Radius of convergence of Taylor series of $\frac{1}{1+z^2}$ around $z_0$ Define $f(z)=\frac{1}{1+z^2}, z\ne \pm i$. Prove that if $z_0=x_0+iy_0$ where $z_0\ne\pm i$ then the radius of convergence of the Taylor series of $f$ around $z_0$ is $\sqrt{x_0^2+(1-|y_0|)^2}$.
Here is what I've got so far:
In general the Taylor series expansion is of the form $\sum_{n=0}^{\infty}\frac{f^{(n)}(z_0)}{n!}(z-z_0)^n$. With $f(z)=\frac{1}{1+z^2}$ I am not sure how to compute this. If I knew this, I would then apply the ratio test and solve for $R$ but at the moment I am stuck.

$f$ is defined on $D= \mathbb C \setminus \{i,-i\}$ and $f$ has poles of order $1$ in $i$ and $-i$ . If $z_0 \in D$, then it is known that the radius of convergence of the Taylor series of $f$ around $z_0$ is given by
$$R= dist (z_0, \partial D).$$
Here: $\partial D=\{i,-i\}$ , hence
$$R= \min\{|z_0-i|,|z_0+i|\}.$$
Can you proceed ?
