Radius of convergence of the Taylor series of $ f(z)=\Large\frac{3}{z^2-3iz-2}$

Question

Find the radius of convergence of the Taylor series of $$ f(z)=\frac{3}{z^2-3iz-2}$$ expanded about $z=1$.

Separating into partial fractions $$f(z)= \frac{3i}{z-i}-\frac{3i}{z-2i}$$

which has poles $z=i$ and $z=2i$ and thus is analytic on $\mathbb C \setminus \{-3,2 \}$

I'm a bit confused about the next part, I know the answer is $\sqrt 2 $ but I'm trying to understand why...

Taking $\min\{(z_0-z_1),(z_0-z_2)\} =\min\{(1-i),(1-2i)\} $

which gives $|1-i|=\sqrt 2$ and $|1-2i|=\sqrt 5$, the smaller being $\sqrt 2$, hence the radius of convergence? Is this logic correct or does the $\min$ part mean something else?

Would be grateful for any help!

Answer 1

The function $f$ is analytic on $\mathbb{C}\setminus\{i,2i\}$. Therefore, the radious of convergence of the Taylor series of $f$ centered at $1$ is$$\min\{|1-i|,|1-2i|\}=\min\{\sqrt2,\sqrt5\}=\sqrt2.$$