Complex contour integral Problem

Question

Show that

$$\oint_{|z|=1} \dfrac {(z+w)(z^{n-1})} {z-w}dz=0$$ using Residue calculus, where $n<0$ and $|w|<1$.

Answer 1

There are $2$ poles inside the unit circle: the one at the origin of order $1-n$, and the residue at this pole is $$\begin{align}\frac1{(-n)!}\lim_{z\rightarrow0}\frac {d^{-n}}{dz^{-n}}\left(z^{1-n}\frac{(z+w)z^{n-1}}{z-w}\right)&=\frac1{(-n)!}\lim_{z\rightarrow0}\frac{d^{-n}}{dz^{-n}}\left(1+\frac{2w}{z-w}\right)\\ &=\frac1{(-n)!}\lim_{z\rightarrow0}\frac{2w(-1)^n(-n)!}{(z-w)^{1-n}}\\ &=-2w^n\end{align}$$ And the residue at $z=w$ is $$\lim_{z\rightarrow w}\left((z-w)\frac{(z+w)z^{n-1}}{z-w}\right)=2w^n$$ So the sum of the residues is $-2w^n+2w^n=0$.