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.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

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

#2**+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

#1**-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

PartialMathematician Dec 16, 2018

#2**+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