Taylor Series of $f(z) = \frac{1}{z^2+i}$ centered at $z=0$

Question

I know how to find Taylor Series for functions like $\frac{1}{z+w}$ for some $w$, but the $z^2$ is tripping me up here. And trying to write in the form of $f(z)=\sum_{n=0}^\infty \frac{f^{(n)}(z_0)}{n!} (z-z_0)^n= \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} z^n$ seems convoluted because of the $z^2$.

Answer 1

$$ f(z) \; = \; -i(1 - iz^2)^{-1} \; = \; \sum_{n=0}^\infty (-i)^{n+1}z^{2n} $$ valid for $|z| < 1$. We just have an infinite GP with common ratio $-iz^2$.

More generally, there is no problem with considering the Taylor series of a function $$ f(z) \; = \; \sum_{n=0}^\infty a_n z^n $$ and deducing from it the Taylor series $$ f(z^2) \; = \; \sum_{n=0}^\infty a_nz^{2n} $$ without chasing all the bothersome derivatives of $f(z^2)$.