<s>[INST] Radius of convergence of $f(z)=\frac{z^4e^z}{(6-z)(2i-z)} $ The Question is:

Compute the radius of convergence for the expansion at the point
$z=4+3i$ of the function $$f(z)=\frac{z^4e^z}{(6-z)(2i-z)}. $$

In my text book the solution is written like this:
$\sqrt{(4-6)^2+3^2}$,  but I don't understand how they calculate it. Can someone please explain it?
Thanks!
 [/INST] The function $f$ is holomorphic on $\mathbb{C} \setminus \{6, 2i\}$. The Radius of Convergence is given by the minimum of the distances from $4+3i$ to $z = 6$ or $z = 2i$. That's due to the fact that we can write $f$ as a power series on $D_R(4+3i)$ (meaning the open disc of radius $R$ around $4+3i$). You also might fancy a look at Radius of Convergence.
The distance from $4 + 3i$ to $z = 2i$ is given by $\sqrt{4^2 + 1^2}$. The distance from $4 +3i$ to $z = 6$ is given by $\sqrt{2^2 + 3^2}$. This will immediately become clear if you draw a figure and use Pythagoras.
Since
\begin{align}
\sqrt{17} = \sqrt{4^2 + 1^2} > \sqrt{2^2 + 3^2} = \sqrt{13}
\end{align}
and $R = \operatorname{min}\{\sqrt{4^2 + 1^2}, \sqrt{2^2 + 3^2}\}$ we find $R = \sqrt{2^2 + 3^2}$ or as the text book has written it: $R = \sqrt{(4-6)^2 + 3^2}$.
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