Complex Variable %5bCompatibility Mode%5d

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MATHEMATICS III:: B.TECH. SEMESTER III

Complex Variable


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FUNCTION OF A COMPLEX VARIABLE:

A complex variable w is said to be a function of complex z,

if to every value of z in certain domain D, there correspond

one or more definite values of w.

Thus if w is function of z it is written as w = f(z).

If w = u + iv and z = x + iy, then u and v both are function of

x, y ; so we write them as u(x, y) and v(x, y).

So we can write w= u(x,y) +iv(x, y).


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Note: (1) If w takes the only one value for each value of z in

the region D, then w is said to be a uniform or single valued

function of z.

2 If there corres ond two or more value of w for some or

all values of z in the given region D, then w is called a many

valued or multi-valued function of z.


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CONTINUITY OF A FUNCTION:

Let f(z) be a single valued function of z defined in some

neighborhood of the point z0 including the point z0. Then

f(z) is said to be continuous at the point z0, if for a given

real number


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DIFFERENTIABILITY:

Let f(z) be a single valued function defined in a domain D.

The function f(z) is said to be differentiable at a point z0, if

Exists. This limit is called the derivative of f z at z = z and

is denoted by f (z0). Therefore, we have

f/(z0) =

Note: (a) A function which is not continuous at point z =

z0can not be differentiable at z = z0.


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(b) A function which is continuous at a point z = z0 , may

or may not be differentiable at z= z0 .

(c) The rules of differentiation of a function of a real

variable x holds also for a function of a complex variable

.

(d) If the function g(z) is differentiable at z and the

function f(z) is differentiable at g(x), then the compositefunction w = f(g(x)) is differentiable at z.


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Example 1: Show that the function f(z) = z is continuous

at the point z = 0 but not differentiable at z= 0.

Solution: Let z = x + i y. Then z = x i y and f(z) = z = x

i y,


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Now consider the path y = mx. We have,


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Example2: Using the definition find the derivative of

at z = -i

Solution:

Example3: Using the definition find the derivative of

f(z) = 3z2 + 4zi 5 + i ; at z = 2

Solution:


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ANALYTIC FUNCTIONS:

A function f(z) of a complex variable z is said to be

analytic at a point z0, if it is differentiable at the point z0

and also at each point in some neighborhood of the point

z Thus anal ticit at a oint z means differentiabilit in.

some open disk about z0. A function f(z) is said to be

analytic in a domain D, if it is analytic at every point in D.

Note: (1) Analyticity implies differentiability but not vice

versa.


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(2) An analytic function is also called holomorphic function.

(3) A function f(z) which is analytic at every point of the

finite complex plane is called an entire function.

(4) A function f(z) is said to be analytic at z =


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CAUCHY RIEMANN EQUATIONS:

Necessary conditions for a function to beanalytic: Suppose that the function f(z) = u(x, y)

+ i v(x, y) is continuous in some neighborhood of

.

Then, the first order partial derivatives of u(x, y)

and v(x, y) exist and satisfy the equations

ux = vyand uy= -vx (1)

At the point z.


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Equation (1) is called Cauchy- Riemann Equation.

Sufficient conditions for a function to be analytic:

Suppose that the real and imaginary parts u(x, y) and

v(x, y) of the function f(z) = u(x, y) +iv(x, y) are

derivatives in a domain D. If u and v satisfy the Cauchy _

Riemann equations at all points in D, then the function

f(z) is analytic in D and

f/ (z) = ux + ivx= vy iuy (2)


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The equation (2) represents the derivative of analyticfunction.

Exmple1: Show that the function

Satisfies the Cauchy-Riemann equations at z = 0 but f/(0)

does not exist.


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Solution: f(z) = u(x, y) + iv(x, y), then we get,

and , (x, y)(0,0)

Since f(0) = 0, we have u(0, 0) = v(0, 0) = 0.

Now z


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Therefore , at z = 0, ux = vyand uy = - vx. Thus, the Cauchy

Riemann equations are satisfied at z = 0.

Now we have,


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Choosing the path y = mx, we get


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Which depends on m. Therefore, the limit does not exist.

Hence f/(0) does not exist.

satisfied for the function .

Solution:


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POLAR FORM OF THE CAUCHY RIEMANNEQUAIONS:

Let f(z) = u(r,


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Using the Cauchy Riemann equation in Cartesiancoordinates ux = vy and uy= -vx, we can write equations (3)

and (4).

; By (2)


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And

; By (1)

Therefore , the Cauchy Riemann equations in polar

coordinates are

The derivative of f(z) in polar form can be deduce as

follows;


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Example: 1 Write f(z) = zn , n any positive integer, in polar

form and verify that the Cauchy Riemann equations are

satisfied. Hence, show that the function f(z) is

differentiable and f/(z) = nz(n-1).


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HARMONIC FUNCTIONS:

A real valued function


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Theorem: If f(z) = u(x, y) + i v(x, y) is analytic in a

domain D, then the real valued functions u(x, y) and v(x,

y) satisfy the Laplace equation

respect ve y n , t at s u x, y an v x, y are armon c

in D.

Proof:


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Example 1: Show that the real and imaginary parts of an

analytic function, f(z) = u(r,


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Example 2: If u and v are harmonic in a region R, then

prove that is analytic in R.

Solution:


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If real part of the complex function f(z) is given to us then

the imaginary part can be calculated and viz versa by thefollowing relations:

f/(z) = ux(z, 0) iuy(z, 0), under the condition z = .

And

f/(z) = vy(z, 0) + ivx(z, 0), under the condition z = .


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Example 1: Prove that u = e-x(x.siny y.cosy) is harmonicfunction . Find v such that f(z) = u + iv is analytic.

Solution:v = e-x

(x.cosy + y.siny) +c

Example2: Find the analytic function f(z) of which ex(x.cosy y.siny) is real part.

Solution: f(z) = ex(x.cosy y.siny) + i ex(y.cosy + x.siny +c)


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COMPLEX INTEGRATION:

Some Important Definitions:

(1)Partition: Let [a, b] be a closed interval where a, b are realnumbers. Then the set of the points p = { t0, t1, tn},where a = t0


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(4) Jordan arc: A continuous arc without multiple points is

called a Jordan arc. A con t inu ou s Jordan curve consists

of a chain of finite number of continuous arc.

(5) Contour: By contour we means a Jordan curve

regular arc.

NOTE: (a) If A be the starting point of the first arc and B

the end point of the last arc, then the integral along such a

curve is written as .


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(b) If the starting point A of the arc coincides with the end

point B of the last arc then the contour AB is said to be

closed.

(c) The integral along such closed contour is written as

.

Although does not indicate the direction along the

curve, but it is conventional to take the direction positive

which is anticlockwise, unless indicated otherwise.


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COMPLEX LINE INTEGRALS: (Riemanns definition of

Integration):

Let z = z(t) = x(t) + iy(t), a

a=z0

b= zn

z1

z2

Zr-1

zr

Zn-1

e1e2

er

en


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On each arc joining zr-1 to zr choose a point er where r = 1,

2, 3, ., n, i.e. zr-1


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As n


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REAL LINE INTEGRAL:

Let P(x , y) and Q(x , y) be real functions of x and y

continuous at all points of a curve C. Then the real line

integral of (Pdx + Qdy) along the curve C can be defined

n a s m ar manner as comp ex ne n egra an s

denoted by


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CONNECTION BETWEEN THE REAL AND COMPLEXLINE INTEGRALS:

If f(z) = u(x, y) + i v(x, y) = u + i v, where z = x + i y, then

the complex line integral can be expressed in

terms of real line integrals as


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PROPERTIES OF COMPLEX INTEGRALS:

We have seen that a complex integral is the combination of

two real integrals. Hence some elementary properties of

real integrals must hold good in case of complex integral

also. Some essential properties are as follows:

(1)

(2) , where k is a constant.

(3) If c1, c2 are two parts of the curve c i.e. is c, c1, c2

represent the curves from a to b, a to m and m to b

respectively, then we can consider c= c1 + c2 and


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(4)

Where c indicates the direction opposite to thatindicated by c.

Example 1: Evaluate from z = 0 t0 z = 4 + 2i along

the curve c given by the line from z = 0 to z = 2i andthen the line from z = 2i to z = 4 + 2i.

Solution:


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Example2: Evaluate where C is the square with

vertex (0, 0), (1, 0), (1, 1), (0, 1).

Solutions:


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COUCHY INTEGRATION:

Some Important definitions:(a)Connected Region: A region is said to be connected region

if any two points of the region D can be connected by a curve

which lies entirely within the region.

(b)Simply Connected Region: A connected region is said to

be a Simply Connected region if all the interior points of a

closed curve C drawn in the region D are the points of the

region D. In other words, if all the points of the


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area bounded by any single closed curve C drawn in the region D

are the points of the region D, then the region D is said to be

simply connected.

(c) Multi- Connected Region: If all the points of the area bounded

by two or more closed curves drawn in the region D, are the points

o e reg on , en e eg on s ca e u - onnec e

region.

(d)Cross cut (or cut): The lines drawn in a multiple connected

region without, intersecting any one of the curves, which makes a

multi-connected region a simply connected one is called cut.


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(d) Greens Theorem: If P(x, y), Q(x, y), , are all

continuous within a domain D and if C is any closed

contour in D, then

CAUCHY INTEGRAL THEOREM:

Statement: If f is analytic with derivative continuous at allpoints inside and on a simple closed curve, then


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Proof of the statement: Try it now !!!

o e: s c ose curve an 1, 2, 3, n are o erclosed curves which lie inside C, and if function f(z) isanalytic in the region between these curves, andcontinuous on C, then

Where integral along each curve is taken in the anti-clockwise direction


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CAUCHY INTEGRAL FORMULA:

Statement: If f(z) is analytic inside and on the boundary C of

a simply connected region and a is any point within C, then

Proof of the statement:


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Theorem: If the function f(z) is analytic in a region D,

then its derivative at any point z = a is also analytic in D is

given by

Where c is the any closed contour in D surrounding the

point z =a.

Proof:


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COUCHY INTEGRAL FORMULA FOR MULTI-CONNECTED REGION:

Statement: If f(z) is analytic in the region bounded by two

closed curves c and c/ and a is a point in the region , then

Where c is the outer contour.

Proof: Assignment 2(a)


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Example 1: Evaluate if c is the circle .

Solution: Ans: -2


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ZEROS, SINGULARITIES AND RESIDUES:

Zeros of a complex function: Let f(z) be a complex function.A point z = z0 is called a zero of f(z) if f(z) is analytic at z0

and f(z0) = 0.

Note:

(a)If f(z0) = 0,

then z = z0 is called a zero of order m. If m = 1, then f(z) is

said to have a simple zero at z =z0 . Thus, the order of the

first non-vanishing derivative of an analytic function at z =

z0 is the order of the zero.


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For example:

(i) Let f(z) = sinz, then f(z0)= sinz0.

Now f(0) = sin0 = 0= sin(n


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Singular points of a complex function: A point z = z0, at

which the function f(z) is not defined or the function isnot analytic, is called a singular point of f(z). A rational

function P(z)/Q(z) has a singular point at z = z0,

whenever Q(z0) = 0 and P(z0) 0.For Example

Let the function, f(z) =

(i)Has zeros at z = -1 and z = 2.

(ii)Has singular points at z = -2 and z = 3.


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Isolated Singularity: The point z = z0 is called an isolated

singularity or isolated singular point of f(z) if we can find

>0 such that the circle enclosed no singular

point other than z0. If no such can be found, the we call z =

z0, a non-isolated singularity.

For example :1: Let the function, is

analytic anywhere except at z =1, and z = 2.Thus, z = 1 and z

=2 are the only singularities of this function, there are nosingularities of f(z) in the neighborhood of z = 1 and z = 2.

Therefore, z =1 and z =2, are the isolated singularities of this

function.


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Example2: Let the function

The function is not analytic where = 0. i.e. at the

points = n

are thesingularities of the function all of which lie on real axis. All

the singularities are isolated except z = 0 because in the

neighborhood of z = 0 there are infinite number of othersingularities z = 1/n when n is large. Therefore z = 0 is the

non isolated singularity of the function.


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Removable Singularity: If the function f(z) is not defined at

z = z0 , but exists, then the point z = z0 is called aremovable singular point.

LAURENT SERIES:

The sum is called the analytic part of the

Laurent series.

The sum is called the principal part of the

Laurent series.


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Example: Let f(z) = sinz/z has a removable singularity at z

=0, since exists.

POLE:

If the principal part of the Laurent series expansion of the

, ,

the form

Where m is a finite integer and are

all zero, then z = z0 is called a pole of order m.

If m = 1, then z = z0 is called a simple pole.


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Example: Let the function, has

a pole order 5 at z = 2 and a pole order 4 at z = -2.

Essential Singular point: If the principal part of the Laurent

series expansion of the function f(z) has infinite number of

terms, then the point z = z0 is called as essential singularpoint.

Example: z = 0 is an essential singular point of the function

e1/2 , since the principal part of the Laurent series

expansion about z = 0, has infinite number of terms.


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IMPORTANT NOTES:

(1)If , then z = a if pole of f(z).(2) If does not exist, then z = a is an essential

singularity.

(3)Limit point of zeros is an isolated essential singularity.

(4)Limit point of the poles is a non isolated essential

singularity.


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RESIDUE AT A POLE (Definition)

Let f(z) be a function analytic within a circle


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being circle


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COMPUTATION OF RESIDUE AT A FINITE POLE:

(I) Residue at a simple pole z = a:(i) Residue of f(z) at a simple pole z = a is given by

(ii) If f(z) = where


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(II) Residue at a pole of order m:

The residue of f(z) at the pole z = a of order m is

Proof: ASSIGNMENT 2(b)


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(III) Residue at a pole z = a of any finite order ( General

Method):If z = a is a pole of order m (which may be equal to 1) then

the Laurents Expansion of f(z) about z = a is given by

(1)


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b1, the coefficient of 1/(z a ) is called the residue at z =a.

Putting z a =t or z = a + t in (1), we have

1

Laurents expansion of f(a + t), where t is very small.

WORKING METHOD: Put z = a + t in the function f(z)

and expand it in powers of t where t is very small, then

find the coefficient of 1/t which gives the residue of f(z) at

the pole z = a.


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Example1: Find the residue of at its poles.

Solution: Ans. Residue at simple pole z = aResidue at simple pole z = -a

Ans. Residue at simple pole z =

Residue at simple pole z = -

Residue at double pole z =


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Residue at Infinity (Definition):

If f(z) has an isolated singularity at infinity, or is analyticthere, and if C, is a large circle which encloses all the finite

singularities of f(z), then residue of f(z) at z =


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(II) A function f(z) may be analytic at z =


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Example2: Find the residue of

at infinity.

Solution: Ans: - 1


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CAUCHYS RESIDUE THEOREM:

Statement: If f(z) is analytic within and on a closed contourC, except at a finite number of poles inside C, then

where is the sum of residue of f(z) at all its poles

within C.

Proof: ASSIGNMENT 2 (c)


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Note:

(a)If a function f(z) is analytic except at finite number ofpoles, then the sum of the residues of f(z) at all these

poles along with infinity is zero.

Example 1: Use residue theorem to evaluate the integral

Where (i) (ii)

Ans: (i) 0; (ii)


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Evaluation of Real Definite Integrals by Residues:

The evaluation of real definite integrals, can easily beachieved by using Cauchys Theorem of residues. For this

we take closed contour C and find the residues of the

function f(z) at all its poles which lie within C. Then usingCauchys theorem of residues, we have

The process of integration along the contour is called

contour integration.


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(a) Real Definite Integrals Involving TrigonometricFunctions:

Consider an integral of the form

(1)

Where F is a real rational function of sin


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As


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Example1: Evaluate the integral , by

residue theorem.

Solution:


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EXERCISE:

1.Find the kind of singularities of the following functions.

2. Find out the zeros and discuss the nature of singularitiesof the function.


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(3) Show that the function has no singularities.

(4) Find the kind of singularities of the following functions.


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(5) Find the residue at the poles of the function.

(6) Find the residue at its poles of the function

(7) Find the residues at its poles in the finite plane of thefunction

(a)

(b)

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