# Radius of convergence, complex power series

I want to compute the radius of convergence of the series for the function $f(z)=\frac{1}{1+z^2}$. I have written this as $$f(z)=\sum_{n=0}^\infty (-1)^n(z^2)^n$$ and then said $$R=\lim_{n \to \infty} \left \lvert \frac{(-1)^n}{(-1)^{n+1}} \right \rvert = \lim_{n \to \infty} \left \lvert -1 \right \rvert = 1$$

Could someone verify if this is correct or not? My main confusion is with writing $f(z)$ in the form that I did. I thought it needed to be in the form $\sum_{n=0}^\infty c_n(z-a)^n$ but I have $z^2$.

To write the coeficients in the form $\sum_{n=0}^\infty c_n(z-a)^n$ you should take $a=0$ and $c_{2n} = (-1)^n$ and $c_{2n+1} = 0$. Thus, your convergence radius is $$r = \frac{1}{\limsup_{n} \sqrt[n]{|c_n|}} = 1$$ and your answer is now justified.