If $f(z)$ is analytic at $z_0$, show that $f(z)$ has a zero of order $k$ at $z_0$ if and only if $\dfrac 1 {f(z)}$ has a pole of order $k$ at $z_0$.
I solved it but I'm not sure about my solution.
($\Rightarrow$) Since $f(z)$ is analytic at $z_0$, we have a power series expansion $f(z)=\sum_n a_n (z-z_0)^n$ for some nbd of $|z-z_0|
But is it okay to substitute $f(z)$ by power series in the denominator? I feel somewhat careful to deal with power series.