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

I just want to know an example of non-constant complex valued function which is open map.

Is this ok?

$f:\mathbb{C}\rightarrow \mathbb{C}$ given by $f(z)=\bar{z}$? This is just a reflection with respect to $x$-axis, right?

share|cite|improve this question
yes, that will do. – user20266 Jun 29 '12 at 17:24
This is a homeomorphism, so it's certainly open. On the other hand, it isn't holomorphic, so you couldn't appeal to the open mapping theorem from complex analysis. Also note that a constant function $\mathbb C \to \mathbb C$ is not open, since the image of the open set $\mathbb C$ is just a point. – Dylan Moreland Jun 29 '12 at 17:24
or the identity maps $f(z)=z$... (but maybe you really meant "not identity" when you wrote "non-constant") – user31373 Jun 29 '12 at 17:29
Thanks to every one , I was just searching for a non-constant complex valued(need not be analytic) map which is open :) – Un Chien Andalou Jun 29 '12 at 18:18
Why not take any analytic non-constant function? – PAD Jun 29 '12 at 20:28
up vote 2 down vote accepted

Compiling the comments: $f(z)=\bar z$, or $f(z)=z$, or any nonconstant holomorphic function will work.

share|cite|improve this answer

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.