# Powers of complex numbers

1. Feb 6, 2013

### ptolema

1. The problem statement, all variables and given/known data

Let z be a complex number such that z^n=(z+1)^n=1. Show that n|6 (n divides 6) and that z^3=1.

2. Relevant equations

n|6 → n=1,2,3,6

3. The attempt at a solution

The z+1, I think, is what throws me off. Considering z^n=1 by itself, for even n, z=±1 and for odd n, z=1. The (z+1) term, however, contradicts this result and leaves me right back where i started. How should I begin looking at this?

2. Feb 6, 2013

### tiny-tim

hi ptolema!

draw the unit circle …

obviously both z and z + 1 must lie on that circle

and the line joining them must be horizontal and of length 1

sooo … ?

3. Feb 6, 2013

### HallsofIvy

Surely you know better than this. z^n= 1 has real roots 1 and -1 but this is a problem about complex numbers! The fact that z^n= 1 means, as tiny-tim said, that z lies on the circle with center 0 and radius 1. And the same for z+ 1. There are two possible values for z.