$$\int_{\gamma=(i,1)} \frac{z^3}{(z-i)^n} dz$$ for any $n\in\mathbb{N}$.
Can someone please help me answer this question as I cannot seem to get the right answer!
Please note that the Cauchy integral formula must be used in order to solve it.
Many thanks in advance!
On
Hint: Parametrize the contour as $z=i+e^{i\theta}$ for $0\leq\theta\leq 2\pi$. Then, by definition, you have $$\int_0^{2\pi}\frac{(e^{i\theta})^3}{((i+e^{i\theta})-i)^n}\cdot ie^{i\theta}\,d\theta=i\int_0^{2\pi}\frac{(e^{i\theta})^3}{e^{in\theta}}\cdot e^{i\theta}\,d\theta=i\int_0^{2\pi}(e^{i3\theta})e^{i(1-n)\theta}\,d\theta.$$Keep simplifying and you can solve it.
Use the residue theorem. Since $\cfrac{z^3}{(z-i)^n}$ has a pole of order $n$ at $z=i$ and analytic everywhere other than $z=i$ in the domain $|z-i|<1$, by residue theorem, we have $$\int_{\gamma}\frac{z^3}{(z-i)^n}dz=2\pi ig(i),\text{where }g(z)=\frac{1}{(n-1)!}(z^3)^{(n-1)}$$
The residue theorem is obtained from Cauchy Integral formula.
By Cauchy Integral formula, we have $$2\pi if(z)=\int_C\frac{f(\zeta)}{\zeta-z}d\zeta$$ Differentiate both sides with respect to $z$ $n-1$ times, we get $$2\pi if^{(n-1)}(z)=(n-1)!\int_C\frac{f(\zeta)}{(\zeta-z)^n}d\zeta$$