Range of a holomorphic function on the disc I have a very simple question that had driven me mad these last hours : 
Let $f$ be a holomorphic function from the unit disc $\mathbb{D} =\{ z \in \mathbb{C} \ | \ |z| < 1\} $ to $\mathbb{C}$. Can the range of $f$ be all the complex plane ? 
 A: I'll add an example without trigonometry: 
$$f(z)=\left(\frac{z }{ 1-z}\right)^2$$
Here, $z\mapsto z/(1-z)$ maps the unit disk onto the halfplane $\operatorname{Re}z>-1/2$. The image of this halfplane under $z\mapsto z^2$ is the entire plane.
A: The function $\sin\colon\mathbb{C}\to\mathbb{C}$ is holomorphic, surjective and periodic of period $2\,\pi$. This implies that its restriction to the half-plane $\mathbb{H}=\{z:\Re{z}>0\}$ is also onto $\mathbb{C}$. Let $\phi\colon\mathbb{D}\to\mathbb{H}$ be a Möebius transformation between $\mathbb{D}$ and $\mathbb{H}$. Then $\sin(\phi(z))$ is holomorphic on $\mathbb{D}$ and onto $\mathbb{C}$.
A: Yes, I think it can. To see this, consider a function $f: \ ]0, 1[ \rightarrow \mathbb{R}$ whose range is the whole of $\mathbb{R}$. One example would be $f(x) = \tan(\pi x - \pi/2)$. Now let $g: \mathbb{D} \rightarrow \mathbb{C}$. For any $z \neq 0 \in \mathbb{D}$, define $g(z)$ to be the complex number with modulus $f(|z|)$ and argument the same as that of $z$. For $z = 0$, take $g(z) = 0$.
