Question

If $z^2 + z + 1 = 0$ where $z$ is a complex number, then the value of $\left(z+ \frac{1}{z}\right)^{2} + \left(z^{2} + \frac{1}{z^{2}}\right)^{2} + \left(z^{3} + \frac{1}{z^{3}}\right)^{2} $ equals

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

The Correct Option is C

We have,
$z^{2}+z+1=0$
$\Rightarrow z=\frac{-1 \pm \sqrt{1-4}}{2}$
$=\frac{-1 \pm \sqrt{3} i}{2}$
$\therefore z=w$ or $w^{2}$
Let $z=w$, then
$\left(z+\frac{1}{z}\right)^{2}+\left(z^{2}+\frac{1}{z^{2}}\right)^{2}+\left(z^{3}+\frac{1}{z^{3}}\right)^{2}$
$=\left(\omega+\frac{1}{\omega}\right)^{2}+\left(\omega^{2}+\frac{1}{\omega^{2}}\right)^{2}+\left(\omega^{3}+\frac{1}{\omega^{3}}\right)^{2}$
$=\left(\omega+\omega^{2}\right)^{2}+\left(\omega^{2}+\omega\right)^{2}+\left(\omega^{3}+1\right)^{2}\,\left[\because \omega^{3}=1\right]$
$=(-1)^{2}+(-1)^{2}+(1+1)^{2}$
$\left[\because 1+\omega+\omega^{2}=0\right]$
$=1+1+4 = 6$
The value will be same when $z=\omega^{2}$.