# conics in $\mathbb{C}^2$ are always connected

We have been said that every conic in $\mathbb{C}^2$is connected, but I have not understood how to prove that. We have seen that through some transformation, I can always have something like $x^2+y^2=1$, that is a sort of "complex circle" in $\mathbb{C}^2$. That is not bounded so it is not compact either. But how do I prove that it is connected? We have been suggested to study the curve on the projective $P_2(\mathbb{C})$, but I really don't know what to do. If I have understood correcly (as the projective space is quite new to me), the solutions (x,y) can be seen in $P_2(\mathbb{C})$ as elements like $(1,\frac{x_1}{x_0},\frac{x_2}{x_0})$ with $x_0,x_1,x_2 \in \mathbb{C}$ and $x_0$ obviously different from 0. Then if I write $x^2+y^2=1$ with the projective coordinates, and then multiply for $x_0^2$ in order to have an homogeneous polynomial, I get $x_{1}^2+x_{2}^2=x_{0}^2$.Now I have the previous solutions and also the solutions for $x_0=0$, that should represent the "point at infinity". So until now I have created an omeomorphism between the curve on $\mathbb{C}^2$ and the points like $(1,\frac{x_1}{x_0},\frac{x_2}{x_0})$ in the projective space, is it right? Now it starts getting really confusing for me. It says I should create an omeomorphism between the points of the curve in $P_2$ and $P_1(\mathbb{C})$, using a stereographic projection. Then I could conclude that my curve is omeomorphic to the Riemann sphere (that is, $P_1(C)$) minus one or two points given by the "points at infinity", and as the sphere is connected, my curve is connected.

I don't get what the projection should be like, how do I do the stereographic projection with $\mathbb{C}$ as field? How should I work with $P_2(\mathbb{C})$ and $P_1(\mathbb{C})$ to get an explicit omeomorphism? Thanks for the help

P.S. I've had an idea, maybe the reasoning is completely wrong (as already said, I'm still not very used to this subject), but I'll give it a try. When I send the solutions of (for instance) $x^2+y^2=1$ in $P_2(\mathbb{C})$ I get something like $(1,\frac{x_1}{x_0},\frac{x_2}{x_0})$, as already said. Now (I think?) I can see $P_2(\mathbb{C})$ as $P_4(\mathbb{R})$ as $\mathbb{C}$ is (topologically speaking) a $\mathbb{R}^2$. Then my points in $P_2(\mathbb{C})$ would now be something like $(1,a,b,c,d)$, with $x=a+ib$ and $y=c+id$. But $y=\sqrt{1-x^2}$, and in $\mathbb{C}$ this gives me two possible values for $y$, one the opposite of the other. So for every $x$ I'd have solutions that in $P_4(\mathbb{R})$ would look like $(1,a,b,Re(\sqrt{1-x^2}),Im(\sqrt{1-x^2}))$ and $(1,a,b,-Re(\sqrt{1-x^2}),-Im(\sqrt{1-x^2}))$. These two subsets are both homeomorphic to $\mathbb{R}^2$ (in fact I'm not 100% sure of this), and through a stereographic projection I have two Riemann spheres. I should now verify that the union of these two spheres is connected, that is path-connected as in $\mathbb{R}^n$ it's the same thing. But it should be sufficient to show that the two spheres are not disjoint, and for $x=1$ I have $y=0$, and I can conclude that the point $(1,1,0,0,0)$ belongs to both subsets, and thus I can conclude.

As already stated I'm not sure at all of this "proof", in particular I'm quite sure that the functions I would like to use are bijections, but I'm not really sure about them bein homeomorphisms.