Is this set in $\Bbb{C}^3$ compact? 
Is $\{(x,y,z) \in \Bbb{C^3} | z_1^2+z_2^2+z_3^2=1\}$ in the Euclidean topology compact?

We know in $\Bbb{R}^n$ compact iff closed and bounded , but is it true for complex also?
 A: Consider the map $i:(\mathbb{R}^2)^n \to \mathbb{C}^n$ given by
$i((x_1,y_1),...,(x_n,y_n)) = (x_1+i y_1,...,x_n+iy_n)$, with
the norm (squared)  $\|(x_1,y_1),...,(x_n,y_n))\|^2 = x_1^2+y_1^2+...+x_n^2+y_n^2$ on $(\mathbb{R}^2)^n$
and
the norm  $\|(z_1,...,z_n)\|^2 = |z_1|^2+...+|z_n|^2$ on $\mathbb{C}^n$.
We see that $i$ is an isometry, hence $C \subset (\mathbb{R}^2)^n$ is compact iff $i(C) \subset \mathbb{C}^n$ is compact.
In particular, this shows that a subset of $\mathbb{C}^n$ is
compact iff it is closed and bounded.
The above example is closed since $(z_1,z_2,z_3) \mapsto z_1^2+z_2^2+z_3^3$ is continuous, but since $(t,it,1)$ is in the set for all $t$ we see that it is not bounded, hence not compact.
As an aside, we see that the unit ball in $\mathbb{C}^n$ is compact,
since $\{(z_1,z_2,z_3)| \|(z_1,z_2,z_3)\| = 1 \}$ is closed and bounded.
A: It depends on how exactly you define the set.
If you are interested in

$\{(x,y,z)\in\mathbb{C} | x^2 + y^2 + z^2 = 1 \}$

then no, this set is not bounded, hence non-compact.  On the other hand I don't think people would typically refer to that as a "unit sphere" (let alone a "unit circle").  More likely they would ask about

$\{(x,y,z)\in\mathbb{C} | |x|^2 + |y|^2 + |z|^2 = 1 \}$

where $|x|$ denotes the norm, defined by $|x|^2 = \overline{x}x$.  That set is closed and bounded, hence compact.
A: Note that the set defined in $\Bbb C^n$ by the "unit sphere equation" $z_1^2+\cdots+z_n^2=1$ is a very different set from the one defined by the corresponding equation in $\Bbb R^{2n}$ (with $2n$ variables).
Since $\Bbb C$ is algebraically closed, the "unit sphere" in $\Bbb C^3$ is not bounded. For instance, for any $z_1,z_2\in \Bbb C$, the equation $z_1^2+z_2^2+z=1$ has at least one solution.
