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

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

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

Guest Dec 16, 2018

Rom · Answer 1 · 2018-12-18T06:46:33

Let me echo Melody's sentiments.

If you don't know how to solve a problem don't guess at it. It just confuses and frustrates the posters.

$\text{This is a subtle problem with an obvious solution of }n=1 \\ \text{and not so obvious solutions of } n=6k,~k \in \mathbb{N}$

$\text{A simple bit of algebra shows that }\\ \left(z + \dfrac 1 z\right)^n = 1 \Rightarrow \Large z = e^{\pm i \frac \pi 3}$

$\text{On the other hand the roots of }z^n = 1 \text{ are }\\ \Large z = e^{i \frac{2\pi}{n}k},~\normalsize k \in \mathbb{Z}$

$\text{So we are looking for }n \text{ such that }\\ 2\pi \dfrac k n = \dfrac \pi 3\\ \dfrac{6k}{n}=1\\ n=6k$

PartialMathematician · Answer 2 · 2018-12-16T04:43:00

Let's try z = anything but 0, so $z \ne 0$ , and n = 0.

$(z+\dfrac{1}{z})^n$ will equal 1 if $(z+\dfrac{1}{z})$ is a positive number (not 0 or less). Any positive number to the power of 0 is 1.

$(z+\dfrac{1}{z})^n$ will be negative if $z < 0$ .

$(z+\dfrac{1}{z})^n$ will equal 0 when $z = \pm\sqrt{-1}$ , but since this is not real, we do not need to consider it.

Only works when $\boxed{n=0}$ .

Hope this helps,

- PM