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$\begingroup$

I mean $$ \text{z1}+\text{z2}+\text{z3}+\text{z4}=0\land | \text{z1}| =| \text{z2}| \land | \text{z3}| =| \text{z2}| \land | \text{z3}| =| \text{z4}|$$ in $\text{z1},\text{z2},\text{z3},\text{z4}$ over the complexes.

Here are my unsuccessful trials in 13.2 on Windows 10.

(i)

Solve[z1 + z2 + z3 + z4 == 0 && Abs[z1] == Abs[z2] && Abs[z3] == Abs[z2] && Abs[z3] == Abs[z4], 
{z1, z2, z3,z4}, Complexes]

Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information.

Solve::svars: Equations may not give solutions for all "solve" variables.

{{z2->z1,z3->-z1,z4->-z1},{z2->-z1,z3->z1,z4->-z1},{z2->-z1,z3->-z1,z4->z1},{z1->0,z2->0,z3->0,z4->0}}

Most of the solutions are lost.

(ii) The command

Reduce[z1 + z2 + z3 + z4 == 0 && Abs[z1] == Abs[z2] && Abs[z3] == Abs[z2] && 
Abs[z3] == Abs[z4], {z1, z2, z3, z4}, Complexes]

almost crashes my comp: "Kernel connection is lost".

(iii)

With the additional constraint Abs[z1]==1,

Solve[z1 + z2 + z3 + z4 == 0 && Abs[z1] == Abs[z2] && Abs[z3] == Abs[z2] && 
Abs[z3] == Abs[z4] && Abs[z1] == 1, {z1, z2, z3, z4}, Complexes]

Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information.

{{z1 -> 1, z2 -> 1, z3 -> -1, z4 -> -1}, {z1 -> 1, z2 -> -1, z3 -> 1, z4 -> -1}, {z1 -> -1, z2 -> 1, z3 -> 1, z4 -> -1}, {z1 -> 1, z2 -> -1, z3 -> -1, z4 -> 1}, {z1 -> -1, z2 -> 1, z3 -> -1, z4 -> 1}, {z1 -> -1, z2 -> -1, z3 -> 1, z4 -> 1}}

Most of the solutions are lost.

(iv)

Reduce[z1 + z2 + z3 + z4 == 0 && Abs[z1] == Abs[z2] &&  Abs[z3] == Abs[z2] &&
Abs[z3] == Abs[z4] && Abs[z1] == 1, {z1, z2, z3, z4}, Complexes]

is running without any response on my comp for hours. The resources of my comp are not exhausted. Likely an infinite loop is formed.

(v) The switching to the reals by

Reduce[x1 + x2 + x3 + x4 == 0 && y1 + y2 + y3 + y4 == 0 && 
x1^2 + y1^2 == x2^2 + y2^2 && x3^2 + y3^2 == x2^2 + y2^2 && 
x3^2 + y3^2 == x4^2 + y4^2 && x1^2 + y1^2 == 1, {x1, x2, x3, x4, y1,y2, y3, y4},Reals]

does not help. The command is running without any response on my comp for hours. The resources of my comp are not exhausted. Likely an infinite loop is created.

$\endgroup$
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  • 3
    $\begingroup$ If the absolute values of all the complexes are the same, it should help to reformulate the problem as z1 = r Exp[I phi1] etc. and impose that r >= 0. $\endgroup$ Apr 5, 2023 at 15:28
  • 1
    $\begingroup$ As I know it, the geometric interpretation of the solutions is the following: z1, z2, z3, z4 are vertices of a rectangle (maybe, a degenerate one) in the complex plane centered at the origin $\endgroup$
    – user64494
    Apr 5, 2023 at 17:21
  • 2
    $\begingroup$ @SjoerdSmit, you can go even further by taking r==1 $\endgroup$
    – mikado
    Apr 5, 2023 at 19:35
  • 2
    $\begingroup$ This is a 2-parameter family if we restrict length to be 1. $\endgroup$ Apr 5, 2023 at 22:58
  • 1
    $\begingroup$ The geometric interpretation is more general: z1,z2,z3,z4 form a rhombus! $\endgroup$ Apr 7, 2023 at 15:42

1 Answer 1

2
$\begingroup$

By taking z1==1, you can find a solution set that you can scale by any complex constant.

Decomposing into real and imaginary components I have

eqn = 1 + c1 + c2 + c3 == 0 && s1 + s2 + s3 == 0 && c1^2 + s1^2 == c2^2 + s2^2 == c3^2 + s3^2 == 1;

This can be solved easily using

Reduce[eqn, Reals]

Alternatively, you find a few interesting values, e.g.

FindInstance[1 + c1 + c2 + c3 == 0 && s1 + s2 + s3 == 0 && 
c1^2 + s1^2 == c2^2 + s2^2 == c3^2 + s3^2 == 1, {c1, c2, c3, s1, s2, s3}, Reals, 8]
$\endgroup$
4
  • $\begingroup$ Thank you. I find this approach advanced. Sum spots on the Sun. (i) "By taking z1==1, you can find a solution set that you can scale by any complex constant" should be elaborated. $\endgroup$
    – user64494
    Apr 6, 2023 at 3:04
  • $\begingroup$ (ii) Your result is a=((c3 == 0 && ((c2 == 0 && ((s2 == 1 && 1 + s3 == 0) || (1 + s2 == 0 && s3 == 1))) || (1 + c2 == 0 && s2 == 0 && (s3 == 1 || 1 + s3 == 0)))) || (s3 == 0 && ((1 + c3 == 0 && ((c2 == 0 && (1 + s2 == 0 || s2 == 1)) || (s2 == 0 && (c2 == 1 || 1 + c2 == 0)) || ((-1 < s2 < 0 || 0 < s2 < 1) && (Sqrt[1 - s2^2] == c2 || c2 + Sqrt[1 - s2^2] == 0)))) || $\endgroup$
    – user64494
    Apr 6, 2023 at 3:06
  • $\begingroup$ (1 + c2 == 0 && c3 == 1 && s2 == 0))) || ((-1 < s3 < 0 || 0 < s3 < 1) && ((s2 == 0 && c2 + Sqrt[c3^2 + s3^2] == 0 && (Sqrt[1 - s3^2] == c3 || c3 + Sqrt[1 - s3^2] == 0)) || (s2 + s3 == 0 && ((c3 + Sqrt[1 - s3^2] == 0 && Sqrt[c3^2 - s2^2 + s3^2] == c2) || (Sqrt[1 - s3^2] == c3 && c2 + Sqrt[c3^2 - s2^2 + s3^2] == 0)))))) && s1 + s2 + s3 == 0 && 1 + c1 + c2 + c3 == 0 . This description leaves much to be desired. For example a /. {c3 -> 1/2, s1 -> 1/4, s3 -> -1/4} results in False. $\endgroup$
    – user64494
    Apr 6, 2023 at 3:11
  • $\begingroup$ (iii) I repeat "As I know it, the geometric interpretation of the solutions is the following: z1, z2, z3, z4 are the vertices of a rectangle (maybe, a degenerate one) in the complex plane centered at the origin". $\endgroup$
    – user64494
    Apr 6, 2023 at 3:18

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