If ( 1 + i 3 1 - i 3 )^n is an integer, then n is - Vidyadip

MCQ

If ${\left( {\frac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right)^n}$ is an integer, then $n$ is

A
$1$
B
$2$
✓
$3$
D
$4$

✓

Answer

Correct option: C.

$3$

c
(c) $\frac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }} = \left( {\frac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right)\,\left( {\frac{{1 + i\sqrt 3 }}{{1 + i\sqrt 3 }}} \right) = \frac{{ - 2 + i2\sqrt 3 }}{4}$
$ = \,\frac{{ - 1 + i\sqrt 3 }}{2} = \omega $
${\left( {\frac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right)^n} = {\omega ^n} = {\omega ^3} = 1 \Rightarrow n = 3$.

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 STD 11 - 4.1 complex nubers→STD 11 - 4.1 complex nubers — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Similar questions

$\operatorname{cosec} 150^{\circ}=?$

$\operatorname{Lim}_{n \rightarrow \infty}\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right) \ldots \ldots\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)\right\}$ is equal to

The term independent of $x$ in the expansion of ${\left( {{x^2} - \frac{1}{x}} \right)^9}$ is

The sum of the series $\frac{1}{{1 + {1^2} + {1^4}}} + \frac{2}{{1 + {2^2} + {2^4}}} + \frac{3}{{1 + {3^2} + {3^4}}} + .........$ to $n$ terms is

The quotient when $1+x^2+x^4+x^6+\ldots+x^{34}$ is divided by $1+x+x^2+x^3+\ldots+x^{17}$ is

If the equation ${x^2} + {y^2} + 2gx + 2fy + c = 0$ represents a circle with $x$ - axis as a diameter and radius $a$, then

A real value of $x$ satisfies the equation $\frac{3-4\text{ix}}{3+4\text{ix}}=\text{a}-\text{ib}\Big(\text{a},\text{b}\in\text{R}\Big),$ if $\text{a}^2+\text{b}^2=$

If the fractional part of the number $\frac{{{2^{403}}}}{{15}}$ is $\frac{k}{{15}}$, then $k$ is equal to

If $\sec\text{x}+\tan\text{x}=\text{k},\cos\text{x}=$

If $\alpha, \beta$ are roots of the equation $x^{2}+5 \sqrt{2} x+10=0, \alpha\,>\,\beta$ and $P_{n}=\alpha^{n}-\beta^{n}$ for each positive integer $\mathrm{n}$, then the value of $\left(\frac{P_{17} P_{20}+5 \sqrt{2} P_{11} P_{19}}{P_{18} P_{19}+5 \sqrt{2} P_{18}^{2}}\right)$ is equal to $....$