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 $|z-z_0|<r$. But since $z_0$ is a zero of order $k$, $a_0=\cdots=a_{k-1}=0$ and $a_k \neq 0$. So $f(z)=\sum_{n=k}^{\infty} a_n (z-z_0)^n$. Since $f(z)$ is analytic on $|z-z_0|<r$, $\dfrac1{f(z)}$ is so on $0<|z-z_0|<r$. Then $$\displaystyle \lim_{z \to z_0}(z-z_0)^k \dfrac1{f(z)}=\lim_{z \to z_0}\dfrac1{a_k+a_{k+1}(z-z_0)+\cdots}=\frac1{a_k}\neq 0,\infty.$$ So $\dfrac1{f(z)}$ has a pole of order $k$ at $z_0$.
But is it okay to substitute $f(z)$ by power series in the denominator? I feel somewhat careful to deal with power series.
Best Answer
Why not simply write $f(z) = (z-z_0)^k\cdot g(z)$ with $g$ analytic and nonzero at $z_0$. Hence $\frac{(z-z_0)^k}{f(z)}=\frac1{g(z)}$ is analytic in a punctured neighbourhood and $\to\frac1{g(z_0)}$ as $z\to z_0$?