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

The argument of $\frac{1+i \sqrt{3}}{\sqrt{3}+i}$,where $i=\sqrt{-1}$,is

If for complex numbers $z_1$ and $z_2$,$\arg(z_1/z_2) = 0$,then $|z_1 - z_2|$ is equal to

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

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:

Convert the given complex number in polar form: $-1+i$

Vedclass Test Series

Exam Paper Generator

Online Exam Module