Let $h$ be the bilinear map: $$h:\mathbb{C}^{n+1}\times \mathbb{C}^{n+1}\to \mathbb{C},\ \ \ \ (z,w)\mapsto -z_0\overline{w_0}+\sum_{i>0} z_i\overline{w_i}$$ and let $\mathfrak{I}:h(z,z)=-1$. There is an obvious left action of $S^1$ on $\mathfrak{I}$ and I wanna prove that $\mathfrak{I}/S^1$ is diffeomorphic to the open ball in $\mathbb{C}^n$.
My attempt
It's pretty clear that every element of $\mathfrak{I}/S^1$ can be written (almost) uniquely as: $$z=(r_0,r_1e^{i\theta_1},...,r_ne^{i\theta_n})$$ so my idea was to define something like $z\mapsto (r_1,...,r_n,\theta_1,...,\theta_n)$, but this clearly works only if we suppose that $r_i\neq 0$ for every $i>0$ (otherwise things get a bit messy, because $\theta_i$ is not even well defined). Is this the right track?
Let $Z=(z_0,z)$ with $z_0 \in \Bbb C$ and $z \in \Bbb C^n$. Then $h(Z,Z) = -1$ if and only if $|z_0|^2 = 1+ \|z\|^2$. It follows that $z_0\neq 0$ whenever $Z \in \Im$ (and even that $|z_0| \geqslant 1$). In this case, its equivalence class satisfies $[z_0,z] = [|z_0|, z z_0^{-1}|z_0|]$, and $(|z_0|, z z_0^{-1}|z_0|)$ is the unique representative whose first element is positive. It follows that we have the following bijection $$ z \in \Bbb C^n \longmapsto \left[\sqrt{1+\|z\|^2}, z\right] \in \Im/S^1. $$ I leave you as an exercise that this is indeed a smooth diffeomorphism. In particular, $\Im/S^1$ is diffeomorphic to $\Bbb C^n$. It is now standard that $\Bbb C^n$ is diffeomorphic to a ball. For instance, the map $u\in B(0;1) \mapsto \frac{u}{\sqrt{1-\|u\|^2}}\in \Bbb C^n$ is a diffeomorphism. Since $$ 1 + \left\|\frac{u}{\sqrt{1-\|u\|^2}}\right\|^2 = \frac{1}{1-\|u\|^2}, $$ it follows that $$ u \in B(0;1) \longmapsto \left[\frac{1}{\sqrt{1-\|u\|^2}}, \frac{u}{\sqrt{1-\|u\|^2}} \right] \in \Im/S^1 $$ is a diffeomorphism.