Apply the Cauchy integral formula to find:
(i) $\displaystyle\int_{|z-2i|=2}(z-i)^{-3}\arcsin(z)dz$
(ii) $\displaystyle\int_{|z-2|=3/2}\frac{\text{Log}(1+z)}{(z^2-1)^2}dz$
(iii) $\displaystyle\int_{|z|=2}\frac{z^k}{(z-1)^{k+1}}dz$ for $k$ a positive integer
(iv) $\displaystyle\int_{\partial Q}\frac{\sin(\pi z)}{z^3+z^2}dz$, where $Q$ is the square whose vertices are $2$, $2i$, $-2$ and $-2i$
My attempt:
(i) Is $f(z)=\text{Arcsin}(z)$ analytic in $|z-2i|=2$? yes because this is no analytic in $z=\pm 1$ and $\pm 1 \notin |z-2i|=2$, then $\int_{|z-2i|=2}\frac{\text{Arcsin}(z)}{(z-i)^3}dz= \frac{f''(i)2\pi i}{2}=\pi i \frac{i}{\sqrt{(1-i^2)^3}}=\frac{-\pi}{\sqrt{8}}$
(ii) since $f(z)=\frac{\text{Log}(1+z)}{(z+1)^2}$ is analytic in $|z-2|=3/2$, then $\int_{|z-2|=3/2}\frac{\text{Log}(1+z)}{(z^2-1)^2}dz=\int_{|z-2|=3/2}\frac{\text{Log}(1+z)}{(z+1)^2(z-1)^2}dz=2\pi i f'(1)=\pi i \frac{1-2\text{Log}(2)}{4}$
(iii) we know $f(z)=z^k$ is a entire function, then $\int_{|z|=2}\frac{z^k}{(z-1)^{k+1}}dz=\frac{2\pi i f^{(k)}(1)}{k!}=\frac{2\pi i}{k!}$
(iv) We know that $f(z)=\sin(\pi z)$ is a entire function, then $\int_{\partial Q}\frac{\sin(\pi z)}{z^3+z^2}dz=\int_{\partial Q}\frac{\sin(\pi z)}{z^2}+\int_{\partial Q}\frac{\sin(\pi z)}{z+1}-\int_{\partial Q}\frac{\sin(\pi z)}{z}=2\pi i f'(0)+2\pi i f(-1)-2\pi i f(0)=2\pi^2 i$
Is all this well? Thank you very much.