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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

Best Answer 

 #2
avatar+4396 
+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.

 

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

 

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

 

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

 

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

.
 Dec 18, 2018
edited by Rom  Dec 18, 2018
 #1
avatar+679 
-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
avatar+4396 
+3
Best Answer

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

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