Integrate $z^2/(z^2 - 1)$ by Cauchy's formula counterclockwise around the circle

Question

Integrate $z^2/(z^2 - 1)$ by Cauchy's formula counterclockwise around the circle for the equation $|z + 1| = 1$

I know that the solution is $2πz^2/(z - 1) = πi$ (with $z = -1$). But how is it solved?

Answer 1

HINT: the circle defined by $|z+1|=1$ is just the circle on the complex plane of radius one centered around the point $z=-1$, that is, it is $\partial\Bbb D(-1,1)$.

Then for $f(z):=\frac{z^2}{z^2-1}=\frac{z^2}{(z+1)(z-1)}$ we have that $1\notin\overline{\Bbb D}(-1,1)$ so the integral reduces to...

Because $\frac{z^2}{z-1}$ is holomorphic on $\overline{\Bbb D}(-1,1)$ by the Cauchy integral formula we have that $$\int_{\partial\Bbb D(-1,1)}f(z)dz=2\pi i\left[\frac{z^2}{z-1}\right]_{z=-1}$$