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  • $\begingroup$ 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. $\endgroup$ Commented 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)))) || $\endgroup$ Commented 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. $\endgroup$ Commented 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". $\endgroup$ Commented Apr 6, 2023 at 3:18