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

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Homework Help Overview

The discussion revolves around the equation involving complex numbers, specifically focusing on the expression \( z^n + \frac{1}{z^n} \). Participants explore the implications of the solutions derived from a quadratic equation related to this expression.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the nature of the solutions to the quadratic equation and their relationship, questioning why only one argument is considered in the context of the problem.

Discussion Status

Some participants have offered insights into the equivalence of the solutions and their implications for the function \( f(z) = z^n + \frac{1}{z^n} \). The discussion reflects an exploration of the properties of the solutions without reaching a definitive consensus.

Contextual Notes

There is an emphasis on understanding the relationship between the solutions and the nature of complex arguments, with participants questioning the assumptions underlying the consideration of only one argument.

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.
 
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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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