Biholomorphism between Riemann Surfaces. Let's consider the homeomorphism $\psi : \mathbb C \rightarrow B_{1}(0), z \rightarrow \frac {z}{|z|+1}$.
Let $Z$ be be the Riemann Surface with topological space $\mathbb C$ induced by the chart $( \mathbb C,\psi)$.
I have to show that $Z$ is biholomorphic $B_{1}(0)$ .
I think that this means that I have to find a holomorphic function $g$ such that :
$\psi \space \circ \space g \space \circ \space id_{\mathbb{C}}^{-1}$ is a holomorphism and
$id_{\mathbb{C}} \space \circ \space g^{-1} \space \circ \space \psi^{-1}$ is also a holomorphism,
but I got stuck finding such function. Any ideas?
Thank you in advance.

One point of importance here is that $g$ need only be holomorphic as a map $B_1(0) \to Z$, not to the usual complex structure on $\mathbb{C}$. For this it suffices to check that $g$ is a diffeomorphism and $\varphi \circ g$ is holomorphic for all coordinate charts $\varphi$ in an atlas for the Riemann surface structure on $Z$. (We only need check holomorphicity one way, which comes from the fact that the inverse of a holomorphic map with nonvanishing derivative is holomorphic).
Since $Z$ is given with $\psi$ as the only chart in an atlas, we only need to show there exists a diffeomorphism $g$ with $\psi \circ g$ holomorphic. Such a map is given by the inverse map of $\psi: \mathbb{C} \to B_1(0)$, which one can work out to be $$g(z) = \frac{z}{1-|z|},$$ which is a diffeomorphism $B_1(0) \to \mathbb{C}$ and satisfies $\psi \circ g(z) = z$ for all $z \in B_1(0)$, which is holomorphic. This completes the proof.
