I need to solve the equation

z1^n + z2^n == 1

with the parametrisation

z1 = Exp[2 π k1] I Cos[θ - ξ]^(2/n)
z2 = Exp[2 π k1] I Sin[θ - ξ]^(2/n)

Under the following assumptions:

k1, k2 ∈ N
k1 >= 0
k2 <= n − 1
0 <= θ <= π/2
Abs[ξ] <= max ξ 
n = 2

The end result would look something like this and the coordinates {x, y, z} would end up as {Re[z1], Im[z1], Re[z2]}

enter image description here

As I have understood this has something to do with the Riemann sphere.

  • 1
    $\begingroup$ Ahem. You now that you are on a Q&A site that focuses excusively on the programming language Mathematica? $\endgroup$ May 22, 2018 at 21:10
  • $\begingroup$ I missed that "detail", sorry. Anyhow the answer helped me figure it out in Python. $\endgroup$
    – Silvio
    May 22, 2018 at 22:01
  • $\begingroup$ I'm glad to hear that. =D $\endgroup$ May 22, 2018 at 22:04
  • $\begingroup$ @Silvio Could you go post a reference to the problem at hand, because some information is missing (e.g. k2 is nowhere to be found in both z1 and z2)? Or go go ahead and add every relevant detail to the question itself. $\endgroup$
    – Sektor
    May 23, 2018 at 8:16

1 Answer 1


By reverse engineering: With $z_1 = x + \operatorname{i} y$ and $z_1 = z + \operatorname{i} w$, $x$, $y$, $z$, $w \in \mathbb{R}$, the equation $\Im(z_1^2 + z_2^2) = 0$ leads to an one real equation for w, at least in the case $n=2$. Substituting the solution w of this equation into $\Re(z_1^2 + z_2^2) = 1$ leads to a real equation in $\mathbb{R}^3$. Its solution set can be plotted like this.

z1 = x + I y;
z2 = z + I w;
n = 2;
eq = ComplexExpand[ReIm[z1^n + z2^n] - {1, 0}];
wsol = Solve[eq[[2]] == 0, w, Reals];
eq3D = Expand[z^2 eq[[1]] /. wsol[[1]]];
R = 5;
ContourPlot3D[eq3D == 0, {x, -R, R}, {y, -R, R}, {z, -R, R},
 RegionFunction -> Function[{x, y, z}, x^2 + y^2 + z^2 <= 25],
 PlotPoints -> 60,
 Mesh -> 25

enter image description here


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