$\int_{|z-2i|=1}\frac{(z-2i+\frac{1}{2})^2 \sin(2i\pi z)}{\overline{z}^{2} (z-2i)^4} \ dz$ Let $C$ be the circle $|z -2i|=1$ 
How Can I Compute this Integral :
$$\int_{C}\frac{(z-2i+\frac{1}{2})^2 \sin(2i\pi z)}{\overline{z}^{2} (z-2i)^4} \ dz$$
Thank you 
 A: Let $I$ be given by the contour integral 
$$I=\int_{|z-2i|=1}\frac{(z-2i+1/2)^2 \sin(2\pi iz)}{\overline{z}^{2} (z-2i)^4} \ dz \tag 1$$
First, we enforce the substitution $w=z-2i$ in $(1)$ to obtain
$$I=\int_{|w|=1}\frac{(w+1/2)^2\sin(2\pi i w )}{(\bar w -2i)^2w^4}\,dw\tag2$$
Next, we note that on the circle $|w|=1$ we have $\bar w=1/w$.  Therefore, we can write $(2)$ as
$$\begin{align}
I&=\int_{|w|=1}\frac{(w+1/2)^2\sin(2\pi i w )}{(1/w -2i)^2w^4}\,dw\\\\
&=\int_{|w|=1}\frac{(w+1/2)^2\sin(2\pi i w )}{(1 -2iw)^2w^2}\,dw\tag3
\end{align}$$
The integral on the right-hand side of $(3)$ can be evaluated using the residue theorem.  There are poles inside $|w|=1$.  There is a first order pole at $w=0$ and a second order pole $w=-i/2$.
Can you finish now?
A: I have tried solving this using cauchy integral formula as follows :
$|z-2i|= 1\Rightarrow  (z-2i)(\overline{z}+2i)=1 \Rightarrow (z-2i+\frac{i}
{2}-\frac{i}{2})(\overline{z}+2i)=1$
Now we have : $$(z-2i+\frac{i}{2}-\frac{i}{2})(\overline{z}+2i)=1 \Rightarrow(z-2i+\frac{i}{2})(\overline{z}+2i)=\frac{i \overline{z}}{2}$$
Which implies that :$$\frac{(z-2i+\frac{i}{2})^2}{\overline{z}^2}=\frac{-1}{4(\overline{z}+2i)^2}$$
hence the integral become :
$$\begin{align}\int_{|z-2i|=1}\frac{(z-2i+\frac{1}{2})^2 \sin(2i\pi z)}{\overline{z}^{2} (z-2i)^4} \ dz&=\frac{-1}{4}\int_{|z-2i|=1}\frac{\sin(2i\pi z)}{(\overline{z}+2i)^2 (z-2i)^2(z-2i)^2} \ dz \\ \\
&=\frac{-1}{4}\int_{|z-2i|=1}\frac{\sin(2i\pi z)}{(z-2i)^2} \ dz \\ \\
&=\frac{-1}{4} 2\pi i \cos(4\pi) \\ \\
&=0
\end{align}
$$
