Complex integration problem via Cauchy's integral formula

Question

I want to integrate the following :$$\int_{|z|=2} \frac{dz}{z^{2}-1}$$ in the positive direction.

So my idea is two split the integral into a sum of two integral , something like $$\int_{|z|=2} \frac{dz}{z-1}+\int_{|z|=2} \frac{dz}{z+1}$$ since $$(z-1)(z+1)=z^{2}-1$$ and then proceed by applying Cauchy's integral formula since the poles of each function in the sum of integral is inside the circle $|z|=2$.

The only problem is how to correctly split the function into a sum of two functions? Is my idea right in order to solve the integral? Thanks

Answer 1

HINT: $$\frac{1}{z^2-1} = \frac{1}{2} \frac{2}{(z-1)(z+1)} = \frac{1}{2} \frac{(z+1)-(z-1)}{(z+1)(z-1)} = \frac{1}{2} \Bigg(\frac{\color{red}{z+1}}{\color{red}{(z+1)}(z-1)}-\frac{\color{blue}{z-1}}{(z+1)\color{blue}{(z-1)}}\Bigg) = \frac{1}{2} \Bigg(\frac{1}{z-1}-\frac{1}{z+1}\Bigg) $$

Now you can proceed as you stated.

But what is Cauchy Index Theorem? I haven't heard of it. I assume you meant Cauchy Integral Theorem?

Answer 2

This is by partial fractions. Here is the general method:

You want $A,B$ with $$\frac{A}{z-1}+\frac{B}{z+1}=\frac{1}{(z-1)(z+1)}$$ Then we have:

$A(z+1)+B(z-1)=1\Rightarrow (A+B)z+A-B=1$ for all $z$ then $A=-B$ and $A-B=1\Rightarrow -2B=1$ and $B=-1/2$ and $A=1/2$. And now you have the integral you would like to evaluate.