Convert the complex number frac{-16}{1+i sqrt | vedclass

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

Question

(A) The given complex number is $z = \frac{-16}{1+i \sqrt{3}}$.Multiply the numerator and denominator by the conjugate $1-i \sqrt{3}$:$z = \frac{-16(1

Vedclass · Accepted Answer

$8\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right)$

Vedclass · Answer

(A) The given complex number is $z = \frac{-16}{1+i \sqrt{3}}$.Multiply the numerator and denominator by the conjugate $1-i \sqrt{3}$:$z = \frac{-16(1-i \sqrt{3})}{(1+i \sqrt{3})(1-i \sqrt{3})} = \frac{-16(1-i \sqrt{3})}{1^2 - (i \sqrt{3})^2} = \frac{-16(1-i \sqrt{3})}{1+3} = \frac{-16(1-i \sqrt{3})}{4} = -4 + i 4 \sqrt{3}$.Let $z = r(\cos 	heta + i \sin 	heta)$,where $r = \sqrt{(-4)^2 + (4 \sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8$.Then $\cos 	heta = \frac{-4}{8} = -\frac{1}{2}$ and $\sin 	heta = \frac{4 \sqrt{3}}{8} = \frac{\sqrt{3}}{2}$.Since $\cos 	heta  0$,$	heta$ lies in the second quadrant.$	heta = \pi - \frac{\pi}{3} = \frac{2 \pi}{3}$.Thus,the polar form is $8\left(\cos \frac{2 \pi}{3} + i \sin \frac{2 \pi}{3}ight)$.

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

Similar Questions

If $arg(z) = \theta$,then $arg(\overline{z}) = $

If $\frac{\pi}{2} < \alpha < \frac{3\pi}{2}$,then the modulus and argument of $(1 + \cos 2\alpha) + i \sin 2\alpha$ are respectively:

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:

Let $z_1$ and $z_2$ be two complex numbers with $\alpha$ and $\beta$ as their principal arguments such that $\alpha + \beta > \pi$,then the principal argument of $z_1 z_2$ is given by:

The argument of the complex number $\sin \frac{6\pi}{5} + i(1 + \cos \frac{6\pi}{5})$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module