( -1+i3 2 ) ^ 20 + ( -1-i3 2 ) ^ 20 = - Vidyadip

MCQ

${{\left( \frac{-1+i\sqrt{3}}{2} \right)}^{20}}+{{\left( \frac{-1-i\sqrt{3}}{2} \right)}^{20}}=$

A
$20\sqrt{3}i$
B
1
C
$\frac{1}{{{2}^{19}}}$
✓
$-1$

✓

Answer

Correct option: D.

$-1$

D

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 "M.C.Q (1 Marks)" from Complex Numbers→Complex Numbers — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Similar questions

In how many ways $n$ books can be arranged in a row so that two specified books are not together

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

A vertical line passing through the point $(h, 0)$ intersects the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at the points $P$ and $Q$. Let the tangents to the ellipse at $P$ and $Q$ meet at the point $R$. If $\Delta(h)=$ area of the triangle $P Q R, \Delta_1=\max _{1 / 2 \leq h \leq 1} \Delta(h)$ and $\Delta_2=\min _{1 / 2 \leq h \leq 1} \Delta(h)$, then $\frac{8}{\sqrt{5}} \Delta_1-8 \Delta_2=$

$\tan 5x\tan 3x\tan 2x = $

If $\text{a}\cos\theta+\text{b}\sin\theta=\text{c}$ have roots $\alpha$ and $\beta$.Then, what will be the value of $\sin\alpha + \sin\beta$

For two data sets, each of size $5$, the variances are given to be $4$ and $5$ and the corresponding means are given to be $2$ and $4$, respectively. The variance of the combined data set is

$\mathop {\lim }\limits_{x \to a} \frac{{\sqrt {a + 2x} - \sqrt {3x} }}{{\sqrt {3a + x} - 2\sqrt x }} = $$(a \ne 0)$

The foot of the perpendicular drawn from the origin, on the line, $3x + y = \lambda \,\left( {\lambda \ne 0} \right)$ is $P$. If the line meets $x-$ axis at $A$ and $y-$ axis at $B$, then the ratio $BP : PA$ is

If $\sin A = \frac{1}{{\sqrt {10} }}$and $\sin B = \frac{1}{{\sqrt 5 }},$ where $A$ and $B$ are positive acute angles, then $A + B = $

$n^2 < 2^n$ for all natural numbers: