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Contour Integration - D2

The document discusses Laurent series and Taylor series expansions, types of isolated singularities including poles and essential singularities, methods for finding residues, and Cauchy's residue theorem. The residue theorem can be used to solve definite integrals by evaluating the sum of residues of a function enclosed by a contour.

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Sherlcok Holmes
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0% found this document useful (0 votes)
128 views33 pages

Contour Integration - D2

The document discusses Laurent series and Taylor series expansions, types of isolated singularities including poles and essential singularities, methods for finding residues, and Cauchy's residue theorem. The residue theorem can be used to solve definite integrals by evaluating the sum of residues of a function enclosed by a contour.

Uploaded by

Sherlcok Holmes
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Contents
 Introduction to Laurent and Taylor series
 convergence of series
 residue at a pole, residue at infinity
 methods of finding residues
 residue theorem
 solving of definite Integrals by contour integration

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residue
In Laurent expansion of any function f(z), the co-efficient of (z-z0)-1 about
isolated singularity z=z0 is defined as the residue of f(z) at z=z0.

R A
f(z) = + Z0

General term
(contains positive Principal term
power of z) (contains negative
power of z)

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Types of Isolated singularity
Isolated Singulariy

Removable Singulariy Poles Essential Singulariy


• If principal part of f(z) does not exist • If there are finite no. of terms in • If there are infinite no. of terms in
• b1=b2=…=bn = 0 principal part of f(z) principal part of f(z)
• Lt f(Z) exist & finite • bm≠ 0 and bm+1= bm+2 = bm+3 =…= 0, • Lt f(Z) does not exist
Z → Z0
Z → Z0
• In the expansion of f(z) all the then z = z0 is called pole of order m
• Limit point of zeros (means at which
powers are non-negative • If m = 1, then it is simple pole
value the function becomes zero) is
• We can remove the singularity by • Lt f(Z) exist & infinite
Z → Z0 essential singularity
redefining the function

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residue
 If a function has removable singularity at z=a, the residue of f(z) at z=a is 0
 If a function is analytic at z=a, the residue of f(z) at z=a is 0
R
Z0

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Solve

Find the residue of z-4 exp(iaz) at z = 0

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solution
exp(iaz) 1 (iaz ) 2
(iaz ) 3
f(z) = z-4 exp(iaz) =
z4
= 4 (1 + iaz +
z + + ….)
2 6

(ia)3 (a)3
Residue = the co-efficient of (z-0)-1 is = -i
6 6

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Solve

Find the residue of sin z/z at z=0

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solution

sin 𝑧 1
𝑓 𝑧 = = ( )
𝑧 𝑧

Residue = the co-efficient of (z-0)-1 is 0

 The function has removable singularity at z=0, the residue of f(z) at z=0 is 0

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1) Residue – at simple pole

R
Z0

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1) Residue –pole of order ‘m’

R
Z0

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1) Residue –pole of order ‘m’

R
Z0

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Cauchy's residue theorem
If f(z) be analytic within and on a simple closed curve C except for a finite
number of isolated singularities inside C then

R
Z0

Sum of residue of f(z) at all the


poles inside C

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Points to remember

Residue for simple pole

Residue for pole of R


Z0
order n

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Solve

Find the residue of z -4 exp(iaz) at z = 0

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solution

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Solve

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solution

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Solve

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solution

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Solve

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solution

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Solve

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solution

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Solve

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solution

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Solve

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solution

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Solve

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solution

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Solve

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solution

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solution

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