Why is $\;x^2+ y^2 = 1,\;$ where $x$ and $ y$ are complex numbers, a sphere?

If $x =a (1+i) + b(1-i)$ and $y = c (1+i) + d(1-i)$ with $a,b,c,d$ real, the real and imaginary parts of the equation $x^2 + y^2 = 1$ say $ab + cd = 1$ and $a^2 + c^2 = b^2 + d^2$. Now we can't have $a^2 + c^2 = 0$, as then the left side of the first equation is $0$. So with $r = \sqrt{a^2 + c^2}$ we have $a = r \cos(\theta)$, $b = r \sin(\theta)$, $c = r \cos(\phi)$, $d = r \sin(\phi)$, for some angles $\theta$, $\phi$, and the first equation says $r^2 \cos(\theta - \phi) = 1$. 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 $$ \eqalign{a &=\frac{\cos(u+v)}{\sqrt{\cos(u)}}\cr b &= \frac{\cos(v)}{\sqrt{\cos(u)}}\cr c &= \frac{\sin(u+v)}{\sqrt{\cos(u)}}\cr d &= \frac{\sin(v)}{\sqrt{\cos(u)}}\cr}$$

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.

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.