TITLE: How to imagine $X^2+Y^2-1=0$ in $C^2$, $X$, $Y$ both complex
QUESTION [5 upvotes]: Like the title, with a graph if convenient, I was reading Mumford's book, a picture make me confused: 
Could someone explain it to me, thanks.

REPLY [2 votes]: This surface is diffeomorphic to the cotangent bundle of the circle $T^*S^1$ as will be shown in the following
explicit construction: 
The cotangent bundle of the circle can be parameterized as a surface in $\mathbb{R}^4$ as follows:
$ u_1^2+u_2^2 = 1 $
$ u_1v_1+u_2v_2 = 0$
Of course, the first equation is the equation of the circle, while the second one expresses the orthogonality
of the radius and the tangent vectors.
The first equation can be parameterized by: $u_1 = cos(\theta)$, $ u_2 = sin(\theta)$, we may define 
$\rho = v_2 u_1 - u_2 v_1$, then the transformation:
$X = cos(\theta + i \rho)$
$Y = sin(\theta + i \rho)$
lead to the required result: $X^2 + Y^2$ = 1.
In order to see that this transformation is invertible, (thus constitutes a diffeomorphism), note that:
$XY = \frac{sinh(2\rho)}{2}+i cosh(2\rho)u_1 u_2$
The definition of $\rho$ is chosen such that the coordinates $\rho$ and $\theta$ considered as functions on $\mathbb{R}^4 \equiv T^*\mathbb{R}^2$
are canonically conjugate.
Here, the equator corresponds to the zero section $\rho = 0$. The points at infinity correspond to:
$\rho = \pm\infty $