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

Answer: 0

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 $z^{2000}+\frac{1}{z^{2000}}+1$
$e^{i2000\theta }+e^{-i2000\theta }+1$
$=2\cos (2000\theta )+1$
$=2\cos \left(6000^{\circ }\right)+1$ (as $\theta =3^{\circ }$ )
$=2\cos \left(\frac{100\pi }{3}\right)+1$
$=2\cos \left(\frac{4\pi }{3}\right)+1=-1+1=0$