# Open map in complex numbers

Is a function $f: \mathbb{C} \rightarrow \mathbb{C}$ where $f(z) = z^2$ an open map if $\mathbb{C}$ has the metric topology $d(z,a) = |z-a|$ ?

I can think of several reasons why $f$ should map open sets to open sets, but am having a difficult time generating a formal proof.

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you could share your thoughts with us :) –  Dominic Michaelis Mar 12 '13 at 20:30

$z^2$ is holomorphic and not constant and hence open, as you are using the standard topology on $\mathbb{C}$