Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
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 $z = \tilde{z}_{1,2}$.

share|cite|improve this answer
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
@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 $e^{ix}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.