Find all possible integer values of n such that the following system of equations has a solution for z:
\(\begin{align*} z^n &= 1, \\ \left(z + \frac{1}{z}\right)^n &= 1. \end{align*}\)
+6248 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.
\(\text{This is a subtle problem with an obvious solution of }n=1 \\ \text{and not so obvious solutions of } n=6k,~k \in \mathbb{N}\)
\(\text{A simple bit of algebra shows that }\\ \left(z + \dfrac 1 z\right)^n = 1 \Rightarrow \Large z = e^{\pm i \frac \pi 3}\)
\(\text{On the other hand the roots of }z^n = 1 \text{ are }\\ \Large z = e^{i \frac{2\pi}{n}k},~\normalsize k \in \mathbb{Z}\)
\(\text{So we are looking for }n \text{ such that }\\ 2\pi \dfrac k n = \dfrac \pi 3\\ \dfrac{6k}{n}=1\\ n=6k\)
+773 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
+6248 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.
\(\text{This is a subtle problem with an obvious solution of }n=1 \\ \text{and not so obvious solutions of } n=6k,~k \in \mathbb{N}\)
\(\text{A simple bit of algebra shows that }\\ \left(z + \dfrac 1 z\right)^n = 1 \Rightarrow \Large z = e^{\pm i \frac \pi 3}\)
\(\text{On the other hand the roots of }z^n = 1 \text{ are }\\ \Large z = e^{i \frac{2\pi}{n}k},~\normalsize k \in \mathbb{Z}\)
\(\text{So we are looking for }n \text{ such that }\\ 2\pi \dfrac k n = \dfrac \pi 3\\ \dfrac{6k}{n}=1\\ n=6k\)
