1
$\begingroup$

$z \in\mathbb{C}^{*}$ 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.

$\endgroup$
  • $\begingroup$ whatever $a$ is, $z+\frac1z=a$ is a quadratic equation. $\endgroup$ – Lord Shark the Unknown Jul 11 at 19:03
  • 4
    $\begingroup$ Rewrite $z$ as $e^{i\theta}$... $\endgroup$ – Peter Foreman Jul 11 at 19:03
  • 1
    $\begingroup$ Does $\Bbb C^\ast$ denote the set of unit complex numbers? $\endgroup$ – J.G. Jul 11 at 19:10
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ @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
0
$\begingroup$

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

$z^2+\dfrac1{z^2}=\left(z+\dfrac1z\right)^2-2=\cdots=2\cos2t$

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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