# $f(z)$ has a zero of order $k$ iff $1/f(z)$ has a pole of order $k$

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.

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Are you familiar with using the local taylor theorem to show that if $f(z)$ has a zero of order $k$ at $z_0$ then $f(z)=g(z)(z-z_0)^k$ where $g(z)$ is analytic and $g(z_0)\neq 0$? – JSchlather Sep 6 '12 at 16:49
@JacobSchlather Yeah, I know that and also the similar counterpart of the case of poles. I can prove it by using these. But I wonder that my proof above is right. – Gobi Sep 6 '12 at 17:01
I would extend "Since $f(z)$ is analytic on $|z-z_0|<r$, $\dfrac1{f(z)}$ is so on $0<|z-z_0|<r$" to something like "Since $f(z)$ is analytic on $|z-z_0|<r$, then for sufficiently small $r$, $\dfrac1{f(z)}$ is so on $0<|z-z_0|<r$." Otherwise, I don't see a problem with your argument. $f$'s Taylor series converges to $f$ in the small neighborhood, so I don't think you need be concerned at the point you are concerned at. – alex.jordan Sep 6 '12 at 17:17

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 $z\to z_0$?