0
$\begingroup$

Let $\gamma\subset\Bbb C$ be a closed, simple (if $\gamma:[a,b]\mapsto\Bbb C$, $\gamma$ is injective on $(a,b)$, so the curve doesn't intersect except for $\gamma(a)=\gamma(b)$) and piece-wise regular curve and $z_1,z_2,...$ is an infinite set of distinct points all inside the domain (call it $\Omega$) defined by $\gamma$ ($\partial\Omega=\gamma$) such that $z_n$ does not converge towards any limit in $\overline\Omega$.

The analytic continuation principle requires a set a distinct points that converge to a point in the open $\Omega$ such that a certain holomorphic function $f(z_n)=0~\forall n\in\Bbb N$ and $z_n\ne\lim z_n~\forall n$.

And in an exercise about the formula ${1\over 2\pi i}\int_{\gamma}\frac{f'}{f}=\{\text{the number of zeros of } f$ inside $\Omega$ counted with their multiplicity (order)$\}$, the following explentation confused me a little bit:

Let $z_1,...,z_m$ be in $\Omega$ the zeros of $f$ of order $k_1,...,k_m \ge 1$ respectively. Then $m$ is finite because otherwise, by the analytic continuation principle, $f$ would be identically zero, which is not the case because $f$ is not constant.

$\endgroup$
  • $\begingroup$ $\Omega$ is not countable. $\endgroup$ – Kavi Rama Murthy Jan 9 at 10:35
  • $\begingroup$ You don't need analytic continuation for that. If the sequence $(f(z_i))_i$ admits an accumulation point, $f$ is constant and equal to the limit. $\endgroup$ – James Jan 9 at 13:27
  • $\begingroup$ Is that a theorem? What is it called? $\endgroup$ – John Cataldo Jan 9 at 13:28
  • $\begingroup$ My advice would be to prove those things assuming $f$ is analytic. If the $ a_n$ are distinct $a_n \to a, f(a_n) = 0$ and $f$ is analytic at $a$ then from the values of $f(a_n)$ you can find $f^{(k)}(a)$ for every $k$, so $\forall n, f(a_n) =0 \implies \forall k, f^{(k)}(a) = 0$ and hence $f=0$. The alternative is to show show directly that $f$ analytic implies $f(z) = f(z_0) + C(z-z_0)^k+O((z-z_0)^{k+1})$ so the zeros of $f$ are isolated. Finally the same result applies for $f$ holomorphic once you showed that holomorphic $\implies$ analytic. $\endgroup$ – reuns Jan 9 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.