When a holomorphic function is identically 0

Question

I'm trying to prove this theorem (Theorem 4.8, Chapter 2, page 52, Complex analysis by Stein and Shakarchi):

Suppose $f$ is a holomorphic function in a region $\Omega$ that vanishes on a sequence of distinct points with a limit point in $\Omega$. Then $f$ is identically $0$.

First we want to prove that $f$ is identically zero in a small disc containing a limit point. So they use the power series expansion of $f$ in that disc. Then this is the part I'm confused about:

If $f$ is not identically zero, there exists a smallest integer $m$ such that $a_m \neq 0$ but then we can write $$f(z) = > \displaystyle\sum_{n=0}^\infty a_m(z-z_0)^m(1+g(z-z_0))$$ where $g(z-z_0)$ converges to $0$ as $z \rightarrow z_0$.

If someone can clarify whats going on here, that would be much appreciated. Or if you can refer me to another place I can read a proof, that would also be appreciated.

Answer 1

I'll try to provide a skeleton of Rudin's proof:

1. Show that a holomorphic function is analytic: it has can be represented as a power series in a disc around each point in its domain.

2. Take a limit point of its set of zeros. If you examine its power-series about that point, then all coefficients must vanish (otherwise we can represent $f(z)=(z-z_0)^m g(z)$ where $g$ is holomorphic and $g(z_0)\neq 0$, but then $g$'s continuity implies that $f$ doesn't vanish in some punctured disc about $z_0$).

3. This shows that the limits points of $f$'s zet of zeroes form an open set. It's clear that the set of non-limit points is also open, so one of them must be empty (connectedness of the domain). A limit point exists by assumption, and we're done.

Answer 2

This is just the definition of zero of order m at $z_0$. In this case you have

$$ f(z)=a_m(z-z_0)^m[1+\frac{a_{m+1}}{a_m}(z-z_0) + \frac{a_{m+2}}{a_m}(z-z_0)^2 .. ] = a_m(z-z_0)^m[1+\phi(z)] $$

Usually the definition is condensed to

$$f(z)=(z-z_0)^m\psi(z)$$ with the obvious fact

$$\psi(z_0)=1$$

Hope this helps.