Question

If $\alpha$ is a complex number satisfying the equation ${\alpha }^2+\alpha +1=0$, then ${\alpha }^{31}$ is equal to

• A. $\alpha$
• B. ${\alpha }^2$
• C. $1$
• D. $i$

Answer 1

The correct option is A

$\alpha$

Explanation for the correct option:

Step 1. Find the value of ${\alpha }^{31}$:

As we know,

${\omega }^3=1$ (cube root of unity)

$\Rightarrow$ ${\omega }^3-1=0$

$\Rightarrow$$(\omega -1)({\omega }^2+\omega +1)=0$

$\Rightarrow$$\omega =1,({\omega }^2+\omega +1)=0$

$\Rightarrow$ $({\omega }^2+\omega +1)=0$

Step 2. $\alpha$ satisfies the equation ${\alpha }^2+\alpha +1=0$

$\Rightarrow$Roots are $\omega ,{\omega }^2$ and ${\omega }^3=1$

$\begin{array}{rcl}\therefore {\alpha }^{31}& =& {\omega }^{31}\\ & =& \omega {\left({\omega }^3\right)}^{10}\\ & =& \omega \\ & =& \mathit{\alpha }\end{array}$

Hence, Option ‘A’ is Correct.