TITLE: Analytic non constant function
QUESTION [0 upvotes]: Stuck up on something in complex analysis.
Let $f$ analytic function and open $\Omega \subset \mathbb{C}$. Show that if $f$ is not a constant on a neighbourhood of $z_0$, then exist a neighbourhood $V$ of $z_0$ so that 
$z\in \mathbb{V}$ and $f(z)=f(z_0) \Rightarrow z=z_0$.
Note: This should be proven without Cauchy-Riemann because of the axiomatic system of the book.

REPLY [1 votes]: Here's how to see that theorem, if you buy the (true) statement that $f$ can be written locally as a power series $\sum a_n (z-z_0)^n$ about $z_0$. WLOG $z_0 = 0$, $f(z_0) = 0$. As $f$ isn't locally constant, let $a_k$ be the minimal nonzero coefficient, WLOG $a_k = 1$.
Then $f(z) = z^k(1+g)$, where $g$ is just all the remaining terms with $z^k$ factored out, note $g(0) = 0$.
By continuity, $1+g$ is nonvanishing in a neighborhood $U$ of 0, in the punctured nbh $U-0$ we deduce $f$ is nonvanishing.
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