In an exercise I am asked to find a power series for $f(z)=\frac{1}{1+z^2}$ centered at $0$.
My approach was the following:
$f(z)=\frac{1}{1+z^2}=\frac{1-z^2}{(1+z^2)(1-z^2)}=(1-z^2)\frac{1}{1-z^4}=(1-z^2)\sum_{n\geq0}z^{4n}$
But this does not seem right because a power series is defined as $\sum a_n(z-a)^n$ and I have something like: $\sum a_n(z-a)^{kn}$, plus I don't know how to pull that $(1-z^2)$ inside the sum.
So how can I write this function as a power series?
Usually what is the main approach when trying to write a function as a power or as a Laurent series? Because I allays find myself having some trouble doing so.
Edit:
I just realized that we can continue to manipulate this expression: $(1-z^2)\sum_{n\geq0}z^{4n}$
$$(1-z^2)\sum_{n\geq0}z^{4n}=\sum_{n\geq0}z^{4n} - z^2\sum_{n\geq0}z^{4n}=$$
$$\sum_{n\geq0}z^{4n} -\sum_{n\geq0}z^{4n+2}=\sum_{n\geq0}(-1)^n z^{2n}$$
$\frac 1 {1+z^{2}}=1-z^{2}+z^{4}-z^{6}+...$ for $|z| <1$. RHS is a geometric sum and this is special case of the sum of a geometric series.