If is a complex cube root of unity such that ^2+ +1=0, then ^ 31 … - Vidyadip

MCQ

If $\alpha$ is a complex cube root of unity such that $\alpha^2+\alpha+1=0$, then $\alpha^{31}$ is

✓
$\alpha$
B
$\alpha^2$
C
$0$
D
1

✓

Answer

Correct option: A.

$\alpha$

(A)
$\alpha^2+\alpha+1=0$
$\therefore \quad(\alpha-1)\left(\alpha^2+\alpha+1\right)=0$
$\therefore \quad \alpha^3-1=0, \alpha \neq 1$
$\Rightarrow \alpha^3=1$
and consequently $\alpha^{31}=\left(\alpha^3\right)^{10} \cdot \alpha=1^{10} \alpha=\alpha$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "MCQ" from Complex Numbers→Complex Numbers — all question types→Maths STD 11 Science question papers→Maharashtra Board STD 11 Science question papers→Maharashtra Board question paper generator→

Similar questions

The point of contact of the tangent to the circle $x^2+y^2=5$ at the point (1,-2) which touches the circle ,$x^2+y^2-8 x+6 y+20=0$ is

The real values of $x$ and $y$ for which the equation $(x+i y)(2-3 i)=4+i$ is satisfied, are

$A(6, 3), B(-3, 5), C(4, -2)$ and $D(x, 3x)$ are four points. If $\triangle\text{DBC} : \triangle\text{ABC}= 1 : 2,$ then $x$ is equal to :

Let PS be the median of the triangle with vertices P(2, 2) Q(6, - 1) and R(7, 3) The equation of the line passing through (1, -1) and parallel to PS is

$\lim _{x \rightarrow-2}\left(\frac{x^7+128}{x^3+8}\right)=$

Function $f (x)=\frac{1-\cos 4 x}{8 x^2}$, where $x \neq 0$ and $f (x)= k$, where $x=0$ is a continuous function at $x=0$, then the value of k will be

Domain of the function $\sqrt{\log \left\{\left(5 x-x^2\right) / 6\right\}}$ is

The equation of the directrix of the parabola $x^2+4 x+2 y=0$ is

Two tangents of the parabola $y^2=8 x$ meet its tangent at vertex at points P and Q . If $PQ =4$, then the locus of the point of intersection of these two tangents is

For the function $f (x)=\left\{\begin{array}{ll}\frac{ e ^{1 / x}-1}{ e ^{1 / x}+1}, & , x \neq 0 \\ 1 & , x=0\end{array}\right.$, which of the following is correct?