Why Must n Divide 6 When z^n and (z+1)^n Equal 1?

ptolema
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Homework Statement



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.

Homework Equations



n|6 → n=1,2,3,6

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?
 
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hi ptolema! :wink:

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 … ? :smile:
 
ptolema said:

Homework Statement



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.

Homework Equations



n|6 → n=1,2,3,6

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?
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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