- #1

skrat

- 748

- 8

## Homework Statement

Let ##K## be a circle ##|z-2|=2##. Using Cauchy's formula calculate ##\int _K\frac{(z-1)\sin(z)}{z^2-2z-3}dz##

## Homework Equations

Cauchy's formula: ##f(a)2\pi i=\oint _k\frac{f(z)dz}{z-a}##

## The Attempt at a Solution

It's my first time dealing with this kind of problems so I am not really sure that what I am doing is in fact the right way to do it. I would be really happy if one could check my solution.

##\frac{(z-1)\sin(z)}{z^2-2z-3}=\frac{(z-1)\sin(z)}{(z-3)(z+1)}## therefore ##a_1=3## and ##a_2=-1##.

For ##a_1=3##:

##f(z)=\frac{(z-1)\sin(z)}{z+1}##

##f(a)2\pi i=\oint _k\frac{f(z)dz}{z-a}##

##\frac{2\sin(3)}{4}2\pi i=\int _K\frac{\frac{(z-1)\sin(z)}{z+1}}{z-3}dz##

For ##a_2=-1##:

##f(z)=\frac{(z-1)\sin(z)}{z-3}##

##f(a)2\pi i=\oint _k\frac{f(z)dz}{z-a}##

##\frac{-2\sin(-1)}{-4}2\pi i=\int _K\frac{\frac{(z-1)\sin(z)}{z-3}}{z+1}dz##

##-\pi i\sin(1)=\int _K\frac{\frac{(z-1)\sin(z)}{z-3}}{z+1}dz##

Since ##K## is a circle with center in 2 and radius 2, point ##a_2=-1## is not inside ##K##. Which means that only integral around ##a_1=3## counts and therefore

##\int _K\frac{(z-1)\sin(z)}{z^2-2z-3}dz=\int _K\frac{\frac{(z-1)\sin(z)}{z+1}}{z-3}dz=\frac{2\sin(3)}{4}2\pi i##.

I HOPE so. Thanks in advance!