Can $\frac1{z^2}$ be integrated on $|z+i|=\frac32$ using Cauchy's theorem? $$
\begin{align}
\int_{|z+i|=\frac{3}{2}}\frac{1}{z^2}dz=0
\end{align}
$$
Is it safe to say the Integral is $0$ due to cauchy's Theorem?
Does this apply for any $z_0$ that lies inside the circle except for the center?
Does the fact that the denomitator is in a power higher than $1$ affect anything?
 A: No, you can't use Cauchy Theorem in this case, because function $f(z)=\frac{1}{z^2}$ isn't analytic inside the circle $|z+i|=\frac{3}{2}$ (point $z_0=0$).
But you can use Residue Theorem very simple way: the Laurent series for $f(z)$ in point $z_0=0$ has only one term, it's $\frac{1}{z^2}$, so the coefficient near $\frac{1}{z}$ is zero, so residue in point $z_0=0$ is zero (there is only one problematic point).
A: The integrand is not analytic at $0$ whichlies inside the contour of integration so we can not use Cauchy's thm.
The main reason of having $0$ as integral value is that the integrand has an antideritave in a domain containing the contour.
If the integrand were 1/z we would get $2\pi i $.(by using Cauchy integral formula extended version--deformation of contours.)
Higher integer powers would give $0$ since they all have antiderivatives.
A: The function $\frac{1}{z^2}$ is bad in the given circle, but it is good out the circle (even at $\infty$). This suggests that, after changing of variable turn outer to inner, then there is a chance to apply the theorem.
