zeroes of anlytic function is isloated .

Question

I want to prove that zeroes of an non constant analytic function is islolated , my process - let S be the zeroes of an analytic function , then first of all that S is a countable set ( i proved this ) i.e zeroes of an non constant analytic function is countable ,let z1 belongs to S is arbitarily so f(z1)=0 , now my goal is to find an open ball centered at point z1 that will contain no zeroes of f(z) other than z1, since set S is countable, so any open ball centered at z1 belongs to S is uncountable but contains countable many points of set S , ( as S is countable ) so , B(z1,r) cannot be contained in S so , it's interior is phi( plz correct if iam wrong) , so set S is not open , also zeroes of an analytic function is closed ( inverse image of a closed set under a continuos function ) so , S is countable , closed set which is not open , so S contains all of it's limit points, i cannot be able to think , how to show zeroes are isolated , we can use these results to show zeroes are isolated ??

Answer 1

If $f(z_0)=0$, let $n$ be the smallest $n\in\mathbb N$ such that $f^{(n)}(z_0)\neq0$. Such a $n$ must exist or otherwise $f$ would be $0$ in a neighborhood of $z_0$. So, the Tayllor series of $f$ near $z_0$ is of the type$$a_n(z-z_0)^n+a_{n+1}(z-z_0)^{n+1}+\cdots,$$with $a_n\neq0$. Therefore, near $z_0$,$$f(z)=a_n(z-z_0)^n\left(1+\frac{a_{n+1}}{a_n}(z-z_0)+\cdots\right)$$and $1+\frac{a_{n+1}}{a_n}(z-z_0)+\cdots\neq0$ if $z$ is close enough to $z_0$. So, $f(z)\neq0$ if $z$ is close enough to $z_0$ but $z\neq z_0$.