Complex Integration

MODULE II

COMPLEX INTEGRATION

Line Integral

Let f(z) be a continuous function of the complex variable z=x+i y defined at all points of a curve C having end points at A and B. Divide the curve C I

into n parts at A= p0,p1,p2,...... pi'...... pn=B

Let δzi = zi - zi-1 where zi = value if z at pi and ξi be any point on the arc pi-1pi.

Then, Lt ∑ f(ξi )δzi , if exist is called the line integral of f(z along C. It is

n-> ∞ i=1

δzi->0

denoted by .

Note:-

If A=p0 and B=pn coincide so that C is a closed curve then is called contour integral and is denoted by .

Let f(z) = U(x,y) +i V(x,y). then =

c c

= .Hence evaluation of line integral of complex function can be obtained by the evaluation of two line integrals of real functions.

When the same path of integration is used, then

b a

(i) and

a b

b c b

(ii) = +where C is a point on the arc joining a and b.

a a c

Analytic Function:- A single valued function f(z) having a unique derivate at every point of a region R is called an analytic function or regular function or holomorphic function of z in R.

The necessary and sufficient conditions for the function f(z)= U + i V where U and V are real ,single valued functions of x and y to be analytic in R are,

I) Ux,Uy,Vx,Vy are continous functaions of x and y in R.

II) Ux = Vy and Uy = - Vx (C-R Equation)

Remark

Every polynomial is analytic every where

The value of the line integral is independent of the path of integration if the integrand is analytic and it depends on the path of the integrand is not analytic.

If f(z) is analytic, it can be integrated in the ordinary way as if z is a real variable.

z2 z2

Eg.f(z) = z2 then = []

z1 z1

PROBLEMS

3+i

Evaluate

(a)along the line y =

(b)along the real axis from 0 to 3 and then vertically to 3+i

Solution:-

= = x2 - y2+ i2xy

Therefore u = x2 – y2 and v = 2xy

Thus ux = 2x , uy = -2y , Vx= 2y ,Vy= 2x

Ux = Vy and Uy = - Vx .Therefore f(z) is analytic

Therefore the value of the integral is independent of the path of integration.

3+i

Let I =

(a) path of integration is y =

Therefore dy = dx

= = =

When z = 0 then x = 0

when z = 3+i then x= 3

dz=dx + idy = dx

3 3

Therefore I = ∫dx =[] =

0 0

(b) Path of integration is OAB

B(3+i)

Therefore I = ∫OA +∫AB

o A

Along OA, y= 0 . dy= 0. Therefore dz = dx

x varies from 0 to 3. and z = x

Therefore = = 9

oA 0

Along AB ,x=3,dx = 0,dz =idy and z =3+iy

y varies from 0 to 1

1 1

Therefore ∫AB = = i∫(9-y2+i6y)dy =

0 0

Therefore I = ∫OA +∫AB =

dx = 6ydy

dz=dx +i dy = (6y+i) dy y2 =

When z = 0 then y = 0

when z = 3+i then y =1

Therefore y varies from 0 to 1

then z= 3y2 + iy

1 1

Therefore I = ∫= ∫=

0 0

2+i

(2)Evaluate ∫ (i) along real axis to 2 and then vertically to 2+i

(ii) along the line 2y = x

Solution:-

(a)=(x-iy)2= x2-y2-2ixy

Path of integration is OAB

2+i B(2,1)

∫=∫OA +∫AB ...... (1)

0 0 A(2,0)

Along OA,y=0,dy=0Therefore dz=dx

x varies from 0 to 2

2 2

∫OA= ∫x2dx = [] = 8/3

0 0

Along AB,x=2,dx=0,dz=idy

y varies from 0 to 1

∫AB = ∫(4-y2-i4y)idy= i(3-1/3-2i)=2+

2+i 0

From (1) ∫ ==

(b)Path of integration is 2y=x . Therefore dx=2dy

dz=dx + idy = ( 2+i) dy

=(x-iy)2= (2y-iy)2=(2-i)2y2

y varies from 0 to 1

∫OB= ∫(2-i)2y2(4+i)dy =

(3)Evaluate ∫|z|dz where C is the left half of the unit circle |z|=1 from z = -i to z = I

Solution:-

circle is |z|=1,i.e; z= eiө

ө varies from - П/2 to П/2

dz= ieiөdө

П/2 П / 2

∫|z|dz = ∫i eiөdө = [eiө ] = 2i

c - П/2 - П / 2

Home work

1+i

Evaluate ∫(x-y+ix2)dz (a) along the straight line from z=0 to z=1+i

0 (b) along the real axis from z=0 to z=1 and then along a line parallel to

imaginary axis from z= 1 to z = 1+i

parallel to real axis from z = i to z= 1+i

Evaluate ∫ |z2| dz around the square with vertices at (0,0),(1,0),(1,1)and (0,1) .
Prove that ∫ = 2 П i and ∫= 0,where C is the circle |z-a|r,if n≠ +1 an
C C

integer.

Simply and Multiply Connected Region

fig.(1) fig(ii) fig(ii)

Simply closed curve Multiply closed curve Multiply connected region

A closed curve which does not cross itself is called a simple closed curve. Otherwise it is called a multiply connected curve.

A region is called a simply connected if every closed curve in the region encloses points of the region only These closed curves can be contracted indefinitely without passing out of it.

If a region is not simply closed then it is multiply connected region. A multiply connected region can be convert to a simply connected region by giving it one or more cuts. Fig(3)

CAUCHY'S INTEGRAL THEOREM

If f(z) is an analytic function and f '(z) is continuous at each point with in and on a simple closed curve C,then

∫ f(z)dz = 0

Cor 1 :- C Q

If f(z) is analytic in the region R and P and Q are two points in R then ∫ f(z) dz is independent of the path joining P and Q and lying entirely in R. P

Proof:-

Let PAQ and PBQ are any two paths joing P and Q

F(z) is analytic in the simple closed curve PAQBP

Therefore by Cauchy's Theorem , = 0 Q

PAQBP B

I.e;+ = 0 P

PAQ QBP A

= -=

PAQ QBP P BQ

Cor 2 :-

If f(z) is analytic in the region bounded by two simple closed curve C1 and C2 then

∫ f(z)dz = ∫ f(z)dz

C1 C2

Proof:-

Let A be a cross cut joining C1 andC2 so that the closed curve APQABRSBA encloses a simply connected region.

Therefore by Cauchy's Theorem , = 0

PAQABRSBA

I.e; ∫ + ∫ + ∫ + ∫= 0

APQA AB BSRB AB

i.e; ∫ + ∫ - ∫ -∫ = 0

C1 AB BSRB AB

i.e; ∫ - ∫ = 0

C1 C2

i.e; ∫ f(z)dz = ∫ f(z)dz

C1 C2

Hence the theorem

Note:-

The theorem can be extended as it a closed curve C contains non intersection closed curves C1,C2,C3, ...... Cn then

∫f(z)dz = ∫f(z)dz+ ∫f(z)dz+ ...... + ∫f(z)dz

C C1 C2 Cn

Cauchy's Integral formula:-

If f(z) is analytic with in and on a closed curve C and if a is a point with in C then

f(a) = ∫dz

Proof:-

Consider the function which is analytic at every point within C except at z=a. Draw a circle C1 with z=a as center and radius r. therefore equation to C1 is z-a = reiө. Ө varies from 0 to 2 П.

dz = rieiөdө

By Cauchy's theorem ∫dz = ∫dz

C C1

2 П 2 П

= ∫ f(a+reiө)ireiө dz = i ∫ f(a+reiө)dӨ ...... (1)

0 reiө 0

In the limiting case as circle C1 shrinks,r 0and circle becomes the point z=a.

2 П

From (1)∫dz = i ∫ 0 f(a) dӨ = 2 П i f(a)

f(a)= ∫dz

Cor:

By Cauchy's theorem f(a)= ∫dz

Differentiating partially w.r.to a, we get

f '(a) = ∫dz

=∫dz

Similarly f ''(a) = ∫dz

f '''(a) = ∫dz

In general f n (a) = ∫dz

Problem1

Evaluate ∫where C is is |z| = 1

Solution

== z2 which is analytic everywhere and hence analytic in and on C

By Cauchy's theorem, ∫= 0

Problem 2

Evaluate ∫dz where c is the circle |z|= 1

Solution

f(z) = z+4 is analytic inside and on C.

Zeros of are z= =-1+ i2

|-1+i2| = >1

Both -1+i2 lie outside C.

By Cauchy's theorem ∫dz = 0

Problem 3

Evaluate ∫where C is the circle |z|= 2

Solution

ez is analytic inside and on C

Zeros of denominator are z = 1 and z= 4.

z= 1 lie inside Cand z= 4 lie outside C

Hence f(z) = is analytic inside and on C

Therefore by Cauchy's theorem, ∫dz = 2 Пi f(1)

That is, ∫= 2 Пi = -2/3 e ПI

Problem 4

If φ (a) = ∫ dz wherae C is the circle = 4, find the values of φ (3) and

φ' (1-i)

Solution

C is =4 that is |z|= 2.which is a circle with center at the origin and radius 2.

φ (3) = ∫ dz

z =3 lie outside the circle C

Therefore by Cauchy's theorem ∫ dz = φ (3) = 0

Let f(z)= which is analytic everywhere.

Therefore by Cauchy's integral theorem, ∫dz = 2 Пi f(a)

where z=a is a piont inside C.

That is φ (a) = 2 Пi f(a) = 2 Пi

Therefore φ '(a) = 2 Пi

therefore φ '(1-i) = 2 Пi = 2 П(6+13i)

Homework

(1)Evaluate ∫where C is the arc of the cycloid x= a(θ - sin θ), y = a(1- cos φθ)

between (0,0) and (2πa,0).

2.Evaluate ∫ where C is the ellipse

Taylors Series

If f(z) is analytic in a circle C with center at z= a ,then for all z inside C,

f(z) = f(a)+ (z-a) f '(a)+ f ''(a)+ f '''(a)+...... (1)

Corollary 1

Putting z=a+h so that z-a=h, we get

f(a+h) = f(a)+ (h) f '(a)+ f ''(a)+ f '''(a)+......

Putting a = 0, in(1)

f(z) = f(0)+ (z) f '(0)+ f ''(0)+ f '''(0)+......

which is known as Maclaurian series.

Laurent's Series

If f(Z) is analytic inside and on the boundary of the ring shaped region R bounded by twp concentric circlesC1 and C2 of radii r1 and r2 respectively (r1>r2) having center at z=a then for all z in R,

f(z) = a0 + a1(z-a) + a2 +a3+...... + b1(z-a)-1 +b2 +b3...... (1)

where an= ∫ ; n=0,1,2,...... and

bn= ∫ ; n=1,2,......

Note 1

If f(z) is analytic inside C2 then bn=0 and an= ∫ =

In this case Laurent's series reduces to Taylors series.

Note 2

in the statement of Laurent's series an= ∫ and this an =.Since f(z) is not given to be analytic inside C2.

Note 3

In practice to obtain Laurent's series, we expand f(z) by Binomial theorem or by some other method.

Problem 1

Expand in the region (1) |z|<1 (2)1<|z|<2 (3)|z|>2 (4)0<|z-1|<1

Solution

Here f(z)=== -

(1)Region of expansion is |z|<1

therefore f(z) =-

= -1/2 +

= {1++++...... }+{1+z+z2+z3+...... }

=+z(1-)+ z2 (1-) + z3 (1-)+......

(2) Region of expansion is 1<|z|<2.Therefore <1 and <1

F(z) =-

= -(1-)-1

= [1++++...... ]---[1+++

+...... ]

(3)Region of expansion is |z|>2.Therefore |1/z|<1 and |2/z|<1

f(Z) = -=-

= [- (1-)-1]

= [1++++...... ] - [1++++...... ]

=+()+()+......

(4) 0<|z-1|<1

Put z-1 = t Therefore z=t+1

Therefore |t|<1

f(z) = -=-

= --= - [1 + t + t 2 + t3 + ...... ] -

= - [+1+(z-1)+(z-1)2+(z-1)3+(z-1)4+......

Problem 2

∞

Show that when |z+1|<1 ,z-2 = 1+ ∑ (n+1)( z+1)n

n=1

Solution

f(z)= z-2 = = = =

= 1 + 2(z+1) + 3(z+1)2 + 4(z+1)3 + ......

∞

=1+ ∑ (n+1)( z+1)n

n=1

Home work

(1)Expand cosz in a Taylor's series about z = π /4

(2)Expand the function sin z / (z- π) about z = π

(3)Expand f(z) =about z= - 2

Singular points and residues

A point at which f(z) is not analytic is called a singular point of f(Z). In other words,if

f(z) = Ψ(z) then zeros of Φ(z), ie the zeros of the denominator are the singular points of f(z),

Φ(z)

provided Ψ(z) and Φ(z) have no common factors.

z=a is called an isolated singular points if there exist a circle with center at z=a which contains no other singular point of f(z).

example:- if f(z) = then z=1 and z= -1 are isolated singular points.

Pole

If z=a is an isolated singular point of f(z) then f(z) can be expanded about z =a by using Laurent's series

∞ ∞

as,f(Z) = ∑ an(z-a ) n + ∑ bn /(z-a) n.

n=0 n=1

if this series contains only finite number of terms with negative powers of (z - a) ,say m,then z = a is called pole of order m.

A pole of order 1 is called a simple pole and a pole of order 2 is called a double pole.

If f(z) has infinite number of terms with negative powers of (z-a) then z =a is called an essential singular point of f(z).

Residue

The coefficient of (z-a)-1 when f(z) is expanded about z =a an isolated singular point of f(Z) is called residue of f(z) at z =a and is denoted as Res{f(z);a}.

Note:

∞ ∞

f(z)= ∑ an(z-a ) n + ∑ bn /(z-a) n.where bn = 1/2πi ∫

n=1 n=1 C

where c is the circle enclosing z= a.

Therefore Res{f(z);a}. = b1= ∫ f(z) dz

∫ f(z) dz = 2 πi Res{f(z);a}

Residue Theorem

If f(z) ia anlytic at all points inside and in a simple closed curve C except at a finite number of isolated singular points within C then ∫ f(z) dz = 2 πi [sum of residue of singular points within C].

Calculation of Residues

(1)if f(z) has a simple pole at z=a then Res{f(z);a}= Lt (z-a)f(z).

z→a

(2) If f(z) has a pole of order m at z=a then Res{f(z);a}= Lt [(z-a)mf(z)]

z→a

(3)If f(z) has a pole at z =a , of any order then Res{f(z);a}=coefficient of t-1 in the expansion of f(z) in powers of t where at = z-a. That is t= z-a in f(z).Expand it in powers of t. then coefficient of

1/t is the residue of f(z) at z=a.

Problems

Determine the poles of the function f(z) =and residue at each pole.Hence evaluate

∫ f(z) dz where C is the circle |z|=2.5

Solution

Poles are the roots of the equation = 0

That is z = 1,double pole and

z = -2 simple pole

Now,Res{f(z);1}= Lt z→1 []

= 5/9

Similarly Res{f(z);-2}=Lt z→-2 = 4/9

Both pole inside C.

Therefore by residue theorem ∫ f(z) dz = 2 πi [sum of residue of singular points within C].

= 2πi ()= 2πi

Problem

Find the residue of at its pole

Solution

Pole is given by,= 0

That implies = -1

But we have ,e π i= -1

Therefore z= π i

Res{f(z); π i}= Lt z→π i = -1

Homework

(1)Evaluate ∫ where C is |z|= 2

(2)Evaluate ∫ dz where C is the circle |z| = 1.5,.

Evaluation of real integrals using residues

TYPE 1

2π

Integrals of the type ∫ F(cosθ , sinθ )dθ where F(cosθ , sinθ ) is a rational function of cosθ and sinθ.

Such inategrals can be reduced to complex line integrals by the substitution z = e iθthat

dz =ie iθ that is dθ =

Also cosθ = eiθ + e-iθ = ()

sinθ = eiθ - e-iθ = ()

As θ varies from 0 to 2π , moves once round the unit circle in the anti-clockwise direction where C is the unit circle |z| = 1.

The integral on the right can be evaluated by using the residue theorem.

Example 1

2π

Evaluate ∫ dθ

Solution

Put z = e iθ so that cosθ = () and dθ =

Then 2π

∫ dθ = ∫= ∫dz

where C is the circle |z| = 1

The poles of the integrand are the roots of = 0 which are z= -2 ±

Of the two poles, only z = -2 ± lies inside the circle C.

Residue at z = -2 + is

= Lt z→α (z- α )

= Lt z→α = 1 =

2 α+4

By residue theorem ∫dz = 2πi () =

2π

∫ dθ = = 2π

0 3

Example 2

Evaluate ∫ dθ where a>|b|

Solution

2π π

∫ dθ = 2 ∫ dθ

0 0

π 2π

or ∫ dθ = ∫ dθ ...... (1)

0 0

Put z = e iθ so that cosθ = () and dθ = we have

2π

∫dθ = ∫

0 C

= ∫ where C is the circle |z| = 1

residue at z = α is = Lt z→α (z- α ) 2

ib(z-α )(z-β)

= 2 = =

ib(α - β)

therefore residue theorem ∫= 2πi = = 2π

From (1) ∫ = π

Type 2

∞

Integrals of the type ∫dx , where f(x) and F(x) are polynomials in x such that →0

-∞

as x→ ∞ and F(x) has no zero on the real axis.

Method

Consider ∫f(z) dz over the closed contour C consisting of the part of real axis from _R to +R, R

Φ(z)

very large and semicircle CR ,|z\= R in the upper half plane. Since R is very large, all the poles having +ve y value lie with in CR.

Further Φ(z) has no zeroes in the real axis.

∫ f(z) dz = 2 πi [sum of residue of ∫ f(z) in the upper half of the region above the real axis.].

Φ(z) Φ(z)

But ∫ f(z) dz = ∫f(z) dz + ∫ f(z) dz

Φ(z) -R Φ(z) CR Φ(z)

= ∫ f(x) dz + ∫ f(z) dz

-R Φ(x) CR Φ(z)

Now Lt z→α = 0.

∫ f(z) dz = 0 as R tends to ∞

Φ(z)

∞

As R tends to ∞, from (1) ∫ f(z) dz = ∫ f(x) dz ...... (3)

Φ(z) -∞ Φ(x)

From (1) and(3)

∞

∫ f(x) dz = 2 πi [sum of residue of f(z) at its poles within C

-∞ Φ(x) Φ(z)

Problem1

∞

Evaluate ∫ dxwhere a > 0 and b > 0

-∞

Solution

The poles of ф(z) = are z = ± ia , ±ib

Of those z= is and z= ib lie in the upper half of the z-plane.

Residue of ф(z) at z=ia is

Lt z→ia (z- ia) = Lt z→ia

= =

Residue of ф(z) at z=ib is

Lt z→ib (z- ib) = Lt z→ib

= =

∞

Therefore ∫ dx =2 πi [sum of residue of ф(z) in the upper half of z-plane].

-∞

= 2 πi[+=π[

= π

a+b

Problem 2

∞

Evaluate ∫

Solution

The poles of ф(z) = are obtained by solving = 0

implies z = (-1)¼(cos π+isin π)1/4 = [cos (2nπ+π) +i sin (2nπ+π)]1/ 4

= cos (2n+1) π +i sin (2n+1) π where n =0,1,2,3.

4 4

When n=0, z = cos π +i sin π = +i

4 4

When n=1, z = cos 3π +i sin 3π = +i

4 4

When n=2, z = cos 5π +i sin 5π = +i

4 4

When n=3, z = cos 7π +i sin 7π = +i

4 4

Of these only the poles corresponding to n =0,1 viz, z = ei π/ 4 and z = e3i π / 4 lie in thae upper half of z-plane.

Residue of ф(z) at z= ei π/ 4 is Lt z→ei π/ 4 z- ei π/ 4

z4 +1

= Lt z→ei π/ 4 = 1 = e -- 3i π / 4

4 e3i π / 4

similarly, residue of ф(z) at z= e3i π/ 4 is e -- 9i π / 4

∞ - ∞

Therefore ∫ = ∫ = 2 πi [sum of residue of ф(z) in the upper half of z-

0 ∞ plane].

=πi [ e -- 3i π / 4+ e -- 9i π / 4]

= πi[Cos(3π / 4) – iSin (3π / 4)+ Cos(9π / 4) – iSin (9π / 4) ]

=πi [ +i ++i]

= πi(-i√2) = π

Homework

- ∞

(1) Evaluate ∫

∞

- ∞

(2) Evaluate ∫ dx

∞

- ∞

(2)Evaluate ∫ dx

∞

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