I am confronted with the following exercise:
Compute the radius of convergence for the expansion at the point $z=4+4i$ for \begin{equation} f(z)=\frac{z^{5}e^{z}}{(2-z)(3i-z)} \end{equation}
I don't really know where to begin to solve this problem. All that seem promising is to use the following identity \begin{equation} a_{k}=\frac{1}{2\pi i}\int_{\Gamma}\frac{f(w)}{(w-z_{0})^{k+1}}dw \end{equation} where $\Gamma$ is the positively oriented boundary of a disc with radius $r$, centered at $z_{0}=4+4i$, together with the Cauchy-Hadamard theorem which gives
\begin{equation} R=\frac{1}{\lim_{k \to \infty}\sup \sqrt[k]{|a_{k}|}} \end{equation}
Is this approach even remotely right, and if not, then how should I proceed?
All you have to do is to calculate the minimum distance of the point of expansion to the poles of the function. For your question note that you have two poles at $z=2$ and $z=2i$ and the distance of the point $z=4+4i$ to them is ${d_1} = \sqrt {{{(4 - 2)}^2} + {4^2}} = 2\sqrt 5 $ and ${d_2} = \sqrt {{4^2} + {{(4 - 2)}^2}} = 2\sqrt 5 $ so that the radius of convergence is equal to $2\sqrt 5 $.