# Cauchy Integral Formula

Cauchy Integral Formula - Multiple Repeated Poles

## Homework Statement

C.I.F is doing my head in.

Evaluate $∫ {\frac {{z}^{2}+1}{ \left( z-3 \right) \left( {z}^{2}-1 \right) }}$

For the closed path |z| = 2

## The Attempt at a Solution

This is a circle of radius 2, with singularities $z = 3, z = -1, z = 1$

Since z = 3 falls outside the circle, we focus on the two singularities inside.

Since the C.I.F is $2\,i\pi \,f \left( z_{{0}} \right)$

f(z) here would be ${\frac {{z}^{2}+1}{z-3}}$

Substituting z = 1, and z = -1

We get -1 and -1/2 from f(z)

So that is -2Pi*i -Pi*i = -3Pi*i

However the answer given is -i*Pi/2

I have been stuck on this for a while and I can't figure out what I'm doing wrong.

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answer is -i*pi/2.why are you leaving the factor z^2-1 in denominator.at z=1 residue is -1/2 and at -1 it is 1/4 and so the answer.

I am sorry, I did not quite understand what you meant. I am not leaving the factor (z^2 - 1) in the denominator.

How did you get those two values (-1/2 and 1/4)?

When z = 1, isn't it (1^2 + 1)/(1 - 3) = 2/-2 = -1 ?

z = -1; (-1)^2 + 1/(-1 -3) = 2/-4 = -1/2

I think I see what you mean, I must also include the factor (z -1) and (z + 1) with (z - 3) every time I evaluate at one of the singularity. Now I get that correct answer.

Thank you.

congratulations.

Lol ^