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# 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*}

Dec 16, 2018

#2
+5095
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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.

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

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

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

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

.
Dec 18, 2018
edited by Rom  Dec 18, 2018

#1
+701
-1

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

Dec 16, 2018
edited by PartialMathematician  Dec 16, 2018
#2
+5095
+3

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.

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

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

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

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

Rom  Dec 18, 2018
edited by Rom  Dec 18, 2018