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

In summary, the conversation discusses finding solutions to a quadratic equation and determining the modulus and argument for each solution. It is noted that the two solutions are essentially the same, and this is why only one argument is considered. This can be verified by computing the function for both solutions.
  • #1
Magnetons
18
4
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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  • #2
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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  • #3
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:
 
  • #4
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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FAQ: Find z^n+ 1/z^n: Why Consider Only One Argument?

1. What is z^n+ 1/z^n?

z^n+ 1/z^n is an expression in mathematics that represents the sum of z^n and 1/z^n. This expression is commonly used in complex analysis and has many applications in physics and engineering.

2. Why is it important to consider only one argument?

Considering only one argument in this expression allows us to simplify the problem and find a general solution. By keeping the argument constant, we can focus on the behavior of the expression and make meaningful conclusions.

3. What is the significance of z^n and 1/z^n in this expression?

z^n and 1/z^n are complex conjugates of each other, meaning they have the same magnitude but opposite signs in their imaginary components. This relationship is important in simplifying the expression and finding a general solution.

4. How is this expression related to De Moivre's theorem?

De Moivre's theorem states that for any complex number z and integer n, (cos(z) + isin(z))^n = cos(nz) + isin(nz). By substituting z for z^n and considering only one argument, we can apply De Moivre's theorem to simplify the expression.

5. Can this expression be used in real-world applications?

Yes, z^n+ 1/z^n has many real-world applications in fields such as electrical engineering, signal processing, and quantum mechanics. It is also useful in solving differential equations and modeling physical systems.

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