝑧) = 0 + 𝑖 (2𝑧 + (𝑧 2)2) = 𝑖 (2𝑧 − 𝑧 2)

1 1
∴ 𝑓(𝑧) = 𝑖 ∫ (2𝑧 − 2
) 𝑑𝑧 + 𝑐 = 𝑖 (𝑧 2 + ) + 𝑐
𝑧 𝑧
1
Thus 𝑓(𝑧) = 𝑖 (𝑧 2 + 𝑧) + 𝑐

To find ɸ,separate the R.H.S into real and imaginary parts

1
i.e., ɸ + 𝑖𝛹 = 𝑖 {(𝑥 + 𝑖𝑦)2 + 𝑥+𝑖𝑦} + 𝑐

𝑥 − 𝑖𝑦
= 𝑖 {(𝑥 2 + 𝑖 2 𝑦 2 + 2𝑥𝑖𝑦) + }+𝑐
(𝑥 + 𝑖𝑦)(𝑥 − 𝑖𝑦)

𝑥 − 𝑖𝑦
= 𝑖{(𝑥 2 − 𝑦 2 ) + 2𝑥𝑖𝑦} + 𝑖 { }+𝑐
𝑥2 + 𝑦2

𝑖𝑥 𝑦
= 𝑖(𝑥 2 − 𝑦 2 ) − 2𝑥𝑦 + + +𝑐
𝑥2 + 𝑦2 𝑥2 + 𝑦2
𝑦 𝑥
i.e. ɸ + 𝑖𝜓 = (−2𝑥𝑦 + 𝑥 2 +𝑦 2) + 𝑖 (𝑥 2 − 𝑦 2 + 𝑥 2 +𝑦 2) + 𝑐.

equating the real and imaginary parts we observe that the imaginary part 𝛹is same as the
𝑦
given problem and the required ɸ = −2𝑥𝑦 + 𝑥 2+𝑦 2 + c.

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6. In a two dimensional fluid flow, the complex potential is w(z) = u + iv show that
velocity potential 𝑢 = 𝑒 𝑥 (𝑥 cos 𝑦 − 𝑦 sin 𝑦) is harmonic and find its harmonic conjugate.

Solution: 𝑢 = 𝑒 𝑥 (𝑥 cos 𝑦 − 𝑦 sin 𝑦)

𝑢𝑥 = 𝑒 𝑥 . cos 𝑦 + (𝑥 cos 𝑦 − 𝑦 sin 𝑦)𝑒 𝑥

𝑖. 𝑒., 𝑢𝑥 = 𝑒 𝑥 (cos 𝑦 + 𝑥 cos 𝑦 − 𝑦 sin 𝑦)

Now 𝑢𝑥𝑥 = 𝑒 𝑥 . cos 𝑦 + (cos 𝑦 + 𝑥 cos 𝑦 − 𝑦 sin 𝑦)𝑒 𝑥

𝑖. 𝑒., 𝑢𝑥𝑥 = 𝑒 𝑥 (2 cos 𝑦 + 𝑥 cos 𝑦 − 𝑦 sin 𝑦) … . . (𝑖)

Also 𝑢𝑦 = 𝑒 𝑥 (−𝑥 sin 𝑦 − [𝑦 cos 𝑦 + sin 𝑦])

= −𝑒 𝑥 (𝑥 sin 𝑦 + 𝑦 cos 𝑦 + sin 𝑦)

Now 𝑢𝑦𝑦 = −𝑒 𝑥 (𝑥 cos 𝑦 + [−𝑦 sin 𝑦 + cos 𝑦] + cos 𝑦 )

𝑖. 𝑒., 𝑢𝑦𝑦 = −𝑒 𝑥 (2 cos 𝑦 + 𝑥 cos 𝑦 − 𝑦 sin 𝑦) … . . (𝑖𝑖)

(𝑖) + (𝑖𝑖) gives 𝑢𝑥𝑥 + 𝑢𝑦𝑦 = 0 ∴ u is harmonic.

To find 𝑓 ′ (𝑧)

consider 𝑓 ′ (𝑧) = 𝑢𝑥 + 𝑖𝑣𝑥 , but 𝑣𝑥 = −𝑢𝑦 ( C − R equation)

i.e., 𝑓 ′ (𝑧) = 𝑢𝑥 − 𝑖𝑢𝑦 putting 𝑥 = 𝑧 and 𝑦 = 0 then

integrating & seperating real & imaginary parts,

𝑣 = 𝑒 𝑥 (𝑥 sin 𝑦 + 𝑦 cos 𝑦) + 𝑐.

7. Find f(z) given that u – v = x3 + 3x2y – 3 xy2 – y3.

Solution: Partially differentiating w. r. t. x and y

ux – vx = 3x2 + 6xy – 3y2 ________________(1)

uy – vy = 3x2 - 6xy – 3y2 ________________(2)

𝐵𝑦 𝐶 − 𝑅 equation, the second equation reduces to

-vx – ux = 3x2 - 6xy – 3y2 ________________(3)

(1) +(3) gives vx= 3y2-3x2

(1) - (3) gives ux = 6xy

consider 𝑓 ′ (𝑧) = 𝑢𝑥 + 𝑖𝑣𝑥 = 6𝑥𝑦 − 𝑖(3x 2 − 3y 2 )

putting 𝑥 = 𝑧 and 𝑦 = 0

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𝑓 ′ (𝑧) = 0 − 𝑖(3𝑧 2 )

Thus 𝑓(𝑧) = −𝑖(𝑧 3 ) + 𝑐

Exercise:

1. Find the orthogonal trajectories of the following family curves.

(i) x 4 + 𝑦 4 − 6𝑥 2 𝑦 2 = constant.
(ii) 𝑥𝑦 + 𝑒 −𝑥 𝑐𝑜𝑠𝑦 = 𝑐

Ans: (i) x 3 𝑦 − 𝑥𝑦 3 = constant (ii) y 2 − x 2 − 2𝑒 −𝑥 𝑠𝑖𝑛𝑦 = 𝑐

2. Construct the analytic function whose imaginary part is

(i) –sinx sinhy (ii) ex [(x 2 − y 2 )cosy − 2xysiny]

Ans: (i) cosz + c (ii) i𝑒 𝑧 𝑧 2 + 𝑐

2 −𝑦 2
3. Construct the analytic function whose real part is e𝑥 𝑐𝑜𝑠2𝑥𝑦
2
Ans: e 𝑧 + 𝐶
𝑦
4. In a two dimensional fluid flow if the stream function is given by tan−1 𝑥 determine
the velocity potential.
5. The Laplace equation governs steady heat conduction in a rectangular plate. Prove
that the temperature function u = x 3 − 3𝑥𝑦 2 + 3𝑥 2 − 3𝑦 2 + 1satisfies Laplace
equation and determine the corresponding analytic function u+ iv.
6. If the potential function is log (𝑥 2 + 𝑦 2 ), find the flux function and the complex
potential function.
7. An electrostatic field in the xy-plane is given by the potential function ∅ = 𝑥 2 − 𝑦 2 ,
find the stream function.

Equation of a curve in complex plane:

A pair of equations of the form x = x(t), y = y(t), where ‘t’ is a parameter represents a curve
C in the x-y plane. The equation z = z(t) = x(t) + i y(t) represents the equation of the curve C
in the complex form, where ‘t ’varies over an appropriate interval.
Example: z = a eit , 0 ≤ t ≤ 2π represents equation of a circle centred at origin, radius a.

Simple curve:
The curve C is said to be simple if it does not intersect itself.
The curve C given by z(t) = x(t) + iy(t), a ≤ t ≤ b is simple if z(t1 ) ≠ z(t 2 ) for any two
different
values t1 and t2 in (a ,b).
Example: circle, semicircle etc.,

Simple closed curve:

A curve is said to be a simple closed curve if it is simple and the end points coincide.

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Example: circle, triangle, rectangle etc.,

Smooth curve:
A curve is said to be smooth if there exists a unique tangent at each of points(differential
curves)
A smooth curve does not contain sharp corners.
Example: Arc of a circle is a smooth curve. Triangle is not a smooth curve.

Contour:
A continues chain of finite number of smooth curves is called a contour.
Example:
Circle is contour (single smooth curve.)
Triangle is contour (chain of three lines)
Rectangle is a contour (chain of four lines)

Positively oriented curve:

A curve is called a positive orientated if any point z(t) on the curve
z(t) = x(t) + iy(t), a ≤ t ≤ b varies from the initial point to the terminal point as t varies
from a to b.

Simply connected region:

A region D is said to be simply connected if every simple closed curve lying entirely in D can
be shrunk a point without crossing the boundary of D.
A region which is not simply connected is called a multiply connected ( region having holes).

Line integral of a complex function:

The line integral of a complex function f(z) from z1 to z2 along a curve C is defined by

∫C f(z)dz = ∫C (u + iv)(dx + idy) = ∫C (udx − vdy) + i ∫C (vdx + udy) , where C is

called the path of integration.

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Taylor’s Theorem:
If a complex function f(z) is analytic at all points inside and on the circle C: |z − a| = r .Then
at each point inside C, we have
f′′ (a) f(n) (a)
f(z) = f(a) + (z − a)f ′ (a) + (z − a)2 + ⋯ = f(a) + ∑∞
n=1 (z − a)n
2! n!

the above series is called Taylor’s series expansion for the function f(z) about the point z = a.

f′′ (a) f(n) (0) n

In particular, if a = 0, f(z) = f(0) + zf ′ (0) + z 2 + ⋯ = f(0) + ∑∞
n=1 z is called
2! n!
Maclaurin’s series for the function f(z).

Examples:
z2 z3
1. Exponential series : ez = 1 + z + 2! + 3! …
z2 z3
2. Logarithmic series: log(1 + z) = z − + …
2 3
z3 z5
3. Sine series: sinz = z − 3! + 5! …
z2 z4
4. Cosine series: cosz = 1 − 2! + 4! …

Problems:
z−1
1. Expand f(z) = z+1 in Taylor’s series about the point (i) z = 0 (ii) z = 1.

z−1 2
Solution: f(z) = z+1 = 1 − z+1

(i) Taylor’s series expansion of f(z) about the point z = 0 is

f(n) (0) n
f(z) = f(0) + ∑∞
n=1 z ------ (1)
n!
(−1)n n!
f(0) = −1, f n (z) = −2 ((z+1)n+1 ) , f n (0) = 2(−1)n+1 n!
Substituting in (1),
z−1 zn
= −1 + ∑∞
n=1 2 n! (−1)
n+1 (n!)
= −1 + ∑∞
n=1 2(−1)
n+1 n
z
z+1

(ii) Taylor’s series expansion of f(z) about the point z = 1 is

f(n) (1)
f(z) = f(1) + ∑∞
n=1 (z − 1)n ----(2)
n!

z−1 2
f(z) = z+1 = 1 − z+1 , f(1) = 0

n (z)
(−1)n n! n (1)
(−1)n n! (−1)n+1 n!
f = −2 ( ),f = −2 n+1 =
(z + 1)n+1 2 2n

Substituting in (2),
z−1 (z−1)n
= ∑∞
n=1 (−1)n+1
z+1 2n

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2z3 +1
2. Find the Taylor’s series expansion of f(z) = about the point z = i.
z2 +z

2z3 +1 2z+1 1 z
Solution: f(z) = = (2z − 2) + z2+z = 2(z − 1) + z + z+1
z2 +z

By Taylor series expansion of f(z) about the point z = i is

f(n) (i)
f(z) = f(i) + ∑∞
n=1 (z − i)n ------ (1)
n!

i 3
f(i) = −
2 2
1 1 i
f ′ (z) = 2 − + , f ′ (i)
= 3 +
z 2 (z + 1)2 2

(−1)n n! (−1)n n! 1 1
f n (z) = n+1
+ n+1
, n ≥ 2, f n (i) = (−1)n n! { n+1 + },n ≥ 2
z (z + 1) i (1 + i)n+1

Substituting in (1),
∞
i 3 i 1 1
f(z) = − + 3 + (z − i) + ∑(−1)n n! { n+1 + } (z − i)n
2 2 2 i (1 + i)n+1
n=2

Laurent’s Theorem:
A complex function f(z) is analytic inside and on the boundary of the annular region D
bounded by two concentric circles C1 and C2 centred at a with C1 as outer circle and C2 as
inner circle, then for all z in D,

b
f(z) = ∑∞ n ∞ n
n=0 a n (z − a) + ∑n=1 (z−a)n ------ (1)

The above series is known as Laurent’s series of f(z) about the point z = a, where
1 f(w)
an = 2πi ∫c dw, n = 0,1,2,3 … . ----- (2)
1 (w−a)n+1

1 f(w)
bn = 2πi ∫c dw, n = 1,2,3 … . ----- (3)
2 (w−a)−n+1

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Note:

1. The first part in the R.H.S of (1) is called the ANALYTIC part of f(z) and the second
part is called the PRINCIPAL part of f(z).
2. R.H.S of (1) is a power series when it contains non-negative powers of (z-a).
3. The evaluation of the coefficients an, bn by using (2) and (3) is more complex. Usually
a rational function is expanded by using known expansions like Binomial,
Exponential and Logarithmic etc.,

Problems:
z2 −1
1. Obtain the power series expansion of the function f(z) = z2+5z+6 in the following regions.

(i) |z| < 2 (ii) 2 < |z| < 3 (iii) |z| > 3
z2 −1 5z+7 5z+7 3 8
Solution: f(z) = z2+5z+6 = 1 − = 1 − (z+2)(z+3) = 1 + z+2 − z+3
z2 +5z+6

z z
(i) |z| < 2 ⟹ | | < 1and | | < 1
2 3
3 8 3 8
f(z) = 1 + − = 1+ −
z+2 z+3 2(1 + ⁄2) 3(1 + z⁄3)
z
3 z −1 8 z −1
= 1 + 2 (1 + 2) − 3 (1 + 3)
3 z z 2 z 3 8 z z 2 z 3
= 1 + 2 {1 − 2 + (2) − (2) + − − −} − {1 − 3 + (3) − (2) − − −}
3
2 z
(ii) 2 < |z| < 3 ⟹ |z| < 1and |3| < 1
3 8 3 8
f(z) = 1 + − = 1+ − z
z+2 z+3 z(1 + 2⁄z) 3(1 + ⁄3)
3 2 −1 8 z −1
= 1 + z (1 + z) − 3 (1 + 3)
3 2 2 2 2 3 8 z z 2 z 3
= 1 + z {1 − z + (z) − (z) + − − −} − {1 − 3 + (3) − (2) − − −}
3
3 2
(iii) |z| > 3 ⟹ | | < 1and | | < 1
z z

3 8 3 8
f(z) = 1 + z+2 − z+3 = 1 + z(1+2⁄ ) − z(1+3⁄
z z)

3 2 −1 8 3 −1
= 1 + z (1 + z) − z (1 + z)
3 2 2 2 2 3 8 3 3 2 3 3
= 1 + z {1 − (z) + (z) − (z) + − − −} − z {1 − (z) + (z) − ( z) + − −}

z
2. Find the power series expansion of f(z) = (z2 +1)(z2+4) in the following regions.

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(i) |z| < 1 (ii) 1 < |z| < 2 (iii) |z| > 2
z z 1 1
Solution: f(z) = (z2 +1)(z2+4) = 3 {z2 +1 − z2+4}

z 2
(i) |z| < 1 ⟹ |z|2 < 1and | | < 1
2
z z 1 1
f(z) = = { 2 − }
+ 1)(z + 4) 3 z + 1 4(1 + (z⁄ )2 )
(z 2 2
2
z 1 2 −1
= 3 (1 + z 2 )−1 − 4 [1 + (z⁄2) ]
z z z 2 z 4 z 6
=3 (1 − z 2 + z 4 − − −) − 12 {1 − (2) + (2) − (2) + − − −}

1 1 2 z z 2
(ii) 1 < |z| < 2 ⟹ |z| < 1and |z| < 1, |2| < 1and |2| < 1

z z 1 1
f(z) = = { − }
(z 2 + 1)(z + 4) 3 z 2 (1 + 1 )
2 z 2
z2 4(1 + (2) )
−1
1 1 −1 z z 2
= [1 + z2 ] − 12 [1 + (2) ]
3z
1 1 1 1 2 1 3 z z 2 z 4 z 6
= 3 [z {1 − z2 + (z2) − (z2 ) + − −} − 4 {1 − (2) + (2) − (2) + − −}]
2 2 2 1 1 2
(iii) |z| > 2 ⟹ | | < 1and | | < 1, | | < 1and | | < 1
z z z z

z z 1 1
f(z) = (z2+1)(z2+4) = 3 { 1 − 4 }
z2 (1+ 2 ) z2 (1+ 2 )
z z

z 1 1 −1 1 4 −1
= { [1 + z2 ] } − z2 [1 + z2]
3 z2
1 1 1 2 4 4 2
= [{1 − z2 + (z2) − − −} − {1 − z2 + (z2 ) − − −}]
3z

1
3. Expand f(z) = (z+1)(z+3) in Laurent’s series valid for
(i) 0 < |z + 1| < 2, (ii) |z + 1| > 2
1
Solution: f(z) = (z+1)(z+3)

Let u = z + 1, then

u
(i) 0 < |z + 1| < 2 ⇒ |u| < 2 or |2| < 1

1 1 1 u
f(z) = u(u+2) = u = 2u (1 + 2)−1
2u(1+ )
2

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1 u u 2 u 3
= 2u {1 − (2) + (2) − (2) + − − −}

1 1 1 u u2
=2 {u − 2 + 22 − 23 + − − −}

1 1 1 z+1 1 z+1 2
=2 {z+1 − 2 + ( 22 ) + 2 ( ) − − − −}
2

2
(ii) |z + 1| > 2 ⟹ |u| > 2 , | | < 1
u

1 1 1 2
f(z) = u(u+2) = 2 = u2 (1 + u)−1
u2 (1+ )
u

1 2 2 2 2 3
= u2 {1 − (u) + (u) − (u) + − − −}

1 2 22 23
=(z+1)2 − (z+1)3 + (z+1)4 − (z+1)5 + − − − −

2z2 −3z+4
4. Expand f(z) = (z−1)(z+2)2 in Laurent’s series valid for
(i) 1 < |z| < 2 , (ii) |z + 1| > 2
2z2 −3z+4 1 1 5 1 6
Solution: f(z) = (z−1)(z+2)2 = 3 (z−1) + 3 (z+2) − (z+2)2

1 z
(i) 1 < |z| < 2 ⟹ | | < 1and | | < 1
z 2

1 1 5 1 6
f(z) = 3z (1−1⁄ ) + 3 2 (1+z − 2
z ⁄2) 4(1+z⁄2)

1 −1 5 −1 3 −2
= 3z [1 − 1⁄z] + 6 [1 + z⁄2] − 2 [1 + z⁄2]

1 1 5 (−1)n n 3 (−1)n (n+1) n

= 3 ∑∞ ∞
n=0 zn+1 + 6 ∑n=0 z − 2 ∑∞
n=0 z
2n 2n

2 1
(ii) |z + 1| > 2, 𝑙𝑒𝑡 𝑢 = 𝑧 + 1𝑠𝑜 𝑡ℎ𝑎𝑡 |u| > 2 𝑜𝑟 | | < 1 𝑎𝑛𝑑 | | < 1
u u
1 1 5 1 6
f(z) = 3 (u−2) + 3 (u+1) − (u+1)2
1 1 5 1 6
= 3u (1−2⁄ ) + 3 u(1+1⁄ ) − 2
u u u2 (1+1⁄u)

1 −1 5 −1 6 −2
= 3u [1 − 2⁄u] + 3u [1 + 1⁄u] − u2 [1 + 1⁄u]

1 2 n 5 (−1)n (−1)(n+1)
= ∑∞ ( ) + ∑∞ − 6 ∑∞
3u n=0 u 3 n=0 un+1 n=0 un+2

1 2 5 (−1)n (−1)n (n+1)

= ∑∞
n=0 + 3 ∑∞ ∞
n=0 (z+1)n+1 − 6 ∑n=0
3 (z+1)n+1 (z+1)n+2

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e2z
5. Expand f(z) = (z−1)3 in Laurent’s series about the point z = 1.
Solution: Let u = z-1
e2(u+1) e2 e2 (2u)n 2n un−3 2n (z−1)n−3
f(z) = = u3 e2u = u3 ∑∞
n=0 = e 2 ∑∞
n=0 = e 2 ∑∞
n=0 .
u3 n! n! n!

Singular point:
A point ‘a’ at which a complex function f(z) fails to be analytic is called singular point of
f(z).
Example: z = 0 is a singular point for f(z)=1/z.

Singularities of an analytic function:

Isolated Singular point:

A singular point ‘a’ of f(z) is said to be an Isolated singular point if there exists a neighbourhood of ‘a’
which contains no other singular point of f(z).
Example: z = 0 is a Isolated singular point of f(z) =1/z.

Removable Singularity:
Laurent’s expansion of f(z) about the point‘a’ is
1
f(z) = ∑∞ n ∞
n=0 a n (z − a) + ∑n=1 bn (z−a)n

If all the negative powers of (z-a) [Principal part] in the above series are zero, then such a
singularity is called removable singularity. The singularity can be removed by defining f(z)
such that it becomes analytic at z = a. If lim f(z) exists finitely, then z = a is a removable
z→a
singularity.
z−sinz
Example: z = 0 is a removable singularity of f(z) = .
z2
Zeros:
A point ‘a’ is called a zero of an analytic function f(z) if f(a) =0

Pole:
Laurent’s expansion of f(z) about ‘a’ is
1
f(z) = ∑∞ n ∞
n=0 a n (z − a) + ∑n=1 bn (z−a)n

If the principal part terminates at n = m where m ≥ 1 so that bm+1= bm+2 = ⋯ = 0 ,

then f(z) is given by
b b b
f(z) = ∑∞ n 1 2 m
n=0 a n (z − a) + z−a + (z−a)2 + ⋯ + (z−a)m

Then the point z = a is called a pole of order m of f(z).

Pole of order one called simple pole, pole of order two is called double pole, pole of order
three is called triple pole.

Determination of poles:

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∅(z)
If f(z) = (z−a)m
where ∅(z) is analytic and not zero at the point ‘a’, then ‘a’ is a pole of
order m of f(z). The poles of f(z) may be obtained by solving the denominator of f(z)=0.
i,e (equating the denominator of f(z) to zero).
Examples:
1 π
1. f(z) = cosz−sinz , z = is a simple pole of f(z).
4
cosz
2. f(z) = (z−1)3 , z = 1 is a triple pole of f(z).
z2
3. f(z) = (z+1)2(z2+1) , z = −1 is a double pole and z = ±i are simple poles.

Essential Singularity: Laurent’s expansion of f(z) about ‘a’ is

1
f(z) = ∑∞ n ∞
n=0 a n (z − a) + ∑n=1 bn (z−a)n

If the number of negative powers of (z –a) in the above series [principal part] is infinite, then
z = a is called an essential singularity. In such case lim f(z) does not exist.
z→a
1 3 1 1
Example: If f(z) = (z + 1)sin z−2 = 1 + z−2 − 6(z−2)2 − 2(z−2)3 + … Since there are infinite
number of terms in the negative powers of (z-2), z = 2 is an essential singularity.
Problems:
Find the nature of the singularities of the following functions:
1 e2z
(i) f(z) = 1−ez (ii) f(z) = (z−1)4

Solution:
1
(i) f(z) = 1−ez f(z) has simple pole at 𝑧 = 2πi.
e2z 1 2 2 4 2 4
(ii) f(z) = (z−1)4 = e2 (z−1)4 + (z−1)3 + (z−1)2 + 3(z−1) + 3 + 15 (z − 1) + ⋯
Since there are 4 terms containing negative powers of (z-1), z = 1 is a pole of 4th
order.
1
Residue: The coefficient of in the Laurent’s expansion of f(z) is called Residue of f(z)
z−a
at the pole z = a.

Determination of a residue:
1 dm−1
If ‘a’ is a pole of order m ≥ 1 of f(z) then residue of f(z) at ‘a’ is R = (m−1)! [lim dzm−1 {(z −
z→a
m
a) f(z)}]
Note:

1. If m = 1(simple pole), Residue = lim(z − a)f(z)

z→a

1 d
2. If m = 2(double pole), Residue = 1! lim dz{(z − a)2 f(z)}
z→a

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1 d2
3. If m = 3(triple pole), Residue = 2! lim dz2{(z − a)3 f(z)}
z→a

Problems:

For the following functions find the poles and residues at each pole.
2z+1 2z+1
1. f(z) = z2−z−2 = (z+1)(z−2)

Solution: z = -1, z = 2 are simple poles.

1
Residue at z = -1 is lim (z + 1)f(z) = 3
z→−1

5
Residue at z = 2 is lim(z − 2)f(z) = 3
z→2

z2
2. f(z) = (z−1)2(z+2)

Solution: z = 1 double pole, z = -2 simple pole.

d 5
Residue at z = 1 is lim dz{(z − 1)2 f(z)} = 9
z→1

4
Residue at z = -2 is lim (z + 2)f(z) = 9
z→−2

zez
3. f(z) = (z−1)3

Solution: z = 1 is a triple pole.

1 d2 3e
Residue at z = 1 is 2! lim dz2{(z − 1)3 f(z)} = 2
z→1

Cauchy’s Residue theorem:

Let C be a simple closed curve and f(z) be analytic within and on C except at a finite number
of poles a1, a2, …, an which lie inside C ,then

∫C f(z)dz = 2πi(R1 + R 2 + ⋯ +R n )

Where R1 , R 2 … R n are the residues of f(z) at a1,a2,…an respectively.

Problems:
e2z
1. Using Cauchy’s residue theorem, evaluate the integral ∫C (z+1)(z−2)
dz where C is the
circle|z| = 3.
e2z
Solution: Given f(z) = (z+1)(z−2)

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z = -1 and z = 2 are simple poles both lie inside C
e−2
Residue at z = -1 is lim (z + 1)f(z) = = R1
z→−1 −3

e4
Residue at z = 2 is lim(z − 2)f(z) = = R2
z→2 3

e2z e−2 e4
∫C (z+1)(z−2)
dz = 2πi(R1 + R 2 ) = 2πi [ −3 + 3 ]

zcosz
2. Using Cauchy’s residue theorem, evaluate the integral ∫C dz where C is the
(z−π⁄2)3
circle
|z − 1| = 1.
zcosz π
Solution: Given f(z) = (z−π , z = 2 is a triple pole which lie inside the circle C
⁄2 )3

π 1 d2 3
Residue at z = = 2! lim dz2 {(z − π⁄2) f(z)} = −1 = R1
2 π
z→
2
zcosz
∫C dz = 2πiR1 = −2πi.
(z−π⁄2)3

1
3. Evaluate the integral ∫C dz where C is the circle |z − 2i| = 3 using Cauchy’s
z2 (z2 +4)
residue theorem.
1
Solution: Given f(z) = z2 (z2+4), z = 0 double pole, z = 2i &-2i are simple poles.

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z = 0 & z = 2i lie inside the circle |z − 2i| = 3
d
Residue at z = 0 is lim dz{z 2 f(z)} = 0 = R1
z→0

i
Residue at z = 2i is lim(z − 2i)f(z) = 16 = R 2
z→2

1 π
∫C dz = 2πi(R1 + R 2 ) = − 8 .
z2 (z2 +4)

4. Evaluate the integral ∫C tanz dz where C is the circle |z| = 2 Using Cauchy’s residue
theorem.
𝜋 3𝜋 5𝜋
Solution: f(z) = tanz has simple poles at 𝑧 = ± 2 , ± ,± …
2 2

π
z = ± 2 lie inside the circle.

π π
Residue at z = 2 is limπ (z − 2 ) f(z) = −1 = R1
z→
2

π π
Residue at z = − 2 is limπ (z + 2 ) f(z) = −1 = R 2
z→−
2

∫C tanzdz = 2πi(R1 + R 2 ) = 2πi(−1 − 1) = −4πi

Exercise:

1. If C is the circle |z − a| = r (constant), prove the following:

dz
(i) ∫C = 2πi (ii) ∫C (z − a)n dz = 0, where n is an integer ≠ −1.
z−a

2. If C is the circle |z| = 1 , verify Cauchy’s theorem for (i) f(z) = z3, (ii) f(z) = ze-z.

3. Find the Laurent’s series expansion of

1
(i) f(z) = z2 −4z+3 about z = 0 in the region 1 < |z| < 3.

z2 −6z−1
(ii) f(z) = (z−1)(z−3)(z+2) in the region 3 < |z + 2| < 5.

7z−2
(iii) f(z) = (z+1)(z−2)z about z = −1.

Ans:
z𝑛−1 1
(i) 𝒇(𝒛) = − ∑∞
n=1 ( 2.3𝑛 + 𝑧 𝑛 )
3𝑛 (𝑧+2)𝑛 1
(ii) 𝒇(𝒛) = ∑∞
n=0 ((𝑧+2)𝑛+1 + 5𝑛+1
) + 𝑧+2

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 Department Of Mathematics
𝟑 𝟓 𝟐 𝟐 𝟐
(iii) 𝒇(𝒛) = − 𝒛+𝟏 − 𝟑 − (𝟏 + 𝟑𝟐 ) (𝒛 + 𝟏) − (𝟏 + 𝟑𝟑 ) (𝒛 + 𝟏)𝟐 − (𝟏 + 𝟑𝟒 ) (𝒛 + 𝟏)𝟑 …

4. Determine the poles of the following functions and the residues at each pole:
sinz z2 −2z
(i) (ii) (z+1)2(z2+4)
𝑧2

Ans: (i) z = 0 is a pole of order 2 and Residue = 1

(ii) z = -1 is a double pole , Residue = -14/25,

7+i 7−i
z = ±2 is simple pole. Residue ,
25 25

5. Evaluate the following integrals using Cauchy’s residue theorem.

1−2z z+4
(i) ∫C dz, C: |z| = 1.5 (ii) ∫C dz, C: |z + 1 − i| = 2
z(z−1)(z−2) z2 +2z+5

2z+1 dz
(iii) ∫C (2z−1)2
dz, C: |z| =1 (iv) ∫C dz, C: |z − i| = 2
(z2 +4)2

π π
Ans: (i) 3πi (ii) 2(3+2𝑖) (iii) πi (iv) 16

Video Links:

https://www.youtube.com/watch?v=ie9hvxq_I54

https://www.youtube.com/watch?v=0Won5Vs_65E

https://www.youtube.com/watch?v=zQ1IxVLi8SA

https://www.youtube.com/watch?v=bIY6ahHVgqA

https://www.youtube.com/watch?v=c8oDYqoMiGA

MA231BT (Statistics, Laplace Transform and Numerical Methods for PDE)