CVT Sajid Ali Complex Integration Complex Integration...

Sajid Ali

ComplexIntegration

Simply ConnectedDomains

Complex LineIntegrals

Cauchy IntegralTheorem

Cauchy IntegralFormula

Residues

Complex Integration

Sajid Ali

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October 2, 2017

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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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.

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..

Sajid Ali

ComplexIntegration

Residues

Complex Integration

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

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

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

F ′(z)dz = F (z1)− F (z0)

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ComplexIntegration

Residues

Complex Integration

F ′(z)dz = F (z1)− F (z0)

Example-1: ∫ 1+i

0z2dz =?

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ComplexIntegration

Residues

Complex Integration

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

Sajid Ali

ComplexIntegration

Residues

Complex Integration

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

⇒∫ 1+i

0z2dz =

3|1+i0

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ComplexIntegration

Residues

Complex Integration

F ′(z)dz = F (z1)− F (z0)

Practice:

∫ πi

−πicos zdz =?

zdz =?.

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ComplexIntegration

Residues

Complex Integration

Way-2: Parameterize the curve

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 =

af (z)z(t)dt

where z = x + i y .

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 =

af (z)z(t)dt

where z = x + i y .

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

zdz =?

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ComplexIntegration

Residues

Complex Integration

zdz =?

Step-1 As C is a unit circle so parameterize

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ComplexIntegration

Residues

Complex Integration

zdz =?

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

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ComplexIntegration

Residues

Complex Integration

zdz =?

Step-2 Differentiate z = ie it

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ComplexIntegration

Residues

Complex Integration

zdz =?

Step-3 Calculate f (z(t)) = e−it

Sajid Ali

ComplexIntegration

Residues

Complex Integration

zdz =?

Step-3 Calculate f (z(t)) = e−it

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

∫ 2π

0ie−ite itdt

= 2πi Ans.

Sajid Ali

ComplexIntegration

Residues

Complex Integration

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

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?

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?

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∫

z − 1dz =?

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∫

z − 1dz =?

Q-5 C is a unit circle∫C

(z + z−1)dz =?

Sajid Ali

ComplexIntegration

Residues

Complex Integration

Practice:Evaluate the same integral

∫C 1/zdz over more paths !!!

Sajid Ali

ComplexIntegration

Residues

Complex Integration

1. For an ellipse

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

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 =?

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.

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

Sajid Ali

ComplexIntegration

Residues

Cf (z)dz = 0

Practice: For an arbitrary closed curve C

∫Cezdz =?, (ii)

cos(z)dz =?, (iii)

∫Czndz =?,

Sajid Ali

ComplexIntegration

Residues

Cf (z)dz = 0

Practice: For an arbitrary closed curve C

∫Cezdz = 0, (ii)

cos(z)dz = 0, (iii)

∫Czndz = 0,

Sajid Ali

ComplexIntegration

Residues

Cf (z)dz = 0

Practice: For an arbitrary closed curve C∫C

sec (z)dz = 0,

z2 + 4dz = 0, C = Unit Circle

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?

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 !!!

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

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∫

z2dz = 0 C = Unit Circle

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

Sajid Ali

ComplexIntegration

Residues

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

Sajid Ali

ComplexIntegration

Residues

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

Sajid Ali

ComplexIntegration

Residues

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.

Sajid Ali

ComplexIntegration

Residues

Complex Integral Formula

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

z − z0dz = 2πif (z0)

where C is taken counterclockwise and contains z0.

Sajid Ali

ComplexIntegration

Residues

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

z − 2dz =

Sajid Ali

ComplexIntegration

Residues

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

z − 2dz = 2πi |ez |z=2

= 2πiez |z=2, (as ez is analytic)

= 2πie2

Sajid Ali

ComplexIntegration

Residues

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

z2 − 4dz =?

Sajid Ali

ComplexIntegration

Residues

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

Sajid Ali

ComplexIntegration

Residues

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

z2 + 1

z2 − 1dz =?

Sajid Ali

ComplexIntegration

Residues

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).

Sajid Ali

ComplexIntegration

Residues

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

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

Sajid Ali

ComplexIntegration

Residues

z2 + 1

z2 − 1dz =?

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

Sajid Ali

ComplexIntegration

Residues

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

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).

Sajid Ali

ComplexIntegration

Residues

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

z2 + 1

z2 − 1dz =?

z2+1z−1z + 1

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

Sajid Ali

ComplexIntegration

Residues

Laurent Series

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

(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

∫Cf (z) dz

Sajid Ali

ComplexIntegration

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,

Sajid Ali

ComplexIntegration

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).

Sajid Ali

ComplexIntegration

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)).

Sajid Ali

ComplexIntegration

Residues

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

Sajid Ali

ComplexIntegration

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

5!− ...

z3− 1

5!− ...

Therefore,

Resz=z0

(f (z)) = −1

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

6= −πi

Sajid Ali

ComplexIntegration

Residues

How to compute residues?For a function of the form

f (z) =p(z)

Sajid Ali

ComplexIntegration

Residues

f (z) =p(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

Sajid Ali

ComplexIntegration

Residues

f (z) =p(z)

Resz=z0

(f (z)) =p(z0)

q′(z0).

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

Sajid Ali

ComplexIntegration

Residues

f (z) =p(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?

Sajid Ali

ComplexIntegration

Residues

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

zdz =?

Sajid Ali

ComplexIntegration

Residues

zdz =?

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

Resz=0

(f (z)) =p(0)

q′(0)=

Sajid Ali

ComplexIntegration

Residues

zdz =?

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

Resz=0

(f (z)) =p(0)

q′(0)=

Therefore, ∫C

zdz = 2πi re-verified

Sajid Ali

ComplexIntegration

Residues

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

cos zdz =?

Sajid Ali

ComplexIntegration

Residues

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.

Sajid Ali

ComplexIntegration

Residues

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.

Therefore, ∫C

zdz = −2πi

Sajid Ali

ComplexIntegration

Residues

Practice:3. What about residues of

f (z) =sin z

at z = ±π/2,±3π/2, ...?

Sajid Ali

ComplexIntegration

Residues

Practice:4. Find the residues of

f (z) =1

z2 sin z

Sajid Ali

ComplexIntegration

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

Sajid Ali

ComplexIntegration

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)).

Sajid Ali

ComplexIntegration

Residues

Example-1:1. Find the integral ∫

z2 − 5z + 6

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

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ComplexIntegration

Residues

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

Sajid Ali

ComplexIntegration

Residues

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.

Resz=3

(f (z)) =p(3)

q′(3)=

Sajid Ali

ComplexIntegration

Residues

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∫

zdz = 2πi

(Resz=2

(f (z)))

= 2πi

Sajid Ali

ComplexIntegration

Residues

Practice:2. Find the integral ∫

z2 + 1, a ∈ R

over C which is a circle of radius 2.

3. Find the integral ∫C

z2 − 1, a ∈ R

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

Sajid Ali

ComplexIntegration

Residues

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

(zeπz

z4 − 16+ zeπ/z

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

cosh z

z2 − 3izdz .

CVT Sajid Ali Complex Integration Complex Integration...

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