I am given this problem: show that all solutions of (z+1)100=(z-1)100 satisfy Re(z)=0. This is what I've done to try to solve the problem:
Let z1 be the argument of z+1, and x2 be the argument of z-1
Taking the 100-th root:
And if the real part is equal to the imaginary part, it seems as if both would be equal to zero, which isn't the case.
Any advice on what I did wrong or an alternate pathway to the solution?