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

Convert the complex number $\frac{-16}{1+i \sqrt{3}}$ into polar form.

If $z=1+i \sqrt{3}$ then $|\operatorname{Arg} z|+|\operatorname{Arg} \bar{z}|$ is equal to

If $\frac{3+i \sin \theta}{4-i \cos \theta}, \theta \in [0, 2 \pi],$ is a real number,then an argument of $\sin \theta + i \cos \theta$ is

$\operatorname{Arg}\left(\frac{4+2 i}{1-2 i}+\frac{3+4 i}{2+3 i}\right)$ lies in the interval

$x$ and $y$ are two complex numbers such that $|x|=|y|=1$. If $\operatorname{Arg}(x)=2 \alpha$,$\operatorname{Arg}(y)=3 \beta$,and $\alpha+\beta=\frac{\pi}{36}$,then $x^6 y^4+\frac{1}{x^6 y^4}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module