Evaluate the integral $\int_C \frac{1}{z^3} dz$

Question

Where $C$ is the square centered in the origin with edges in $-1+i$, $-1-i$, $1+i$ and $1-i$.

The first step I made is $$\displaystyle \int_C \frac{1}{z^3} dz = \int_C \frac{\frac{1}{z^2}}{z} dz$$

But the problem is on how could I apply the Cauchy's Integral Formula. Any help?

And using the same $C$ as above, how would it work for $\displaystyle \int_C \frac{sin(z)}{z^4}dz$? If it's possible, just a hint.

I saw a few examples here, but I still do not understand the process to solve this kind of integrals.

Answer 1

In the region enclosed in the square, the integrand function has one pole $z =0$. By residue theorem $$\int_C \frac{1}{z^3}dz = 2\pi i \operatorname{res}_{z=0}\frac{1}{z^3} = 0$$, i.e. the integral is zero.

Answer 2

Since the square is homotopic to the boundary of the unit disk $D$, we have that $$ \int_C \frac{1}{z^3}dz=\int_{\partial D} \frac{1}{z^3}dz $$ Now use the parametrization $\theta\mapsto e^{i\theta}$, with $0\le\theta\le 2\pi$.