If z_1 = -sqrt{3} + i and z_2 = -sqrt{3} - i, | vedclass

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

Question

(C) Given $z_1 = -\sqrt{3} + i$ and $z_2 = -\sqrt{3} - i$. We know that the argument of a quotient is given by $	ext{arg}(\frac{z_1}{z_2}) = 	ext{ar

Vedclass · Accepted Answer

$-\frac{\pi}{3}$

Vedclass · Answer

(C) Given $z_1 = -\sqrt{3} + i$ and $z_2 = -\sqrt{3} - i$. We know that the argument of a quotient is given by $	ext{arg}(\frac{z_1}{z_2}) = 	ext{arg}(z_1) - 	ext{arg}(z_2)$. For $z_1 = -\sqrt{3} + i$,the point lies in the second quadrant. $	ext{arg}(z_1) = \pi - 	an^{-1}(\frac{1}{\sqrt{3}}) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$. For $z_2 = -\sqrt{3} - i$,the point lies in the third quadrant. $	ext{arg}(z_2) = -(\pi - 	an^{-1}(\frac{1}{\sqrt{3}})) = -(\pi - \frac{\pi}{6}) = -\frac{5\pi}{6}$. Therefore,$	ext{arg}(\frac{z_1}{z_2}) = \frac{5\pi}{6} - (-\frac{5\pi}{6}) = \frac{10\pi}{6} = \frac{5\pi}{3}$. Since the principal amplitude must lie in the interval $(-\pi, \pi]$,we subtract $2\pi$: $\frac{5\pi}{3} - 2\pi = -\frac{\pi}{3}$. Thus,the principal amplitude is $-\frac{\pi}{3}$.

If $z_1 = -\sqrt{3} + i$ and $z_2 = -\sqrt{3} - i$,then the principal amplitude of the complex number $\frac{z_1}{z_2}$ is

Similar Questions

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

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

The amplitude of $\sin \frac{\pi}{5} + i(1 - \cos \frac{\pi}{5})$ is:

If complex numbers $z_1$ and $z_2$ are such that $|z_1| = \sqrt{2}$,$|z_2| = \sqrt{3}$ and $|z_1 + z_2| = \sqrt{5 - 2\sqrt{3}}$,then the value of $|Arg(z_1) - Arg(z_2)|$ is

If $z = 1 - \cos \alpha + i \sin \alpha $,then $\text{amp } z$ =

Vedclass Test Series

Exam Paper Generator

Online Exam Module