$z \in\mathbb{C}^{*}$ 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.
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whatever $a$ is, $z+\frac1z=a$ is a quadratic equation. – Angina Seng Jul 11 '19 at 19:03
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4Rewrite $z$ as $e^{i\theta}$... – Peter Foreman Jul 11 '19 at 19:03
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1Does $\Bbb C^\ast$ denote the set of unit complex numbers? – J.G. Jul 11 '19 at 19:10
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Let $\theta:=\frac{\pi}{2018}$ so$$z^2-(e^{i\theta}+e^{-i\theta})z+1=0\implies z=e^{\pm i\theta}\\\implies z^{2018}+z^{-2018}=2\cos\pi=-2.$$
J.G.
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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 '19 at 19:29
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2@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 '19 at 19:34
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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
$$z^{n+1}+\dfrac1{z^{n+1}}=\left(z^n+\frac1{z^n}\right)\left(z+\frac1z\right)-\left(z^{n-1}+\frac1{z^{n-1}}\right) =2\cos nt\cdot2\cos t-2\cos(n-1)t$$
$$z^{n+1}+\dfrac1{z^{n+1}}=2[\cos(n-1)t+\cos(n+1)t]-2\cos(n-1)t$$
$$\implies z^{n+1}+\dfrac1{z^{n+1}}=2\cos(n+1)t$$
lab bhattacharjee
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