Recently, I read the notes "Vector bundles on Riemann surfaces" by Sabin Cautis (http://www-bcf.usc.edu/~cautis/classes/notes-bundles.pdf). On the sixth page of these notes, there is a statement without any explanation to it: "For example, $z_0^2+z_1^2-1 = 0 $ is isomorphic (as a complex manifold) to $\mathbb{C}$." We are dealing with the zero set of this polynomial in $\mathbb{C}^2$.
Since I didn't manage to find a proof of this statement in the time available, I discussed it with some colleagues over a few beers, one evening. Even though approaches were attempted, we didn't quite get anywhere. The approach that looked most promising is to prove the given zero set is simply connected and then invoke the uniformisation theorem. By determining the automorphism group, we would then be able to find out that it's indeed $\mathbb{C}$. Would this approach be able to solve the question? (i.e. can anyone fill in the details?) And if not, how could the statement be proven differently?
Remark. This zero-set of course looks very sphere-like at first sight, yet an absence of $|$ prevents this.
(I wouldn't be surprised if this question is around on this site already, in which case I couldn't find it and would appreciate a link t)
The statement is simply false. As requested, here is an extension of my above comment.
I denote by $Z(z_0^2+z_1^2-1) \subset \Bbb{C}^2$ the vanishing locus of the polynomial $z_0^2+z_1^2-1$.
Consider the linear change of coordinates $$ z_0 \mapsto \frac{1}{2} \left(\frac{w_0}{1+i}+w_1 \right) $$ $$ z_1 \mapsto \frac{-i}{2}\left(\frac{w_0}{1+i}-w_1 \right) $$ This gives an isomorphism of the complex variety $Z(z_0^2+z_1^2-1)$ onto the hyperbola $Z(w_0w_1-1)$. Now we clearly have $$ Z(w_0w_1-1)=\{(t,t^{-1}) \ \; \vert \; t \in \Bbb{C}^*\} $$ Projection onto the first factor gives the desired isomorphism of $Z(w_0w_1-1)$ with $\Bbb{C}^*$, and $\Bbb{C}^*$ is clearly not isomorphic to $\Bbb{C}$, e.g. $\Bbb{C}$ is simply connected, while $\Bbb{C}^*$ isn't. Hence we can conclude that $Z(z_0^2+z_1^2-1)$ is not isomorphic to $\Bbb{C}$ as well.