If {left( {frac{{1 + isqrt 3 }}{{1 - isqrt 3 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $z = \frac{1 + i\sqrt{3}}{1 - i\sqrt{3}}$.Multiply the numerator and denominator by the conjugate of the denominator $(1 + i\sqrt{3})$:$z = \f

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Let $z = \frac{1 + i\sqrt{3}}{1 - i\sqrt{3}}$.Multiply the numerator and denominator by the conjugate of the denominator $(1 + i\sqrt{3})$:$z = \frac{(1 + i\sqrt{3})(1 + i\sqrt{3})}{(1 - i\sqrt{3})(1 + i\sqrt{3})} = \frac{1 + 2i\sqrt{3} + (i\sqrt{3})^2}{1^2 + (\sqrt{3})^2} = \frac{1 + 2i\sqrt{3} - 3}{1 + 3} = \frac{-2 + 2i\sqrt{3}}{4} = \frac{-1 + i\sqrt{3}}{2}$.This is the value of $\omega$,the complex cube root of unity.Thus,$z^n = \omega^n$.For $\omega^n$ to be an integer,$n$ must be a multiple of $3$ because $\omega^3 = 1$.The smallest positive integer $n$ is $3$.

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

Similar Questions

Given $z = (1 + i\sqrt{3})^{100}$,then $\frac{\text{Re}(z)}{\text{Im}(z)}$ equals

If $\omega$ is a complex cube root of unity,then $\sin \left\{\left(\omega^{10}+\omega^{23}\right) \pi-\frac{\pi}{4}\right\}=$

If $\alpha, \beta$ are the roots of the equation $x^2-4x+8=0$,then for any $n \in N$,$\alpha^{2n}+\beta^{2n}$ equals

Let $z = \cos \theta + i \sin \theta$. Then,the value of $\sum_{m=1}^{15} \text{Im}(z^{2m-1})$ at $\theta = 2^{\circ}$ is

Consider a regular $10$-gon with its vertices on the unit circle. With one vertex fixed,draw straight lines to the other $9$ vertices. Call them $L_1, L_2, \ldots, L_9$ and denote their lengths by $l_1, l_2, \ldots, l_9$ respectively. Then,the product $l_1 \times l_2 \times \ldots \times l_9$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module