Computing the integral with residues

Question

I need to compute this integral $\int\limits_{|z-i|=2}\bar{z}\cos(z)\,dz$ with residue theorem and so on. So I can represent it as $\int\limits_{|z-i|=2}\frac{|z|^2\cos(z)}{z}\,dz$ what should I do next? I can’t realize what $|z|$ exactly is.

Answer 1

Note that\begin{align}\overline z\cos(z)&=\left(\overline z+i\right)\cos(z)-i\cos(z)\\&=\frac{\lvert z-i\rvert^2}{z-i}\cos(z)-i\cos(z)\\&=\frac{4\cos(z)}{z-i}-i\cos(z),\end{align}when $\lvert z-i\rvert=2$. So, by the residue theorem, your integral is equal to $8\pi i\cos(i)$, since $\operatorname{res}_{z=i}\frac{4\cos(z)}{z-i}=4\cos(i)$ and $i\cos(z)$ is entire.