How can I solve this integral with complex number

Question

$$\int_{|z-1| = 1}\frac{1}{(1-z^2)}dz$$

I tried to do this by residue calculus $$I=2\pi*iRes(f(z),1)$$ but I coudn't get the answer..

I would be grateful if you could give a clue.

Additional question)

$$\int_{|z| = 3}\frac{z}{(1-z^2)}dz$$ Is $$I=2\pi*i{Res(f,0)}+Res(f,1)$$ and the answer is 0 right?

Answer 1

(1) We have $\int_{|z-1| = 1}\frac{1}{(1-z^2)}dz=2\pi iRes(f(z);1)$

$Res(f(z);1)=\displaystyle \lim_{z\to1}\frac{z-1}{(1-z^2)}=\frac {-1}2$

$\int_{|z-1| = 1}\frac{1}{(1-z^2)}dz=2\pi i\times\frac{-1}{2}=-\pi i$

(2)$\int_{|z| = 3}\frac{z}{(1-z^2)}dz=2\pi i(Res(f;1)+Res(f;-1))$

$Res(f(z);1)=\displaystyle \lim_{z\to1}\frac{(z-1)z}{(1-z^2)}=\frac {-1}2$

$Res(f(z);-1)=\displaystyle \lim_{z\to-1}\frac{(z+1)z}{(1-z^2)}=\frac {-1}2$

$\int_{|z| = 3}\frac{z}{(1-z^2)}dz=2\pi i(\frac{-1}{2}-\frac{1}{2})=-2\pi i$

Answer 2

HINT:

Note that

$$\frac{1}{1-z^2}=\frac{1/2}{1-z}+ \frac{1/2}{1+z}$$

The poles are at $z=\pm 1$ and the residues are $\pm 1/2$. Now, which poles, if any are contained in $|z-1|<1$? Which are contained in $|z-1| <3$?