If z = sqrt{2} sqrt{1 + sqrt{3} i} represents | vedclass

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

Question

(B) Given $z = \sqrt{2} \sqrt{1 + \sqrt{3} i}$.We can write $1 + \sqrt{3} i = \frac{1}{2} (2 + 2 \sqrt{3} i) = \frac{1}{2} ((\sqrt{3})^2 + i^2 + 2 \sq

Vedclass · Accepted Answer

$2 \left[ \cos \left( \frac{-5 \pi}{6} \right) + i \sin \left( \frac{-5 \pi}{6} \right) \right]$

Vedclass · Answer

(B) Given $z = \sqrt{2} \sqrt{1 + \sqrt{3} i}$.We can write $1 + \sqrt{3} i = \frac{1}{2} (2 + 2 \sqrt{3} i) = \frac{1}{2} ((\sqrt{3})^2 + i^2 + 2 \sqrt{3} i) = \frac{1}{2} (\sqrt{3} + i)^2$.Thus,$z = \sqrt{2} \cdot \frac{1}{\sqrt{2}} (\sqrt{3} + i) = \pm (\sqrt{3} + i)$.Since $z$ lies in the third quadrant,both real and imaginary parts must be negative.Therefore,$z = -\sqrt{3} - i$.The modulus is $|z| = \sqrt{(-\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2$.The argument $	heta$ in the third quadrant is given by $	heta = -(\pi - 	an^{-1}(\frac{1}{\sqrt{3}})) = -(\pi - \frac{\pi}{6}) = -\frac{5 \pi}{6}$.Thus,the polar form is $z = 2 \left[ \cos \left( \frac{-5 \pi}{6} ight) + i \sin \left( \frac{-5 \pi}{6} ight) ight]$.

If $z = \sqrt{2} \sqrt{1 + \sqrt{3} i}$ represents a point $P$ in the Argand plane and $P$ lies in the third quadrant,then the polar form of $z$ is

Similar Questions

Let the two values of $z = \sqrt{\frac{1-i}{1+i}}$ be $z_1$ and $z_2$. If $-\frac{\pi}{2} < \operatorname{Arg}(z_1) < \operatorname{Arg}(z_2) < \pi$,then $\arg(z_1) + \arg(z_2) = $

The amplitude of ${e^{{e^{ - i\theta }}}}$ is equal to

Find the modulus and the argument of the complex number $z = -\sqrt{3} + i$.

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module