How to solve $x^2+y^2+z^2=0$ in $\mathbb C^3$? Firstly, $(0,0,0)$ is trival.
Then, we can easily check that those in the form of $x^2+y^2=-z^2$, $x=iz\sin\theta,y=iz\cos\theta$ is a kind of solution.
However, are there any other "kinds" of solutions? That is the true question.
 A: In some sense, the set of solutions $S = \{(x,y,z)\in\mathbb C^3\colon \ x^2+y^2+z^2=0\}$ is a "complex elliptic cone" by analogy with the real case of elliptic cones, and for some obscure reasons from algebraic geometry, this should have a parameterization (or at least a family of parameterizations...) of the form $U \subset\mathbb C^2 \to S$ (for instance, the real case has these parameterizations).
If you take both $z$ and $\theta$ to be arbitrary complex numbers, all the points of $S$ can actually be expressed in that way. First of all, the map you have defined $\varphi(z,\theta) = (iz\sin\theta, iz\cos \theta, z)$ lies in $S$, because $(iz\sin\theta)^2+ (iz\cos \theta)^2+ z^2 = 0$ since $\sin^2 z + \cos^2 z = 1$ for every complex number $z$. So far so good! Now we have to prove that for every point $P_0 = (x_0,y_0,z_0)\in S$, there is a $(z, \theta)\in \mathbb C^2$ s.t. its image is $P_0$, the harder part.

Let's suppose that, for an arbitrary point, there is such a $(z,\theta)$, then $z = z_0$. I'll drop the subscripts from the coordinates of $P_0$ for simplicity. There will appear a lot of letters, but they are just to keep track of the "free variables" (things that depend on $x$ and $y$) to finally construct the hardest part ($\theta$) from the original data.
We have the conditions
$$\begin{cases}x &= iz \cos\theta\\y &= iz \sin\theta\end{cases} \iff\begin{cases}\cos\theta &= -i\frac{x}z =: p\\\sin\theta &= -i\frac{y}{z} =:q\end{cases}.$$
These two equations imply restrictions on $\cos\theta = \frac{e^{i\theta}+e^{-i\theta}}{2}$ and $\sin\theta = \frac{e^{i\theta}-e^{-i\theta}}{2i}$; by simple manipulation, we have that $$\begin{cases}e^{i\theta} &= p+iq\\e^{-i\theta} &= p-iq\end{cases}.$$ But if $\theta = a+bi$, analyzing only the second equation, we have that $e^{-i\theta} = e^b e^{i(-a)}$ yields $b = \ln |p-iq|$ and $a = -\arg (p-iq)+2k\pi$, where $k \in \mathbb Z$. We found a candidate to $\theta$, i.e.  $\theta = \ln|p-iq|-\left(\arg(p-iq)+2k\pi\right)i$. Obviously this gives the desired result when plugged in $e^{-i\theta}$; we have to check that the calculations work out for $e^{i\theta}$ - and they, in fact, do. Since $p^2 + q^2 = 1$ (by the choice of $(x,y,z) \in S$), then$ (p+iq)(p-iq) = 1$ and thus $\ln|p+iq| = -\ln|p-iq|$ and, for the same reason, $\arg(p+iq) = -\arg(p-iq) + 2\ell \pi$, so, in the end, such a choice for $\theta_k$ in fact yields the desired choices for $e^{i\theta}, e^{-i\theta}$, then $\cos \theta, \sin \theta$ and finally we backtrack to the desired choices of $x$ and $y$.
Explicitly, given $(x,y,z) \in S$, then $\varphi(z, \ln|\frac{y-ix}{z}| - \arg(\frac{y-ix}{z})i+2k\pi i) = (x,y,z)$, so $\varphi$ is surjective and every solution of $x^2+y^2+z^2 = 0$ can be expressed by your parameterization. Note that $\varphi$ is definitely not injective, since there's $|\mathbb Z| = \infty$ many choices of $\theta$ for every $P_0 \in S$. However, leaving behind a few weird points, this is locally an (analytic? meromorphic?) homeomorphism. Yay!
A: I would suggest you the following algebraic construction.

Let
$$
\begin{align}x&=a+bi\\ y&=c+di\\
z&=m+ni\end{align}
$$
We want to find all $a,b,c,d,m,n\in\mathbb R$, such that $x^2+y^2+z^2=0$.
We have,
$$
\begin{align}&(x-iy)(x+iy)=-z^2\\
&\begin{cases}x-iy=-\alpha,\,\alpha\in\mathbb C\setminus \{0\} \\ x+iy=\frac {z^2}{\alpha}\end{cases} \\
\implies &x=\frac {z^2-\alpha^2}{2\alpha}\\ \implies &y=\frac {z^2+\alpha^2}{2i\alpha}\end{align}
$$
Then, putting $z=m+ni$ and setting $\alpha=p+qi$, we conclude that:
$$
\begin{align}x&=\frac {(m+ni)^2-(p+qi)^2}{2(p+qi)}\\
y&=\frac {(m+ni)^2+(p+qi)^2}{2i(p+qi)}\\
z&=mi+n\end{align}
$$
where $m,n,p,q\in\mathbb R$.
Finally, multiply the numerator and denominator of $x$ by the complex conjugate $p-qi$, and the numerator and denominator of $y$ by the complex conjugate $i(p-qi)$ and expand the expressions you get.  Now, you have obtained the numbers $a,b,c,d\in\mathbb R$, in terms of $m,n,p,q\in\mathbb R$ and you have the solution you want to achieve.
