UNIT3
FUNCTION OF COMPLEX VARIABLE
In the narrow sense of the term, the theory of function of a complex variable is the theory of analytic functions (cf. Analytic function) of one or several complex variables. As an independent discipline, the theory of functions of a complex variable took shape in about the middle of the 19th century as the theory of analytic functions.
Complex function
x + iy is a complex variable which is denoted by z
If for each value of the complex variable z = x + iy in a region R, we have one or more than one values of w = u + iv, then w is called a complex function of z.
And it is denoted as
w = u(x , y) + iv(x , y) = f(z) 
Neighbourhood of
Let a point in the complex plane and z be any positive number, then the set of points z such that
<ε 
Is called ε neighbourhood of
Limit of a function of a complex variable
Suppose f(z) is a single valued function defined at all points in some neighbourhood of point 
The
Example1: Find 
Sol. Here we have Divide numerator and denominator by , we get 
Continuity A function w = f(z) is said to be continuous at z = , if
Also if w = f(z) = u (x, y) + iv (x, y) is continuous at z = then u (x, y), v (x, y) are also continuous at z = .
Differentiability
Let f(z) be a single valued function of the variable z, then
Provided that the limit exists and has the same value for all the different ways in which approaches to zero.
Example2: if f(z) is a complex function given below, then discuss 
Sol. If z→0 along radius vector y = mx But along , In different paths we get different value of that means 0 and –i/2, in that case the function is not differentiable at z = 0. 
Key takeaways
<ε 2. Limit of a function of a complex variable Suppose f(z) is a single valued function defined at all points in some neighborhood of point  The 3. A function w = f(z) is said to be continuous at z = , if 4. 
In Cartesian form
Theorem; The necessary condition for a function to be analytic at all the
points in a region R are (ii) Provided, 
Proof:
Let be an analytic function in region R. Along real axis Then f’(z), becomes ………… (1)
Along imaginary axis From equation (1) and (2) Equating real and imaginary parts Therefore and These are called Cauchy Riemann Equations. 
CR equation in polar from
CR equations in polar form are
Proof: As we know that x = r cos and u is the function of x and y z = x + iy = r ( cos Differentiate (1) partially with respect to r, we get Now differentiate (1) with respect to , we get Substitute the value of , we get Equating real and imaginary parts, we get Proved 
Key takeaways
(ii) 2. CR equations in polar form are 
A function is said to be analytic at a point if f is differentiable not only at but at every point of some neighborhood at .
Note
1. A point at which the function is not differentiable is called singular point.
2. A function which is analytic everywhere is called an entire function.
3. An entire function is always analytic, differentiable and continuous function (converse is not true)
4. Analytic function is always differentiable and continuous but converse is not true.
5. A differentiable function is always continuous but converse is not true.
The necessary condition for f(z) to be analytic
f(z) = u + i(v) is to be analytic at all the points in a region R are
1. …………. (1) 2. ……...…. (2) Provided exists 
Equation (1) and (2) are known as CauchyRiemann equations.
The sufficient condition for f(z) to be analytic
f(z) = u + i(v) is to be analytic at all the points in a region R are 1. …………. (1) 2. ……...…. (2) are continuous function of x and y in region R. 
Important note
1. If a function is analytic in a domain D, then u and v will satisfy CauchyRiemann conditions.
2. CR conditions are necessary but not sufficient for analytic function.
3. CR conditions are sufficient if the partial derivative are continuous.
State and prove sufficient condition for analytic functions
Statement – The sufficient condition for a function to be analytic at all points in a region R are
1 2 are continuous function of x and y in region R. Proof: Let f(z) be a simple valued function having at each point in the region R. Then CauchyReimann equation are satisfied by Taylor’s Theorem Ignoring the terms of second power and higher power We know CR equation Replacing Respectively in (1) we get
Show that is analytic at Ans The function f(z) is analytic at if the function is analytic at z=0 Since Now is differentiable at z=0 and at all points in its neighbourhood Hence the function is analytic at z=0 and in turn f(z) is analytic at 
Example1: If w = log z, then find . Also determine where w is nonanalytic. Sol. Here we have Therefore and Again Hence the CR conditions are satisfied also the partial derivatives are continuous except at (0, 0). So that w is analytic everywhere but not at z = 0 
Example2: Prove that the function is an analytical function.
Sol. Let =u+iv Let =u and =v Hence CREquation satisfied. 
Example3: Prove that 
Sol. Given that Since V=2xy Now But Hence 
Example4: Show that polar form of CR equations are 
Sol. z = x + iy = U and v are expressed in terms of r and θ. Differentiate it partially w.r.t. r and θ, we get By equating real and imaginary parts, we get 
Key takeaways
 A function is said to be analytic at a point if f is differentiable not only at but at every point of some neighborhood at .
 A point at which the function is not differentiable is called singular point.
 A function which is analytic everywhere is called an entire function.
 If a function is analytic in a domain D, then u and v will satisfy CauchyRiemann conditions.
 CR conditions are necessary but not sufficient for analytic function.
 CR conditions are sufficient if the partial derivative are continuous.
A function which satisfies the Laplace equation is known as a harmonic function.
Theorem if f(z) = u + iv is an analytic function, then u and v are both harmonic functions. 
Proof: Suppose f(z) = u + iv, be an analytic function, then we have Differentiate (1) with respect to x, we get Differentiate (2) with respect to y, we get Add 3 and 4 Similarly So that u and v are harmonic functions. 
Example: Prove that and are harmonic functions of (x, y). 
Sol. We have Now Here it satisfies Laplace equation so that u (x, y) is harmonic. Now On adding the above results We get So that v(x, y) is also a harmonic function. 
Example: Find the harmonic conjugate function of the function U (x, y) = 2x (1 – y). 
Sol. We have, U(x, y) = 2x (1 – y) Let V is the harmonic conjugate of U. So that by total differentiation, Hence the harmonic conjugate of U is 
A point at which a function f(z) is not analytic is known as a singular point or singularity of the function.
Isolated singular point If z = a is a singularity of f(z) and if there is no other singularity within a small circle surrounding the point z = a, then z = a is said to be an isolated singularity of the function f (z); otherwise, it is called nonisolated.
Pole of order m Suppose a function f(z) have an isolated singular point z = a, f(z) can be expanded in a Laurent’s series around z = a, giving
…… (1) In some cases it may happen that the coefficient , then equation (1) becomes Then z = a is said to be a pole of order m of the function f(z). 
Note The pole is said to be simple pole when m = 1.
In this case
Working steps to find singularity
Step1: If exists and it is finite then z = a is a removable singular point.
Step2: If does not exists then z = a is an essential singular point.
Step3: If is infinite then f(z) has a pole at z = a. the order of the pole is same as the number of negative power terms in the series expansion of f(z).
Example: Find the singularity of the function
Sol. As we know that So that there is a number of singularity. is not analytic at z = a (1/z = ∞ at z = 0) 
Example: Find the singularity of 
Sol. Here we have We find the poles by putting the denominator equals to zero. That means 
Example: Determine the poles of the function 
Sol. Here we have We find the poles by putting the denominator of the function equals to zero We get By De Moivre’s theorem If n = 0, then pole If n = 1, then pole If n = 2, then pole If n = 3, then pole 
Key takeaways
If f(z) is analytic in a closed curve C, except at a finite number of poles within C, then 
Cauchy’s integral formula
Cauchy’s integral formula can be defined as
Where f(z) is analytic function within and on closed curve C, a is any point within C.
Example1: Evaluate by using Cauchy’s integral formula. Here c is the circle z  2 = ½ 
Sol. it is given that Find its poles by equating denominator equals to zero. There is one pole inside the circle, z = 2, So that Now by using Cauchy’s integral formula, we get 
Example2: Evaluate the integral given below by using Cauchy’s integral formula 
Sol. Here we have
Find its poles by equating denominator equals to zero. We get There are two poles in the circle Z = 0 and z = 1 So that

Example3: Evaluate if c is circle z  1 = 1. 
Sol. Here we have
Find its poles by equating denominator equals to zero. The given circle encloses a simple pole at z = 1. So that

Key takeaways
Cauchy’s integral formula 
Residue at a pole
If z = a is an isolated singularity of f(z) then f(x) can be expressed expanded in Laurent’s series about z = a
So that
Note the coefficient of which is is called the residue of f(z) at z = a and it is written as
Since So that 
Method of finding residues
 If (z) has a simple pole at z = a, then Res{f(z), a} = .
 If f(z) has a pole of order m at z = a, then
3. 
Example: Find the residue of f(z) = z cos (1/z) at z = 0 
Sol. Which is the Laurent’s series expansion about z = 0 So that By the definition of residue Residue of f(z) at z = 0 is 1/2. 
Cauchy’s residue theorem
If f(z) is analytic in a closed curve C, except at a finite number of poles within C, then
Proof:
Suppose be the nonintersecting circles with centres at respectively.
Radii so small that they lie within the closed curve C. then f(z) is analytic in the multiple connected regions lying between the curves C and
Now applying the Cauchy’s theorem
Example: Find the poles of the following functions and residue at each pole: and hence evaluate

Where c: z = 3. Sol. The poles of the function are The pole at z = 1 is of second order and the pole at z = 2 is simple Residue of f(z) (at z = 1) Residue of f(z) ( at z = 2)

Example: Evaluate Where C is the circle z = 4. 
Sol. Here we have, Poles are given by Out of these, the poles z = πi , 0 and πi lie inside the circle z = 4. The given function 1/sinh z is of the form Its poles at z = a is Residue (at z = πi) Residue (at z = 0) Residue (at z = πi) Hence the required integral is = 
Key takeaways
2. Cauchy’s residue theorem If f(z) is analytic in a closed curve C, except at a finite number of poles within C, then 
References
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 P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
 S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
 W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
 N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
 B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.
 T. Veerarajan, “Engineering Mathematics”, Tata McgrawHill, New Delhi, 2010
 Higher engineering mathematics, HK Dass
 BV ramana, higher engineering mathematics