Argument of frac{1-i sqrt{3}}{1+i sqrt{3}} is | 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 = \frac

Vedclass · Accepted Answer

$240$

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 $(1-i \sqrt{3})$:$z = \frac{(1-i \sqrt{3})(1-i \sqrt{3})}{(1+i \sqrt{3})(1-i \sqrt{3})} = \frac{1 - 2i \sqrt{3} + i^2(3)}{1^2 + (\sqrt{3})^2} = \frac{1 - 2i \sqrt{3} - 3}{1 + 3} = \frac{-2 - 2i \sqrt{3}}{4} = -\frac{1}{2} - i \frac{\sqrt{3}}{2}$.Since the complex number $z = -\frac{1}{2} - i \frac{\sqrt{3}}{2}$ lies in the third quadrant,the argument is given by $	ext{Arg}(z) = -\pi + 	an^{-1}\left|\frac{	ext{Im}(z)}{	ext{Re}(z)}ight|$ or $\pi + 	an^{-1}\left(\frac{\sqrt{3}/2}{1/2}ight)$.$	ext{Arg}(z) = 180^{\circ} + 	an^{-1}(\sqrt{3}) = 180^{\circ} + 60^{\circ} = 240^{\circ}$.

Argument of $\frac{1-i \sqrt{3}}{1+i \sqrt{3}}$ is (in $^{\circ}$)

Similar Questions

If ${z_1}, {z_2}$ and ${z_3}, {z_4}$ are two pairs of conjugate complex numbers,then $arg\left( \frac{z_1}{z_4} \right) + arg\left( \frac{z_2}{z_3} \right)$ equals:

If $|z| = 4$ and $\text{arg}(z) = \frac{5\pi}{6}$,then $z =$

If $z_1 \cdot z_2 \cdot \dots \cdot z_n = z$,then $arg(z_1) + arg(z_2) + \dots + arg(z_n)$ and $arg(z)$ differ by a

If $z$ is a complex number such that $|z - \bar{z}| = 2$ and $|z + \bar{z}| = 4$,then which of the following is always incorrect -

Assertion $(A)$: If the arguments of $\bar{z}_1$ and $z_2$ are $\frac{\pi}{5}$ and $\frac{\pi}{3}$ respectively,then $\arg(z_1 z_2)$ is $\frac{2\pi}{15}$. Reason $(R)$: For any complex number $z$,$\arg(\bar{z}) = \frac{\pi}{2} + \arg(z)$. The correct option among the following is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module