Playing around with complex numbers, I found a cute way of solving the quadratic equation. Let's start with the (monic) equation \begin{equation} z^2+pz+q = 0 \end{equation} where $z$ and the coefficients are allowed to be complex numbers. We know that the two solutions $z_1$ and $z_2$ are numbers in the complex plane; we have also the relations $p=-(z_1+z_2)$ and $q=z_1z_2$.
I wanted to map $z_1$ to the number $1$ and $z_2$ to the number $-1$, and the plan is this: traslate the segment $z_1z_2$ towards the origin so that its midpoint $G$ coincides with $0$, then apply a rotation and a scaling.
Since applying a rotation and a scaling amounts to multipliying by a complex number $\zeta$, the following relations hold: \begin{equation} \zeta(z_1-G)=1\qquad\text{and}\qquad\zeta(z_2-G)=-1 \end{equation} The midpoint $G$ is known from the original equation, because $G=\frac{1}{2}(z_1+z_2)=-\frac{1}{2}p$, so if we can find $\zeta$ from the equation parameters we can rebuild the roots $z_1$ and $z_2$: \begin{equation} z_1 = \frac{1}{\zeta}+G\qquad\text{and}\qquad z_2 = -\frac{1}{\zeta}+G \end{equation} So, let's find it. We have that \begin{equation} \zeta(z_1-G)\cdot\zeta(z_2-G) = 1\cdot(-1) = -1 \end{equation} so that \begin{align} \zeta(z_1-G)\cdot\zeta(z_2-G) &= \zeta^2(z_1z_2-G(z_1+z_2)+G^2)=\\ &=\zeta^2(q+pG+G^2) = \\ &=\zeta^2\left(q+p\left(-\frac{1}{2}p\right)+\left(-\frac{1}{2}p\right)^2\right)=\\ &=\zeta^2\left(q-\frac{1}{4}p^2\right)=\\ &=-1 \end{align} which yields \begin{equation} \zeta = \pm\frac{1}{\sqrt{\frac{1}{4}p^2-q}} \end{equation} (actually not considering the $\pm$ does not change the final result) From here we can rebuild the roots: \begin{equation} z_1 = \frac{1}{\zeta}+G = \sqrt{\frac{1}{4}p^2-q}-\frac{1}{2}p\qquad\text{and}\qquad z_2 = -\frac{1}{\zeta}+G = -\sqrt{\frac{1}{4}p^2-q}-\frac{1}{2}p \end{equation} Now, I find this method cute since it involves mapping the unknown roots of the original equation to the roots of the "basic" equation $z^2=1$ and then does a sort of tracing back onto the transformation to rebuild the original roots.
I'm now trying to extend these ideas to the general cubic equation, trying to map the unknown roots to the cubic roots of unity but there are some difficulties.
My specific question is the following.
We know that a rotation+scaling is equivalent to a multiplication by a complex number. I now need a transformation that sends three complex numbers, whose barycenter $G$ is the origin, i.e. $z_1+z_2+z_3=0$, to the cubic roots of unity. I guess it is not a linear transformation, because if the three numbers are collinear it's not possible to send them to non-collinear points via a linear map, so what kind of transformation is it, and is it possible to express it via complex number operations only, bypassing their coordinate representation?
EDIT: I believe the key to find this function is the fact that any triangle in $\mathbb{R}^3$ can be projected onto an equilateral triangle.