Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 1 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.