Why is $\;x^2+ y^2 = 1,\;$ where $x$ and $ y$ are complex numbers, a sphere? I've heard $x^2 + y^2 = 1,\;$ where $x, y$ are complex numbers, is supposed to be a sphere with two points removed, or also a cylinder.
The problem is I've been trying to wrap my head around this for a while now, and I just don't see it. If it is a sphere, is it like a sphere in $\mathbb{R}^4,\;$ i.e. a $3-$sphere?
I think I see how the stereographic projection works for the extended complex numbers, but if anyone could help me see this complex curve as a sphere, that'd be great.
 A: If $x =a (1+i) + b(1-i)$ and $y = c (1+i) + d(1-i)$ with $a,b,c,d$ real, then the real and the imaginary parts of the equation $x^2 + y^2 = 1$ is $ab + cd = 1/4$ and $a^2 + c^2 = b^2 + d^2$.
Now we can't have $a^2 + c^2 = 0$, as then the left-hand side of the first equation is $0$.  So with $r = \sqrt{a^2 + c^2}$ we have $a = r \cos(\theta)$, $c = r \sin(\theta)$, $b = r \cos(\phi)$, $d = r \sin(\phi)$, for some angles $\theta$, $\phi$, and the first equation
says $r^2 \cos(\theta - \phi) = 1/4$.  Thus we must have $\cos(\theta - \phi) > 0$.  We may
take as parameters $u = \theta - \phi \in (-\pi/2, \pi/2)$ and $v = \phi \in [0, 2 \pi)$ with $v = 0$ and $v = 2 \pi$ identified, and thus
$$ \eqalign{a &=\frac{\cos(u+v)}{2 \sqrt{\cos(u)}}\cr
            b &= \frac{\cos(v)}{2 \sqrt{\cos(u)}}\cr
            c &= \frac{\sin(u+v)}{2 \sqrt{\cos(u)}}\cr
            d &= \frac{\sin(v)}{2 \sqrt{\cos(u)}}\cr}$$
This gives us a homeomorphism from the cylinder $(-\pi/2, \pi/2) \times ({\mathbb R} \mod 2\pi)$
to the solution set of $x^2 + y^2 = 1$ in ${\mathbb C}^2$.
The cylinder is also homeomorphic to the $2$-sphere $S^2$ with two points removed.
A: It is a standard exercise that the solution to the equation $z_{1}^{2} + \ldots + z_{n}^{2} = 1$ is diffeomorphic to the $TS^{n-1}$, the tangent bundle of the $n$-sphere. 
In your case $n = 2$ and we are looking at $TS^{1}$, the tangent bundle of the circle. The circle is parallelizable, being a Lie group, therefore this bundle is trivial and we have $TS^{1} \simeq T \times \mathbb{R}$, a cylinder like you suggested.
