+0

# Complex Numbers (repost)

+2
113
2
+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!

Sep 7, 2019

#1
+6046
+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
+507
+1

Thanks!

Davis  Sep 15, 2019