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**Cauchy Integral Formula - Multiple Repeated Poles**

## Homework Statement

C.I.F is doing my head in.

Evaluate [itex]∫ {\frac {{z}^{2}+1}{ \left( z-3 \right) \left( {z}^{2}-1 \right) }}[/itex]

For the closed path |z| = 2

## The Attempt at a Solution

This is a circle of radius 2, with singularities [itex]z = 3, z = -1, z = 1[/itex]

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

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

f(z) here would be [itex]{\frac {{z}^{2}+1}{z-3}}[/itex]

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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