What happens if we make the equation of a sphere complex instead of real? The equation for a sphere is 
$$
r^2=x^2+y^2+z^2
$$
where $r,x,y,z\in\mathbb{R}$, the invariance group keeping $r^2$ invariant is $O(3)$.
What happens if we keep the same equation but make the terms complex:
$$
z_0^2=z_1^2+z_2^2+z_3^2
$$
where $z_0,z_1,z_2,z_3\in\mathbb{C}$.
What is the geometric structure? What is the invariance group that keeps $z_0^2$ invariant?

I am also interested in the circle $r^2=x^2+y^2\to z_0^2=z_1^2+z_2^2$
 A: An answer a little longer than Captain Lama's. I'm just going to examine the 2D case:
$$z_0^2 = z_1^2 + z_2^2$$
Except for the degenerate case $z_0 = 0$, you can divide by $z_0$ to get 
$$1 = \left(\frac {z_1}{z_0}\right)^2 + \left(\frac {z_2}{z_0}\right)^2$$
or just $$1 = w^2 + z^2$$
While this looks like the equation of a unit circle, it is actually unbounded. The graph is asymptotic to the planes $z = \pm iw$. 
A linear transformation of $(w, z)$ will looks like $$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}w\\z\end{bmatrix} = \begin{bmatrix}aw + bz\\cw +dz\end{bmatrix}$$
If it is to be invariant, the result must still satisfy the same equation:
$$(aw + bz)^2 + (cw +dz)^2 = 1$$
for all $w, z$ satisfying $w^2 + z^2 = 1$. In particular, it must be true when $w = 1, z = 0$, when $w = 0, z = 1$ and when $w = z = \frac {\sqrt 2}2$. The first gives $$a^2 + b^2 = 1$$ the second gives $$c^2 + d^2 = 1$$ and the third then gives $$ab + cd = 0$$
Conversely, a little algebra shows that those three conditions and $w^2 + z^2 = 1$ are sufficient to prove $(aw + bz)^2 + (cw +dz)^2 = 1$. I.e., those conditions are exactly the requirement for $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ to be invariant.
But of course, they are also exactly the equations defining $O_2(\Bbb C)$.
If you want something a bit more like the unit circle, then the equation you need is $$1 = |w|^2 + |z|^2 = w\bar w + z\bar z$$ That will give $U_2(\Bbb C)$ as the invariant group instead of $O_2(\Bbb C)$. 
