# z the root of $z+\frac{1}{z}=2\cos\frac{\pi}{2018}$ then the value of $z^{2018}+\frac{1}{z^{2018}}$ is..

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

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

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

• 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 $2\sqrt{cos^2(a)-1}$ – DaniVaja Jul 11 at 19:29
• how to continue? – DaniVaja Jul 11 at 19:30
• @DaniVaja All we're using here is $2\cos x=e^{ix}+e^{-ix}$, which follows from $e^{\pm ix}=\cos x\pm i\sin x$. – J.G. Jul 11 at 19:34
• Thanks. I delete my comment. – Piquito Jul 11 at 19:35
• Ok, I understood this but I didn;t understood why $z=e^{+-i*theta}$ – DaniVaja Jul 11 at 20:00

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

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

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

using strong induction

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

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

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