Possible analytic images of the unit disc? From Picard's theorem, the image of $\mathbb{C}$ under an analytic function has to the whole plane or $\mathbb{C}$ minus a single point. What about other open sets? This may be too broad, so how about the unit disc?
Let $B$ be the open unit disc. If $f:B\to \mathbb{C}$ is analytic, what can we say about $f(B)$? 
We know that $f(B)$ has to be open. Can something further be said? 
For instance, $f(B)$ need not be simply connected (consider $exp(\alpha z)$ for a large enough $\alpha$). 
 A: In fact $f(B)$ can be any open subset of $\Bbb C$. (Edit: Or rather any connected open set; thanks to Robert Israel for noticing that this is what I meant...)
Bonus: If $V$ is any open connected subset of the plane there exists $f$ with non-vanishing derivative such that $f(B)=V$, and in fact if $V\ne\Bbb C$ we can take $f$ to be a covering map.
Say $V\subset\Bbb C$ is open and connected. Assume first that $\Bbb C\setminus V$ contains more than one point. Now the Uniformiization Theorem shows that there is a simply connected Riemann surface $M$ and a holomorphic covering map from $M$ onto $V$. And $M$ is either the Riemann sphere, the disk or the plane. Except $M$ cannot be the Riemann sphere since $V$ is not compact, and the little Picard theorem shows that $V$ cannot be $\Bbb C$, so $M$ is the disk.
That does it except for $V=\Bbb C\setminus \{0\}$ and $V=\Bbb C$. There is a strip $S$ so that $\exp(S)=\Bbb C\setminus\{0\}$ and a map from the disk onto $S$. And there is a map from the disk onto the region $\Re z>-1$; the square of this map maps $B$ onto $\Bbb C$.
We're done except for the bonus part, and we're almost done with that. If $V\ne\Bbb C$ we've given a holomorphic covering map from $B$ onto $V$. If $V=\Bbb C$ there is no such covering map. But there does exist a holomorphic surjection with non-vanishing derivative:
Say $f$ is the entire function with $f(0)=0$ and $f'(z)=e^{\cos(z)}$. Then $f'(z+2\pi)-f'(z)=0$, so there exists $c$ with $f(z+2\pi)=f(z)+c$. Note that $c=f(0)>0$; in particular $c\ne0$.
Since $f(z+2\pi)=f(z)+c$ with $c\ne0$ the Big Picard theorem shows that $f$ takes every complex value infinitely many times. Hence $f(\Bbb C\setminus\{0\})=\Bbb C$. As above there is a covering map $g$ from $B$ onto $\Bbb C\setminus\{0\}$; now $f\circ g$ is a map from $B$ onto $\Bbb C$ with non-vanishing derivative.
A: Bloch-Landau theorem states: There exists an absolute constant $\ell$ such that, for any holomorphic function $f$ defined in the unit disc, with the property $|f'(0)| =1$, Then:
The image of the unit disc contains a ball with radius greater or equal to $\ell$.
Notice that the condition on the derivative is "just" a normalization constraint.
