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)$i$

Hint: For answering this question we will use the concept of cube roots of unity which states that the cube roots of unity are $1,\omega ,{{\omega }^{2}}$ where there are 1 real root and 2 complex conjugate roots. And find the value of $\alpha $ and use it to derive the value of ${{\alpha }^{31}}$ .

Complete step by step answer:
Now considering from the question we have the equation ${{\alpha }^{2}}+\alpha +1=0$ in which $\alpha $ is a complex number. Here we need to find the value of $\alpha $ and use it and derive ${{\alpha }^{31}}$ .
So by observing the given equation we can say that it is similar to the equation of cube root of unity which is mathematically given as
 $\begin{align}
  & {{x}^{3}}=1 \\
 & \Rightarrow {{x}^{3}}-1=0 \\
 & \Rightarrow \left( x-1 \right)\left( {{x}^{2}}+x+1 \right)=0 \\
\end{align}$.
Here either $x=1$ or $\left( {{x}^{2}}+x+1 \right)=0$ gives the cube roots of unity. Here it has 3 roots one real and 2 complex conjugate roots they are $-\dfrac{1}{2}\pm i\dfrac{\sqrt{3}}{2}$ which are represented as $\omega $ and ${{\omega }^{2}}$ .
Here in this question we have that $\alpha $ is a complex number so it can be $\omega $ .
As we know that ${{\omega }^{3}}=1$ by using it here we can say ${{\alpha }^{3}}=1\Rightarrow {{\alpha }^{30}}=1$ .
As here we need the value of ${{\alpha }^{31}}$ we can write as ${{\alpha }^{30}}{{\alpha }^{1}}$ which is equal to $\alpha $ .

So, the correct answer is “Option A”.

Note: While answering this question we can choose any value for $\alpha $ between $\omega $ and ${{\omega }^{2}}$ . If we choose ${{\omega }^{2}}$ instead of $\omega $ as the value of $\alpha $ we will have ${{\alpha }^{31}}={{\left( {{\omega }^{2}} \right)}^{31}}$ . By simplifying it we will have ${{\alpha }^{31}}={{\omega }^{62}}={{\omega }^{60}}{{\omega }^{2}}$ which can be further simplified as ${{\alpha }^{31}}={{\omega }^{60}}{{\omega }^{2}}={{\omega }^{2}}=\alpha $ . We will have the same answer.