Biholomorphic $f:\{z\in\mathbb{C}:|z|<1,~Im(z)>0\}\rightarrow\{z\in\mathbb{C}:|z|<1\}$

Question

Is there a bijective holomorphic function $$f:\{z\in\mathbb{C}:|z|<1,~Im(z)>0\}\rightarrow\{z\in\mathbb{C}:|z|<1\}$$ such that $f^{-1}$ is holomorphic?

You can give me a composition of functions. For example: If you know a biholomorphic function $$f_0:\{z\in\mathbb{C}:|z|<1,~Im(z)>0\}\rightarrow\{z\in\mathbb{C}:Im(z)>0\}=:H$$ that's enough as the function $$f_1:H\rightarrow\{z\in\mathbb{C}:|z|<1\}$$ is biholomorphic.

Answer 1

The map $z \to \frac{1+z}{1-z}$ takes the upper-half disk to the first quadrant. Now use the squaring map $z \to z^2$ to complete the answer.

Answer 2

What a familiar looking question. Saw this with a one unit pre-translation on a qualifying exam today.

We write a sequence of transformations, each of which is a conformal equivalence (a biholomorphism):

• $\mu_1$: from the open upper half disk to $\Pi^+ \setminus \overline{\Bbb{D}}$, the upper half plane without the closed unit disk: $z \mapsto-1/z$.
• $\mu_2$: from $\Pi^+ \setminus \overline{\Bbb{D}}$ to $\Pi^+$: $z \mapsto \frac{1}{2}\left(z+\dfrac{1}{z}\right)$. One should probably show that this map holds $\Bbb{R} \setminus (-1,1)$ fixed and takes the upper unit semicircle $\{\mathrm{e}^{\mathrm{i}t} : 0<t<\pi\}$ to $(-1,1)$ via the cosine.
• $\mu_1^{-1}$: from $\Pi^+$ to $\Bbb{D}$.

Then $\mu_1^{-1} \circ \mu_2 \circ \mu_1$ maps your set biholomorphically to the open unit disk.

Edit: Fixed the missing $\frac{1}{2}$ in $\mu_2$ and stopped claiming $\mu_2(z) = \frac{1+z^2}{2z}$ was a Möbius transform.