Zeros of Holomorphic Functions

Question

I am studying theory on the zeros of a function; in particular the the zeros of some order $m$ of a holomorphic function. However I do not understand the following proof presented to me in proving that:

A function $f(z)$ has a zero of order $m$ $\iff$ $f(z)=g(z)(z-z_0)^m$ where g(z) is holomorphic at $z_0$ and $g(z_0)\neq0$.

I define a zero of a function of order $m$ at $z_0$ to mean that the first $0,\dots m-1$ derivatives evaluated at $z_0$ are zero and the $m$th derivative is non-zero at $z_0$.

The only if part of the prove leaves me a little unsure as to why the function $g(z)$ as I shall show below is holomorphic. I have searched online but it has mostly been stated that this function is holomorphic so maybe I'm missing something obvious?

The proof goes like this:

The Taylor series for $f(z)$ is: \begin{align*} f(z)&=0+\dots+0+\frac{f^{(m)}(z_0)}{m!}(z-z_0)^m+\frac{f^{(m+1)}(z_0)}{(m+1)!}(z-z_0)^{m+1}+\dots \\&=(z-z_0)^m \Big\{ \frac{f^{(m)}(z_0)}{m!}+\frac{f^{(m+1)}(z_0)}{(m+1)!}(z-z_0)+\dots\Big\} \end{align*}

and now the function in the curly parenthesis is denoted as $g(z)$, which is claimed to be the holomorphic function. It is clear $g(z_0)\neq0$ but could it be pointed out why the function is also a convergent power series ie analytic/holomorphic etc.

Answer 1

If the series after “$f(z)=$” is convergent in the disk $D(z_0,r)$ then after dividing by $(z-z_0)^n$, we still have a convergent power series. And the sum of a convergent power series is always an analytic function.

Answer 2

When defining an analytic function, you only need that the taylor series converges pointwise. It is clear that if we fix $z$, then at at each step $z-z_0=k$ for some $k \in \mathbb C$, so the series after factoring, say $\{f_n(z)\}$, still converges since if it did not, then $\{k f_n(z)\}$ would not either, which is the original series.