Calculate integral using Cauchy's formula

[skrat]

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!

 

[vela]

You have the basic idea down, but what you wrote below is incorrect.

For $a_2=-1$:

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

For the given K and the way you chose $a$ and f, you can't say that
$$f(a)2\pi i=\oint _k\frac{f(z)dz}{z-a}$$ because function f isn't differentiable everywhere inside the region bounded by K. Therefore, the stuff you wrote below isn't correct:

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

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

Note also that when you considered the other pole, you said
$$\int _K \frac{\frac{(z-1)\sin(z)}{z+1}}{z-3}\,dz = \frac{2 \sin 3}{4}2\pi i.$$ You also said
$$\int _K \frac{\frac{(z-1)\sin(z)}{z-3}}{z+1}\,dz = -\pi i \sin 1.$$ You have the same integral equal to two different quantities!

When you calculated what the integral equaled by considering the pole at z=3, you were done. The pole at z=-1 is outside the contour, so you don't need to consider it at all.

 

[skrat]

When you calculated what the integral equaled by considering the pole at z=3, you were done. The pole at z=-1 is outside the contour, so you don't need to consider it at all.

I completely agree with that. I only wanted to check if my idea is right, where I completely forgot that $f$ may not be Holomorphic outside $K$.

So in case I would have two singular points inside given $K$, than the integral would be a sum of two, or not?

 

[vela]

Yes, that's how it will turn out, as you will learn later.

 

[skrat]

Thank you!