# Analytic non constant function

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.

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You take the one tool that makes this easy, and poof. What have you done up until this point then? – Alex Youcis Apr 22 '12 at 2:27
I think it is related to analytical continuation. If the derivative is not equal to zero in some neighbourhood $z_0$ then we can choise enough small $r$ that all $z$'s comply injectivity. – user974514 Apr 22 '12 at 2:31
The relevant theorem is the Inverse Function Theorem. – Ragib Zaman Apr 22 '12 at 2:37
@RagibZaman: the Inverse Function Theorem doesn't help if $f'(z_0) = 0$. Fortunately the question does not ask to show that $f$ is injective on $V$. – Robert Israel Apr 22 '12 at 5:56

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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Thanks for the answer, but I have some question concerning your proof. Why is $1+g$ non vanishing?We excluded the point of $g(0)=0$ but $g$ is $n-k$ degree polynome how can you be sure that there will be no other $z$ in neighbourhood $U$ such that $f(z)=0$ ? – user974514 Apr 22 '12 at 17:17
$1+g := h$ is a continuous function, $h(0) = 1$. Plug in any $\epsilon < 1$ into the definition of continuity, and it spits out a ball of radius $\delta$, $U$, such that if you stay within the ball, $h(z)$ stays within $\epsilon$ of 1, in particular it can't hit 0. – uncookedfalcon Apr 22 '12 at 23:09
also, to be super clear, generically $\sum a_i z^i$ "goes on forever", that is $a_i \neq 0$ for infinitely many $a_i$; if all of the terms past some $n$ are 0, $f$ is just a polynomial! Hope that helps! – uncookedfalcon Apr 22 '12 at 23:11