Determine $z^n+z^{-n}$ if $z+\frac{1}{z}=-2\cos{x}$ with $z \in \mathbb{C}$.
Determine $z^n+z^{-n}$ if $z+\frac{1}{z}=-2\cos{x}$
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algebra-precalculus
complex-numbers
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1Hint: You can solve $z+\frac{1}{z}=...$. – 2011-11-03
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0I'm assuming $x \in \mathbb{R}$? – 2011-11-03
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0I'm curious, do you know how to use induction? – 2011-11-03
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0Yes, x is real. – 2011-11-03
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0There 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.... – 2011-11-03
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0Is there a mistake in this question? Should it be determine z? You seem to be given $z^n+z^{-n}$... – 2011-12-01
2 Answers
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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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2Here 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 :) – 2011-11-03
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0@user9176: Why is $1/z$ also a solution? – 2011-11-03
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1@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 $z$. – 2011-11-03
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0You can also figure it out by looking to the starting equation ;) – 2011-11-04
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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 $e^{ix}$.