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Let $f$ an holomorphic function not bounded. How can I show that $f(\mathbb C)$ is dense in $\mathbb C$ ? I'm sure we have to use Liouville theorem, but I don't see how.

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    $\begingroup$ Note: there is a (difficult) generalization of this, called Picard's theorem : if $f(\mathbb C)$ omits at least two values, then $f$ must be constant $\endgroup$ – Glougloubarbaki Mar 22 '15 at 0:06
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By contradiction, suppose that $f(\mathbb C)$ is not dense in $\mathbb C$. Therefore, there is a $z_0\in\mathbb C$ and a $r>0$ such that $f(\mathbb C)\cap B_r(z_0)=\emptyset$. Therefore $$\frac{1}{f-z_0}<\frac{1}{r}$$ and thus, by Liouville theorem $\frac{1}{f-z_0}$ is constant. This implies that $f$ is also constant which is a contradiction.

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