complex integration-how to solve the given problem

Question

how do we calculate the value

$\frac{1}{2\pi i}\int\frac{\sum_{n=0}^{15}z^n}{(z-i)^3}dz$ in $C$:|z-i|=2 ?

the answer for this is 1+15i.. how to get it?

can someone please explain?

Answer 1

We take $f(z)=\sum_{n=0}^{15} z^n$, and by cauchy integral theorem we know that $$\frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{(z-i)^3}dz=\frac{1}{2}f^{(2)}(i)=\frac{1}{2}\left(\frac{1-16z^{15}+15z^{16}}{(z-1)^2}\right)'|_{z=i}=\frac{1}{2}\left(2\frac{-1+120z^{14}-224z^{15}+105z^{16}}{(z-1)^3}\right)|_{z=i}=52+60i$$ Addendum:

$\sum_{k=0}^{15}z^k=\frac{z^{16}-1}{z-1}$,take the derivative, we have $\frac{16z^{15}}{z-1}-\frac{z^{16}-1}{(z-1)^2}=\frac{1-16z^{15}+15z^{16}}{(z-1)^2}$. By similar computation we get the second derivative.