Question

If $\alpha$ is a non real root of $z=(1{)}^{1/5},$ then the value of $(1+\alpha +{\alpha }^2+{\alpha }^{-2}-{\alpha }^{-1})$ is

• A. $2$
• B. $\frac{2}{\alpha }$
• C. $-2{\alpha }^4$
• D. ${\alpha }^4$

Answer 1

The correct option is C $-2{\alpha }^4$

Let, $1,\alpha ,{\alpha }^2,{\alpha }^3,{\alpha }^4$ be the roots of $z^5=1$ and,
$E=1+\alpha +{\alpha }^2+\frac{1}{{\alpha }^2}-\frac{1}{\alpha }$
$\Rightarrow E=\frac{{\alpha }^2+{\alpha }^3+{\alpha }^4+1-\alpha }{{\alpha }^2}$
$\Rightarrow E=\frac{(1+\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4)-2\alpha }{{\alpha }^2}$
We know that, sum of $n$ roots of unity is zero, so
$\Rightarrow E=\frac{0-2\alpha }{{\alpha }^2}=\frac{-2}{\alpha }$
$\Rightarrow E=\frac{-2}{\alpha }\times \frac{{\alpha }^4}{{\alpha }^4}=\frac{-2{\alpha }^4}{{\alpha }^5}(\because \alpha \text{is the root of}{z}^5=1)$
$\Rightarrow E=-2{\alpha }^4$