Find all possible integer values of n such that the following system of equations has a solution for z: \(\begin{align*} z^n &= 1, \\ \left(z + \frac{1}{z}\right)^n &= 1. \end{align*}\)

I just have no idea what to do.

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

Guest Aug 21, 2020

#1

#2**0 **

That is wrong.

I will make this equation a little easier to read now.

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

Look at what it means if z^n=1. n could be 0, hence z=All integers. z can be 1, hence n could be all integers. The next equation means that if n is 0, z IS all integers. If n is all integers, z=1, but 1+1/1=2!!! 2^integer=not 1, therefore n has to be ZERO (0)

An alternative is that if (z+1/z)^n=1, we do 1 to the nth root and get z+1/z=1, and z+1=z. This is impossible, therefore n has to be 0!

,

Pangolin14
Aug 21, 2020