$z \in\mathbb{C}^{*}$ is a root of the equation $z+\frac{1}{z}=2\cos\frac{\pi}{2018} $ then $z^{2018}+\frac{1}{z^{2018}}$ has the value...the right answer is -2.

  • $\begingroup$ whatever $a$ is, $z+\frac1z=a$ is a quadratic equation. $\endgroup$ – Lord Shark the Unknown Jul 11 at 19:03
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    $\begingroup$ Rewrite $z$ as $e^{i\theta}$... $\endgroup$ – Peter Foreman Jul 11 at 19:03
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    $\begingroup$ Does $\Bbb C^\ast$ denote the set of unit complex numbers? $\endgroup$ – J.G. Jul 11 at 19:10

Let $\theta:=\frac{\pi}{2018}$ so$$z^2-(e^{i\theta}+e^{-i\theta})z+1=0\implies z=e^{\pm i\theta}\\\implies z^{2018}+z^{-2018}=2\cos\pi=-2.$$

  • $\begingroup$ Hi!Thank you for your answer.I didn't learn at school about the fact with $e^{i..}$.Can I use Moivre formula or something else to solve this ? I got a quadratic $z^2-za+1=0$ where $a=2cos(\pi/2018)$ and I got the discriminant $2\sqrt{cos^2(a)-1}$ $\endgroup$ – DaniVaja Jul 11 at 19:29
  • $\begingroup$ how to continue? $\endgroup$ – DaniVaja Jul 11 at 19:30
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    $\begingroup$ @DaniVaja All we're using here is $2\cos x=e^{ix}+e^{-ix}$, which follows from $e^{\pm ix}=\cos x\pm i\sin x$. $\endgroup$ – J.G. Jul 11 at 19:34
  • $\begingroup$ Thanks. I delete my comment. $\endgroup$ – Piquito Jul 11 at 19:35
  • $\begingroup$ Ok, I understood this but I didn;t understood why $z=e^{+-i*theta}$ $\endgroup$ – DaniVaja Jul 11 at 20:00

If $z+\dfrac1z=2\cos t$


Similarly, $z^3+\dfrac1{z^3}=\left(z+\dfrac1z\right)^3-3\left(z+\dfrac1z\right)=\cdots=2\cos3t$

Like Proving that $\frac{\phi^{400}+1}{\phi^{200}}$ is an integer.

using strong induction

$$z^{n+1}+\dfrac1{z^{n+1}}=\left(z^n+\frac1{z^n}\right)\left(z+\frac1z\right)-\left(z^{n-1}+\frac1{z^{n-1}}\right) =2\cos nt\cdot2\cos t-2\cos(n-1)t$$

By Werner Formulas


$$\implies z^{n+1}+\dfrac1{z^{n+1}}=2\cos(n+1)t$$


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