Question

If $z+\frac{1}{z}=2\cos 6^0$, then $z^{1000}+\frac{1}{z^{1000}}+1$ is equal to

• A. 0
• B. 1
• C. -1
• D. 2

Answer 1

The correct option is A 0

$z+\frac{1}{z}=2\cos 6^o$

$z=\cos ]theta+i\sin ]theta$

$z+\frac{1}{z}=\cos \theta +i\sin \theta +\cos \theta -i\sin \theta$

$2\cos 6^o=2\cos \theta$

$\theta =6^o$

$z^{1000}+\frac{1}{z^{1000}}=\cos \left(1000\theta \right)+i\sin \left(1000\theta \right)+\cos \left(1000\theta \right)-i\sin \left(1000\theta \right)+1$

$=2\cos \left(1000\theta \right)+1$

$=2\cos \left({6000}^o\right)+1$

$=2\cos (5760+240)+1$

$2\cos \left({240}^o\right)+1=2\cos \left({120}^o\right)+1$

$=-2\cos 60+1-1+1=0$

$\therefore$ $z^{1000}+\frac{1}{z^{1000}}+1=0$