Question

If $z+z^{-1}=1$ , then $z^{100}+z^{-100}$ is equal to

• A. $i$
• B. $-i$
• C. $1$
• D. $-1$

Answer 1

Answer: (D)

$z+z^{-1}=1\Rightarrow z^2-z+1=0$
$\Rightarrow z=-\omega$ or $-{\omega }^2$ For $z=-\omega$, we have
$z^{100}+z^{-100}=(-\omega {)}^{100},(-\omega {)}^{-100}$
$=\omega +\frac{1}{\omega }=\omega +{\omega }^2=-1$
For $z=-{\omega }^2$
$z^{100}+z^{-100}={\left(-{\omega }^2\right)}^{100}+{\left(-{\omega }^2\right)}^{-100}$
$={\omega }^{200}+\frac{1}{{\omega }^{200}}={\omega }^2+\frac{1}{{\omega }^2}={\omega }^2+\omega =-1$