Question

Find the value of $z^{14}+\frac{1}{z^{14}},$ where $z$ is a root of the equation $\frac{z^2+1}{z}=1.$

• A. $1$
• B. $-1$
• C. $\omega$
• D. ${\omega }^2$

Answer 1

The correct option is A $1$

$z^2+1=z$

$z^2-z+1=0$

Hence the roots are

$z=-w,-w^2$

Hence let

$z=-w$

Now $w^3=1$

Hence

$z^{14}=z^{15}.\frac{1}{z}$

$=-\frac{1}{w}$

Hence

$z^{14}+\frac{1}{z^{14}}$ for $z=-w$

$=-(w+\frac{1}{w})$
$=-\frac{w^2+1}{w}$
$=-\frac{-w}{w}$
$=1$