I have recently learned how to compute complex integrals using the Cauchy integral formula but I have come across one with an nth power involved where I am unsure how to proceed. The integral is:
$$\int_{C_1(i)}\frac{z^3}{(z-i)^n}dz$$
For reference, my definition provided for the Cauchy integral formula is:
$$\bbox[yellow] {f(z_0) = \frac{1}{2πi}\oint_{C_r(\hat z)}{\frac{f(z)}{z-z_0}}dz } $$
EDIT: Have asked my lecturer and apprently it is possible to use Cauchy's integral formula for derivatives but I have had no luck, any help?
Notice that $f(z)=z^3$ is entire. So it is differentiable on $D_{1+\epsilon}(i)$.
Cauchy's integral formula for derivatives tells us that for all $\zeta \in D_{1+\epsilon}(i)$ $$f^{(n-1)}(\zeta)=\frac{(n-1)!}{2\pi i}\int_{C_1(i)}\frac{f(z)dz}{(z-i)^n}$$
Hence $$\int_{C_1(i)}\frac{z^3dz}{(z-i)^n}=\frac{2\pi i f^{(n-1)}(i)}{(n-1)!}$$
Hopefully that is enough for you to determine the integral for various values of $n$
Edit: Set $n'=n-1$ then Cauchy's formula for derivatives states: $$f^{(n')}(\zeta)=\frac{n'!}{2\pi i}\int_{C_1(i)}\frac{f(z)dz}{(z-i)^{n'+1}}$$ Work out this integral and then replace $n'$ with $n-1$ in the final answer