How to prove that surface in $\mathbb{C}^3$ is not homemorphic to $\mathbb{C}^2$ In my homology class I've got a following problem: I need to prove, that $X = \{(x,y,z) \in \mathbb{C}^3: \ x^2 + y^2 + z^2 = 0\}$  is not homemorphic to $\mathbb{C}^2$.
I have tried to delete point zero and show, that $ X \backslash \{0\}$ will not be connected, but I can't understand how this set looks like. I also tried to do something with homology of $X$ or $X \backslash \{0\}$, and I think, that it's an intendend way to solve this problem, but I didn't understand, how to do it.
So I want to ask: how to compute homology of such sets? Or mayby I should be looking in other direction?
 A: Here is the solution I came up with. Let $x = u_1 + iv_1,\ y = u_2 + iv_2, \ z = u_3 + i v_3$. Consider vectors $u = (u_1,u_2,u_3),\ v = (v_1,v_2,v_3) \ \in \mathbb{R}^3$, then equation $x^2 + y^2 + z^2 = 0$ equivalent to the system
\begin{equation} 
   \begin{cases}
       |u|^2 - |v|^2 = 0 \\
       (u,v) = 0
   \end{cases}
\end{equation}
Consider $X \backslash \{0\}$, then this system is equivalent to
\begin{equation} 
   \begin{cases}
       |\tilde{u}| = 1 \\
       (\tilde{u},v) = 0 \\
       v \neq 0
   \end{cases}
\end{equation}
where $\tilde{u} = \frac{u}{|v|}$
So $\tilde{u}$ lie in $\mathbb{S}^2$, and $v$ is tangent nonzero vector to $\tilde{u}$, thus $A \backslash \{0\}$ is homotopically equivalent to $T^1 \mathbb{S}^2 = \{ (u,v) \in T\mathbb{S}^2 \ : \ |v| = 1\}$ and it is known, that unit tangent bundle to $\mathbb{S}^2$ is homeomorphic to $\mathbb{R}P^3$.
Hence $X \backslash \{ 0\}$ is homotopically equivalent to $\mathbb{R}P^3$, but $\mathbb{C}^2 \backslash \{0\}$ is homotopically equivalent to $\mathbb{S}^3$, these spaces are not homotopically equivalent, thus initial spaces are not homemorphic.
I hope this solution will be useful to someone.
