(z+i)^5 + (z-i)^5 = 0
The Attempt at a Solution
I have found the five possible values for z by expanding the original expression into powers of z and i. The imaginary terms cancel exactly, leaving a cubic equation in z which is easily solvable as one of its roots is z=0.
The answers quote the solutions as z = cot[pi(1 + 2n)/10], n = 0, 1, 2, 3, 4. The values I obtained are z=0, z^2=5+sqrt(20), z^2=5-sqrt(20).
My answers are roots of the equation, but the way that the solutions are written suggests that there is a (probably neater) method than the one I used which uses trigonometry. Can anyone see it?