2
$\begingroup$

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 $|z-z_0|$f(z)$ is analytic on $|z-z_0|$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.

  • 0
    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 $g(z_0)\neq 0$?2012-09-06
  • 0
    @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.2012-09-06
  • 0
    I would extend "Since $f(z)$ is analytic on $|z-z_0|, $\dfrac1{f(z)}$ is so on $0<|z-z_0|" to something like "Since $f(z)$ is analytic on $|z-z_0|, then for sufficiently small $r$, $\dfrac1{f(z)}$ is so on $0<|z-z_0|." 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.2012-09-06

1 Answers 1