Question

If $z+\frac{1}{z}+1=0,$ then $z^{2003}+\frac{1}{z^{2003}}$ is equal to

• A. $1$
• B. $-1$
• C. $0$
• D. None of these

Answer 1

Answer: (B)

Given equation can be written as $z^2+z+1=0$
$\Rightarrow$ $z=\omega ,{\left(\omega \right)}^2\left(roots\right)$
Now, $z^{2003}={\omega }^{2003}={\omega }^2$
$z^{2003}+\frac{1}{z^{2003}}={\omega }^2+\frac{1}{{\omega }^2}={\omega }^2+\omega =-1$