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Determine $z^n+z^{-n}$ if $z+\frac{1}{z}=-2\cos{x}$ with $z \in \mathbb{C}$.

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Hint: You can solve $z+\frac{1}{z}=...$. –  N. S. Nov 3 '11 at 16:23
    
I'm assuming $x \in \mathbb{R}$? –  rcollyer Nov 3 '11 at 16:23
    
I'm curious, do you know how to use induction? –  N. S. Nov 3 '11 at 16:26
    
Yes, x is real. –  Daniel Nov 3 '11 at 16:26
    
There was an answer which had a mistake, but which could had been changed to make it work. Unfortunately it was deleted before i could make another comment. So here is another idea: $z=r(\cos(\theta)+i\sin(\theta)$. Sub it in the equation, and instead of looking to the real part , look at the imaginary part. That yields $r$, and then the rest is simple.... –  N. S. Nov 3 '11 at 17:09

2 Answers 2

up vote 2 down vote accepted

Since $z + \frac{1}{z} = - 2 \cos(x)$ is equivalent to $z^2 + 2 z \cos(x) + 1 = 0$, it is solved by $z_{1,2} = -\cos(x) \pm i \sqrt{1-\cos^2(x)}$. Since $1-\cos^2(x) = \sin^2(x)$, these also solve the equation $\tilde{z}_{1,2} = -\cos(x) \mp i \sin(x) = -\exp(\pm i x)$.

Now to find $z^n+z^{-n}$ for $z$ being the solution of $z+\frac{1}{z} = -2 \cos(x)$ subsitute the $z = \tilde{z}_{1,2}$.

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Here is a neat way of solving this equation without using the quadratic formula: $z^2 + 2 z \cos(x) + 1 = 0$ is quadratic, so it has two complex solutions. But then , if $z$ is a solution, so are $\overline{z}$ and $\frac{1}{z}$. Which means that two of $z, \overline{z}, \frac{1}{z}$ have to be equal, and this implies $|z|=1$. Thus, $z_{1,2}= \cos(\theta) \pm i \sin (\theta)$ which leads to $\cos(\theta)=-\cos(x)$ and implicitely $\sin(\theta)= \pm \sin(x)$.... This proof is way too complicated though for pre-calculus :) –  N. S. Nov 3 '11 at 16:40
    
@user9176: Why is $1/z$ also a solution? –  Weltschmerz Nov 3 '11 at 22:34
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@Weltschmerz Because, assuming $z!=0$ $0 = \frac{1}{z^2} \left( z^2 + 2 z \cos(x) + 1\right) = \left(\frac{1}{z}\right)^2 + 2 \left(\frac{1}{z}\right) \cos(x) + 1$, therefore $w =\frac{1}{z}$ solves the same equation as $z$. –  Sasha Nov 3 '11 at 23:12
    
You can also figure it out by looking to the starting equation ;) –  N. S. Nov 4 '11 at 5:54

The hint from user9176 is important: $$ z+\frac1z = -2\cos x $$ Multiply both sides by $z$: $$ z^2 + 1 = -2z\cos x $$ That's a quadratic equation in $z$. Solve it.

Then remember certain identities involving $e^{ix}$.

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