Find z^n+ 1/z^n: Why Consider Only One Argument?

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SUMMARY

The discussion centers on the equation \( z^n + \frac{1}{z^n} \) and the reasoning for considering only one argument, specifically \( \frac{\pi}{6} \), instead of the alternative \( \frac{11\pi}{6} \). The quadratic equation derived is \( z^2 - \sqrt{3}z + 1 = 0 \), yielding solutions \( z = \frac{\sqrt{3} + i}{2} \) and \( z = \frac{\sqrt{3} - i}{2} \). Both solutions have a modulus of 1 and are inverses of each other, leading to the conclusion that they yield the same function value, thus justifying the focus on a single argument.

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Magnetons
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Homework Statement
compute ##z^n+ \frac 1 z^n##
if ##z+\frac 1 z= \sqrt3##
Relevant Equations
No Equation
Firstly I converted the given equation to a quadratic equation which is
##z^2- (\sqrt3)z+1=0##
I got two solutions:
1st sol ##z=\frac {(\sqrt3 + i)} {2}##
2nd sol ## z=\frac {(\sqrt3 -1)} {2}##
Then I found modulus and argument for both solution . Modulus=1 Arguments are ##\frac {\Pi} {6}## and ##\frac {11* \Pi } {6}##

My question is {Why we consider only one argument ##\frac {\Pi} {6}## and not other}
 
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Magnetons said:
My question is \mathbf {Why we consider only one argument \frac {\Pi} {6} and not other}
First, let ##z_0## be a solution to ##z + \frac 1 z = w##. Then, ##\frac 1 {z_0}## is also a solution. The two solutions are, therefore, essentially the same. You can check, if you want, that your two solutions are inverses of each other.
 
PeroK said:
First, let ##z_0## be a solution to ##z + \frac 1 z = w##. Then, ##\frac 1 {z_0}## is also a solution. The two solutions are, therefore, essentially the same. You can check, if you want, that your two solutions are inverses of each other.
Got it:smile::check:
 
And, of course, if you want to compute ##f(z) = z^n + \dfrac 1 {z^n}##, then ##f(z_0) = f(\frac 1 {z_0})##. And you get the same value for both solutions.
 

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