CVT Sajid Ali Complex Integration Complex Integration...

CVT

Sajid Ali

ComplexIntegration

Simply ConnectedDomains

Complex LineIntegrals

Cauchy IntegralTheorem

Cauchy IntegralFormula

Residues

Complex Integration

Sajid Ali

SEECS-NUST

October 2, 2017

CVT

Sajid Ali

ComplexIntegration





Residues

Complex Integration

A complex line integral is given by∫Cf (z)dz ,

where C is the curve along which the integration is carriedout.

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Sajid Ali

ComplexIntegration





Residues

Complex Integration


where C is the curve along which the integration is carriedout. In order to evaluate it we need to characterize domains.

Simply Connected Domains: A domain D is calledsimply connected if every closed path in it can becontinuously deformed (or shrinked) to only points of D.

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Sajid Ali

ComplexIntegration





Residues

Complex Integration


where C is the curve along which the integration is carriedout. In order to evaluate it we need to characterize domains.

Simply Connected Domains: A domain D is calledsimply connected if every closed path in it can becontinuously deformed (or shrinked) to only points of D.

For example, complex plane, interior of a an open disk, etc.Counter example, a unit circle without origin, an annulusregion etc..

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Sajid Ali

ComplexIntegration





Residues

Complex Integration

How to evaluate complex line integral:Way-1: Direct substitution by using fundamental theorem ofcomplex analysis

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Sajid Ali

ComplexIntegration





Residues

Complex Integration

How to evaluate complex line integral:Way-1: Direct substitution by using fundamental theorem ofcomplex analysisWay-2: Identify path and then carry out integration

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Sajid Ali

ComplexIntegration





Residues

Complex Integration

Way-1: Suppose F is analytic in a simply connected domainD then for all paths between z0 and z1 the fundamentaltheorem of complex line integrals states∫ z1

z0

F ′(z)dz = F (z1)− F (z0)

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Sajid Ali

ComplexIntegration





Residues

Complex Integration


z0

F ′(z)dz = F (z1)− F (z0)

Example-1: ∫ 1+i

0z2dz =?

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Sajid Ali

ComplexIntegration





Residues

Complex Integration


z0

F ′(z)dz = F (z1)− F (z0)

Example-1: ∫ 1+i

0z2dz =?

Since the domain of f (z) = z2 is the entire complex planewhich is simply connected and the function is analytic in itand

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Sajid Ali

ComplexIntegration





Residues

Complex Integration


z0

F ′(z)dz = F (z1)− F (z0)

Example-1: ∫ 1+i

0z2dz =?

Since the domain of f (z) = z2 is the entire complex planewhich is simply connected and the function is analytic in itand

d

dz

z3

3= z2

⇒∫ 1+i

0z2dz =

z3

3|1+i0

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Sajid Ali

ComplexIntegration





Residues

Complex Integration


z0

F ′(z)dz = F (z1)− F (z0)

Practice:

1.

∫ πi

−πicos zdz =?

2.

∫ i

−i

1

zdz =?.

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Sajid Ali

ComplexIntegration





Residues

Complex Integration

Way-2: Parameterize the curve

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Sajid Ali

ComplexIntegration





Residues

Complex Integration

Way-2: Parameterize the curveLet C be a piecewise smooth path, represented by z = z(t),where a < t < b. Let f (z) be a continuous function on Cthen ∫

cf (z)dz =

∫ b

af (z)z(t)dt

where z = x + i y .

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Sajid Ali

ComplexIntegration





Residues

Complex Integration

Way-2: Parameterize the curveLet C be a piecewise smooth path, represented by z = z(t),where a < t < b. Let f (z) be a continuous function on Cthen ∫

cf (z)dz =

∫ b

af (z)z(t)dt

where z = x + i y .

Examples:For boundary of a unit circle C evaluate the integral∫

C

1

zdz =?

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Sajid Ali

ComplexIntegration





Residues

Complex Integration


C

1

zdz =?

Step-1 As C is a unit circle so parameterize

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Sajid Ali

ComplexIntegration





Residues

Complex Integration


C

1

zdz =?


z(t) = cos t + i sin t = e it

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Sajid Ali

ComplexIntegration





Residues

Complex Integration


C

1

zdz =?



Step-2 Differentiate z = ie it

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Sajid Ali

ComplexIntegration





Residues

Complex Integration


C

1

zdz =?




Step-3 Calculate f (z(t)) = e−it

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Sajid Ali

ComplexIntegration





Residues

Complex Integration


C

1

zdz =?




Step-3 Calculate f (z(t)) = e−it

Step-4 Evaluate the integral∫Cf (z)dz =

∫ 2π

0ie−ite itdt

= 2πi Ans.

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Sajid Ali

ComplexIntegration





Residues

Complex Integration

Follow-up Questions:What ifQ-1 C is a rectangular box with z = 0 at center?

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Sajid Ali

ComplexIntegration





Residues

Complex Integration

Follow-up Questions:What ifQ-1 C is a rectangular box with z = 0 at center?Q-2 C is a closed curve with z = 0 not at its center?

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Sajid Ali

ComplexIntegration





Residues

Complex Integration

Follow-up Questions:What ifQ-1 C is a rectangular box with z = 0 at center?Q-2 C is a closed curve with z = 0 not at its center?Q-3 C is a closed curve and does not contain z = 0?

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Sajid Ali

ComplexIntegration





Residues

Complex Integration

Follow-up Questions:What ifQ-1 C is a rectangular box with z = 0 at center?Q-2 C is a closed curve with z = 0 not at its center?Q-3 C is a closed curve and does not contain z = 0?Q-4 C is a circle and contain z = 1, so∫

C

1

z − 1dz =?

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Sajid Ali

ComplexIntegration





Residues

Complex Integration

Follow-up Questions:What ifQ-1 C is a rectangular box with z = 0 at center?Q-2 C is a closed curve with z = 0 not at its center?Q-3 C is a closed curve and does not contain z = 0?Q-4 C is a circle and contain z = 1, so∫

C

1

z − 1dz =?

Q-5 C is a unit circle∫C

(z + z−1)dz =?

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Sajid Ali

ComplexIntegration





Residues

Complex Integration

Practice:Evaluate the same integral

∫C 1/zdz over more paths !!!

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Sajid Ali

ComplexIntegration





Residues

Complex Integration



1. For an ellipse

z(t) = a cos t + ib sin t

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Sajid Ali

ComplexIntegration





Residues

Complex Integration



1. For an ellipse

z(t) = a cos t + ib sin t

2. For a closed parabolic path

z(t) = t + it2, 0 < t < 1

z(t) = 1− t + i(1− t), 0 < t < 1

evaluate the integral ∫Cz2 dz =?

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Sajid Ali

ComplexIntegration





Residues

Complex Integral Theorem

We have observed that the answer to a complex line integralalong a closed curve is zero or 2πi that is whether singularityis outside or inside the closed curve. It is the consequence ofa famous theorem in complex analysis.

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Sajid Ali

ComplexIntegration





Residues



Cauchy Integral Theorem:If f (z) is analytic in a simply connected domain D, then forevery closed path C in D∫

Cf (z)dz = 0

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Sajid Ali

ComplexIntegration





Residues




Cf (z)dz = 0

Practice: For an arbitrary closed curve C

(i)

∫Cezdz =?, (ii)

∫C

cos(z)dz =?, (iii)

∫Czndz =?,

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Sajid Ali

ComplexIntegration





Residues




Cf (z)dz = 0

Practice: For an arbitrary closed curve C

(i)

∫Cezdz = 0, (ii)

∫C

cos(z)dz = 0, (iii)

∫Czndz = 0,

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Sajid Ali

ComplexIntegration





Residues




Cf (z)dz = 0

Practice: For an arbitrary closed curve C∫C

sec (z)dz = 0,

∫C

1

z2 + 4dz = 0, C = Unit Circle

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Sajid Ali

ComplexIntegration





Residues


What about the converse of Cauchy integral theorem? i.e., ifthe integral of a complex function is zero does it guaranteethat the function is analytic?

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Sajid Ali

ComplexIntegration





Residues


What about the converse of Cauchy integral theorem? i.e., ifthe integral of a complex function is zero does it guaranteethat the function is analytic? NO !!!

CVT

Sajid Ali

ComplexIntegration





Residues


What about the converse of Cauchy integral theorem? i.e., ifthe integral of a complex function is zero does it guaranteethat the function is analytic? NO !!! For example

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Sajid Ali

ComplexIntegration





Residues


What about the converse of Cauchy integral theorem? i.e., ifthe integral of a complex function is zero does it guaranteethat the function is analytic? NO !!! For example∫

C

1

z2dz = 0 C = Unit Circle

where f (z) = 1/z2 is clearly not analytic at z = 0.

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Sajid Ali

ComplexIntegration





Residues



C

1


where f (z) = 1/z2 is clearly not analytic at z = 0.Therefore Cauchy integral theorem only provide necessarycondition and not the sufficient condition.

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Sajid Ali

ComplexIntegration





Residues



C

1



Independence of path:If f (z) is analytic in a simply connected domain D, then theintegral of f is independent of path in D.

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Sajid Ali

ComplexIntegration





Residues



C

1



Independence of path:If f (z) is analytic in a simply connected domain D, then theintegral of f is independent of path in D.This is an important result from the physical point of view.If f (z) has the sense of a vector field on the plane thenabove theorem implies that the vector field is conservative.

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Sajid Ali

ComplexIntegration





Residues

Complex Integral Formula

If f (z) is analytic in a simply connected domain D, then∫C

f (z)

z − z0dz = 2πif (z0)

where C is taken counterclockwise and contains z0.

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Sajid Ali

ComplexIntegration





Residues



f (z)



Example-1:Suppose C : any contour containing z0 = 2∫

c

ez

z − 2dz =

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Sajid Ali

ComplexIntegration





Residues



f (z)



Example-1:Suppose C : any contour containing z0 = 2∫

c

ez

z − 2dz = 2πi |ez |z=2

= 2πiez |z=2, (as ez is analytic)

= 2πie2

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Sajid Ali

ComplexIntegration





Residues


PracticeSuppose C : is the contours (i) |z | = 1 (ii) |z − 1| = 2 (iii)|z + 1| = 2. Evaluate ∫

c

1

z2 − 4dz =?

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Sajid Ali

ComplexIntegration





Residues


Example-2:Suppose we have C : |z − 1| = 1

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Sajid Ali

ComplexIntegration





Residues


Example-2:Suppose we have C : |z − 1| = 1 then find the value of

I =

∫C

z2 + 1

z2 − 1dz =?

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Sajid Ali

ComplexIntegration





Residues



I =

∫C

z2 + 1

z2 − 1dz =?

Sol. Since C represent a circle of radius one centered at oneunit on the x-axis therefore it includes the point ofsingularity of f (z).

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Sajid Ali

ComplexIntegration





Residues



I =

∫C

z2 + 1

z2 − 1dz =?

Sol. Since C represent a circle of radius one centered at oneunit on the x-axis therefore it includes the point ofsingularity of f (z). But we can write the integrand such that

I =

∫C

z2+1z+1

z − 1dz

in which case ∴ f (z) = z2+1z+1 is analytic on the given

contour, C , as z = −1 is not on C so, f (1) = 1+11+1 = 1

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Sajid Ali

ComplexIntegration





Residues



I =

∫C

z2 + 1

z2 − 1dz =?


I =

∫C

z2+1z+1

z − 1dz

in which case ∴ f (z) = z2+1z+1 is analytic on the given contour,

C , as z = −1 is not on C so, f (1) = 1+11+1 = 1 therefore∫

C= 2πi

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Sajid Ali

ComplexIntegration





Residues


Example-3:Suppose we have C : |z + 1| = 1 then find the value of

I =

∫C

z2 + 1

z2 − 1dz =?

Sol. Since C represent a circle of radius one centered at oneunit on the x-axis therefore it includes the point ofsingularity of f (z).

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Sajid Ali

ComplexIntegration





Residues


Example-3:Suppose we have C : |z + 1| = 1 then find the value of

I =

∫C

z2 + 1

z2 − 1dz =?


I =

∫C

z2+1z−1z + 1

dz

in which case ∴ f (z) = z2+1z−1 is analytic on the given contour,

C , as z = 1 is not on C so, f (−1) = 2−2 = −1 therefore∫

C= −2πi

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Sajid Ali

ComplexIntegration





Residues

Laurent Series

Review:The coefficients an in the Laurent series are determined froma complex line integral

an =1

2πi

∫C

f (z)

(z − z0)n+1dz

where C is a counter-clock wise oriented closed curve lyingin an annulus in which f (z) is analytic. The expansion off (z) is then valid in the annulus.

Note that the power n = −1, yields the complex integral ofthe original function

a(−1) =1

2πi

∫Cf (z) dz

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Sajid Ali

ComplexIntegration





Residues

Residues

The coefficient of the term 1/z in the Laurent series of acomplex function f (z) is known as the residue of f (z) atz = z0,

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Sajid Ali

ComplexIntegration





Residues

Residues

The coefficient of the term 1/z in the Laurent series of acomplex function f (z) is known as the residue of f (z) atz = z0, i.e.,

Resz=z0

(f (z)) = a(−1).

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Sajid Ali

ComplexIntegration





Residues

Residues

The coefficient of the term 1/z in the Laurent series of acomplex function f (z) is known as the residue of f (z) atz = z0, i.e.,

Resz=z0

(f (z)) = a(−1).

Therefore the integral of f (z) over C , a counter-clock wiseoriented closed curve lying in an annulus in which f (z) isanalytic ∫

Cf (z) dz = 2πi Res

z=z0(f (z)).

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Sajid Ali

ComplexIntegration





Residues

Residues

Example-1:Integrate f (z) = z−4 sin z around unit circle.

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Sajid Ali

ComplexIntegration





Residues

Residues

Example-1:Integrate f (z) = z−4 sin z around unit circle.Solution:The Laurent series of f (z) is

f (z) = z−4(z − z3

3!+

z5

5!− ...

)=

1

z3− 1

3!z+

z

5!− ...

Therefore,

Resz=z0

(f (z)) = −1

6.

Hence the integral is∫Cf (z) dz = −2πi × 1

6= −πi

3.

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Sajid Ali

ComplexIntegration





Residues

Residues

How to compute residues?For a function of the form

f (z) =p(z)

q(z)

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Sajid Ali

ComplexIntegration





Residues

Residues


f (z) =p(z)

q(z)

such that p(z0) 6= 0, q(z0) = 0, q′(z0) 6= 0, then the residueof f (z) at z = z0 is given by

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Sajid Ali

ComplexIntegration





Residues

Residues


f (z) =p(z)

q(z)


Resz=z0

(f (z)) =p(z0)

q′(z0).

Note that q(z) has a simple zero at z = z0.

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Sajid Ali

ComplexIntegration





Residues

Residues


f (z) =p(z)

q(z)


Resz=z0

(f (z)) =p(z0)

q′(z0).

Note that q(z) has a simple zero at z = z0.

What if q(z) has a zero of order n?

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Sajid Ali

ComplexIntegration





Residues

Residues

Example-11. For a unit circle centered at origin, compute∫

C

1

zdz =?

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Sajid Ali

ComplexIntegration





Residues

Residues


C

1

zdz =?

Solution:Here q(z) = z which has a simple zero at z = 0 therefore

Resz=0

(f (z)) =p(0)

q′(0)=

1

1= 1.

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Sajid Ali

ComplexIntegration





Residues

Residues


C

1

zdz =?

Solution:Here q(z) = z which has a simple zero at z = 0 therefore

Resz=0

(f (z)) =p(0)

q′(0)=

1

1= 1.

Therefore, ∫C

1

zdz = 2πi re-verified

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Sajid Ali

ComplexIntegration





Residues

Residues

Example-2:2. For a unit circle centered at z = 1, compute∫

C

sin z

cos zdz =?

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Sajid Ali

ComplexIntegration





Residues

Residues


C

sin z

cos zdz =?

Solution:Here q(z) = cos z which has simple zeroes atz = ±π/2,±3π/2, ... but there is only one isolatedsingularity inside C therefore

Resz=π/2

(f (z)) =p(π/2)

q′(π/2)=

1

−1= −1.

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Sajid Ali

ComplexIntegration





Residues

Residues


C

sin z

cos zdz =?

Solution:Here q(z) = cos z which has simple zeroes atz = ±π/2,±3π/2, ... but there is only one isolatedsingularity inside C therefore

Resz=π/2

(f (z)) =p(π/2)

q′(π/2)=

1

−1= −1.

Therefore, ∫C

1

zdz = −2πi

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Sajid Ali

ComplexIntegration





Residues

Residues

Practice:3. What about residues of

f (z) =sin z

cos z

at z = ±π/2,±3π/2, ...?

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Sajid Ali

ComplexIntegration





Residues

Residues

Practice:4. Find the residues of

f (z) =1

z2 sin z

z =?

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Sajid Ali

ComplexIntegration





Residues

Residues

Multiple Residues:If a complex function f (z) is analytic everywhere in domainD except for finitely many points z1, z2, ..., zn then for aclosed simple curve C enclosing all singular points Cauchyresidue theorem states

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Sajid Ali

ComplexIntegration





Residues

Residues

Multiple Residues:If a complex function f (z) is analytic everywhere in domainD except for finitely many points z1, z2, ..., zn then for aclosed simple curve C enclosing all singular points Cauchyresidue theorem states∫

Cf (z)dz = 2πi

∑k=1

Resz=zk

(f (z)).

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Sajid Ali

ComplexIntegration





Residues

Residues

Example-1:1. Find the integral ∫

C

1

z2 − 5z + 6

over C which is a (i) circle of radius 4, (ii) circle of radius5/2.

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Sajid Ali

ComplexIntegration





Residues

Residues


C

1

z2 − 5z + 6

over C which is a (i) circle of radius 4, (ii) circle of radius5/2.Solution (i):Here q(z) = z2 − 5z + 6 which has simple zeroes at z = 2, 3and both isolated singularities are inside C therefore

CVT

Sajid Ali

ComplexIntegration





Residues

Residues


C

1

z2 − 5z + 6

over C which is a (i) circle of radius 4, (ii) circle of radius5/2.Solution (i):Here q(z) = z2 − 5z + 6 which has simple zeroes at z = 2, 3and both isolated singularities are inside C therefore

Resz=2

(f (z)) =p(2)

q′(2)=

1

−1= −1.

Resz=3

(f (z)) =p(3)

q′(3)=

1

1= 1.

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Sajid Ali

ComplexIntegration





Residues

Residues


C

1

z2 − 5z + 6

over C which is a (i) circle of radius 4, (ii) circle of radius5/2.Solution (ii):Since the radius of other circle is 5/2, which does notinclude the singularity 3 therefore∫

C

1

zdz = 2πi

(Resz=2

(f (z)))

= 2πi

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Sajid Ali

ComplexIntegration





Residues

Residues

Practice:2. Find the integral ∫

C

eaz

z2 + 1, a ∈ R

over C which is a circle of radius 2.

3. Find the integral ∫C

tan z

z2 − 1, a ∈ R

over C which is a circle of radius 3/2.

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Sajid Ali

ComplexIntegration





Residues

Residues

Practice:4. Evaluate the following integral, where C is the ellipse9x2 + y2 = 9 ∫

C

(zeπz

z4 − 16+ zeπ/z

).

5. Evaluate the following integral, where C is a unit circle atorigin ∫

C

cosh z

z2 − 3izdz .

CVT Sajid Ali Complex Integration Complex Integration...

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