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Find all possible integer values of n such that the following system of equations has a solution for z:

\(\begin{align*} z^n &= 1, \\ \left(z + \frac{1}{z}\right)^n &= 1. \end{align*}\)
 

 Jan 12, 2019
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n must be even, or else the two bases in the exponent would be equal, which is impossible.

 

if n is even, either z or z+1/z is negative. if z is negative, then -z=z+1/z or 2z=-1/z, and 2z^2=-1 which is imposssible.

 

then z+1/z is negative. then -z-1/z=z, so 2z=-1/z and 2z^2=-1 which is also impossible.

 

 

so from what i get, there are no possible solutions for n such that there is a solution for z.

 

HOPE THIS HELPED!!

 Jan 12, 2019

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