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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?

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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 :) – La Belle Noiseuse Jun 29 '12 at 18:18
Why not take any analytic non-constant function? – PAD Jun 29 '12 at 20:28

1 Answer 1

up vote 2 down vote accepted

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

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