Solve z^3+conjugate(z)) = 0.
Note that z = 0 is a solution, so assume that z not 0.
Let z = R*exp(is) with R>0. Then z^3+conjugate(z) = 0 becomes R^3*exp(i3s) + R*exp(-is) = 0, so R^2*exp(i4s) + 1 = 0.
Thus R^2*exp(i4s) = -1. Taking norms shows R = 1, leaving exp(i4s) = -1 = exp( i (2k+1) pi ) for k any integer.
Thus s = (2k+1) pi / 4 for k some integer.
So z = R*exp(is) = exp(i (2k+1) pi / 4) for some integer k.
Thus solution set is at most { 0, exp(i pi / 4), exp(i 3 pi / 4), exp(i 5 pi / 4), exp(i 7 pi / 4) } = { 0, ( 2^(-1/2) )( +/- 1 +/- i ) }
Check them:
Check exp(i pi / 4): exp(i pi / 4))^3 + exp( (-1)(i pi / 4) ) = exp(i 3 pi / 4) + exp( (-1)(i pi / 4) ) = 0 (They're antipodal on the unit circle, occurring at angles 3pi/4 and -pi/4).
Check exp(i 3 pi / 4): exp(i 3 pi / 4))^3 + exp( (-1)(i 3 pi / 4) ) = exp(i 9 pi / 4) + exp( i (- 3 pi / 4) ) = exp(i pi / 4) + exp( i 5 pi / 4 ) = 0 (Antipodal).
Check exp(i 5 pi / 4): exp(i 5 pi / 4))^3 + exp( (-1)(i 5 pi / 4) ) = exp(i 15 pi / 4) + exp( i (- 5 pi / 4) ) = exp(- i pi / 4) + exp( i 3 pi / 4 ) = 0 (Antipodal).
Check exp(i 5 pi / 4): exp(i 7 pi / 4))^3 + exp( (-1)(i 7 pi / 4) ) = exp(i 21 pi / 4) + exp( i (- 7 pi / 4) ) = exp( i 5 pi / 4) + exp( i pi / 4 ) = 0 (Antipodal).
All work, so the solution set to z^3+conjugate(z)) = 0 is { 0, exp(i pi / 4), exp(i 3 pi / 4), exp(i 5 pi / 4), exp(i 7 pi / 4) }.
Using your method of substituting , we have
Solving the first equation:
Substituting into the second equation: If then
the second of which is impossible since .
And if then
and since we knew that gives , which is impossible since , and if .
So the solution is and every combination of .