I’m looking for an explicit formula for the vertices of a regular tetrahedron that covers four given points. In particular:

Given four distinct real numbers $a_1$, $a_2$, $a_3$, $a_4$, is there a simple formula for four complex numbers $z_1$, $z_2$, $z_3$, $z_4$ such that the four points $(a_i,z_i)$, $i=1,\ldots,4$, in ${\Bbb R}\times{\Bbb C}$ form the vertices of a regular tetrahedron? That is, $|a_i-a_j|^2+|z_i-z_j|^2$ is independent of choice of distinct $i$, $j$.

Same question but with $a_i$ complex.

As an example of the type of thing I’m looking for, for three real numbers numbers $a$, $b$, $c$, $(a,(b-c)/\sqrt 3), (b,(c-a)/\sqrt 3), (c,(a-b)/\sqrt 3)$ are the vertices of an equilateral triangle.

I've tried things like $(a,(b+c\omega+d\overline\omega)/2)$ (where $\omega$ is a third root of unity) which looks to be on the right track but... I just hope someone knows the answer.

Thanks in advance.