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Lovely Professional University

TERM PAPER

OF

MTH-201

TOPIC: What is role of singular points in complex

analysis, Discuss different types of singular points?

Submitted to:- Submitted by :

Miss Nivedita Abhishek bansal

Roll no: RB6802A06

B.TECH-M.TECH (ECE)

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ACKNOWLEDGEMENT

The present piece of work owes its existence to many persons without whose help it would not

have been possible to complete the project.

I would like to thank Miss Nivedita for being my guide from the institute side for the last one

month and for his valuable suggestions and constructive criticism.

I would also like to thank to him for arranging such a good project which gave me a lot of

practical exposure to corporate world and from which I learnt a lot which will definitely help mein the coming times when I join the corporate after finishing my course.

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Complex analysis:-

Singularity is extremelyortant in complex analysis, where they characterize the possible

behaviors of analytic functions. Complex singularities are points in the domain of a function

where fails to be analytic.

Types of Singular Points:-

Role of singular points in complex analysis:-

Complex Integration: First of all, you have to be able to locate and characterize the

singularities of the integrand.

A: Location

To that end, note that if g(z) and h(z) are (simpler) functions of the complex variable z, then f(z)

= g(z)/h(z) diverges (tends to infinity) where g(z) diverges and also where h(z) vanishes. To

discern the behavior of f(z) near a singularity, it suffices to expand g(z) and h(z) in a Laurent

series to lowest (most negative/least positive) order. Thus:

Note that

cos(x+iy) = sin(x) cos(iy) + cos(x) sin(iy) = sin(x) cosh(y) + i cos(x) sinh(y)

so that cos(z) diverges when y = m(z) does, at y ±�. So does, then, cot(z). To sum up, thesingularities of cot(z) are at z = 0,±,±2, . . . and �.

B: Order

Next, we need to determine the severity of each singularity. Since

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where the (infinite) sum clearly defines the analytic part of the function cot(z). That is, in the

corresponding Laurent expansion, aí1 = 1 and an = 0, n < í1. This last fact implies, by

definition, that cot(z) has a pole of the first order at z = 0. The same analysis will verify that

cot(z) has poles of the first order at all of its singularitites, z = 0,±1,±2, . . . ,�; e.g.:

By contrast, the function z cot(z) would have a pole of the second order at z = 0,

but poles of the first order at z = ±1,±2, . . . ,. Continuing in this vein, the function

sec2(z) = [cos(z)]2 has a (double) pole wherever cos(z) = 0, i.e., at z = ±1

2 (2n+1).

Finally, ez has an essential singularity at z = 0. To see this, note that the Taylor Expansion has an infinite radius of convergence, so that the expansion

is valid in the entire complex plan outside z = 0. This turns out to be the actual Laurent

expansion of e1/z , proving that there is a nonzero coefficient an for n < N, regardless how

negative an N we chose: there is no limit to the ³depth´ of the singularity of e1/z at z = 0; it is

essential .

the function g(z) = [f(z)]1 vanishes at z0, and at the rate (zz0)m, i.e., we then know that

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and where the Taylor coefficients bn are related (but are not equal in general) to the Laurent

coefficients aní2m. As a simple but nontrivial example, consider

where f(z) has a pole of order 1 at both z0 = ±2, and g(z) vanishes linearly at the same points (in

all cases, it is the nonzero term of lowest order that determines the order of the pole and the rate

of vanishing).

C: Residues

Finally, we will need the result

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Therefore integration of the function is F=2i(Sum of residues).

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Refrences:-

www.google.com

www.mathworld.com

www.wikipedia.com

www.answers.com

www.ask.com

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