We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+2
70
2
avatar+507 

My first post of this question is now not on the first page, so I thought I would repost it.

 

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

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

Edit: \(z\) can be a complex number, too!

 

Link to first question: https://web2.0calc.com/questions/help-plles

 Sep 7, 2019
 #1
avatar+6017 
+3

\(z = r e^{i \theta}\\ z^n = r^n e^{i n \theta}\\ z^n = 1 \Rightarrow r = 1\\ z = e^{i \theta}\)

 

\(z + \dfrac 1 z = e^{i\theta}+ e^{-i\theta} = 2\cos(\theta)\\~\\ \left(z + \dfrac 1 z \right)^n = 2^n \cos^n(\theta) = 1\\ \cos^n(\theta) = \dfrac{1}{2^n}\\ \cos(\theta) = \dfrac 1 2\\ \theta = \pm \dfrac{\pi}{3}\)

 

\(z = e^{\pm i \pi/3}\)

.
 Sep 8, 2019
 #2
avatar+507 
+1

Thanks!

Davis  Sep 15, 2019

5 Online Users