Question

$z$ is a complex number such that $z+\frac{1}{z}=2 \cos 3^{\circ}$, then the value of $z^{2000}+\frac{1}{z^{2000}}+1$ is equal to

Answer 1

Let $z=\cos \theta+i \sin \theta=e^{i \theta} $;
$ \frac{1}{z}=\cos \theta-i \sin \theta=e^{-i \theta}$
So that $z+\frac{1}{z}=2 \cos \theta\left(\theta=3^{\circ}\right)$
Now $z^{2000}+\frac{1}{z^{2000}}+1$
$e^{i\, 2000 \,\theta}+e^{-i \,2000\, \theta}+1$
$=2 \cos (2000 \theta)+1$
$=2 \cos \left(6000^{\circ}\right)+1$ (as $\theta=3^{\circ}$ )
$=2 \cos \left(\frac{100 \pi}{3}\right)+1$
$=2 \cos \left(\frac{4 \pi}{3}\right)+1=-1+1=0$