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

AI Thread Summary
The discussion focuses on solving the equation z^n + 1/z^n by converting it into a quadratic form, yielding two solutions with the same modulus but different arguments. The key point is that both solutions are inverses of each other, leading to the conclusion that they produce the same result when substituted into the function f(z) = z^n + 1/z^n. Therefore, only one argument, specifically π/6, is considered because both solutions yield the same value. This reinforces the concept that the function is symmetric with respect to the unit circle in the complex plane. Understanding this symmetry clarifies why only one argument is necessary for the analysis.
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.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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