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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.2012-06-29
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    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](http://en.wikipedia.org/wiki/Open_mapping_theorem_(complex_analysis)) 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.2012-06-29
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    or the identity maps $f(z)=z$... (but maybe you really meant "not identity" when you wrote "non-constant")2012-06-29
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    Thanks to every one , I was just searching for a non-constant complex valued(need not be analytic) map which is open :)2012-06-29
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    Why not take any analytic non-constant function?2012-06-29

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