Which of the following is a fourth root of fr | vedclass

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

Question

(B) The given complex number is $z = \frac{1}{2} + i\frac{\sqrt{3}}{2}$.In polar form,this is $z = \cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\

Vedclass · Accepted Answer

$cis\left(\frac{\pi}{12}\right)$

Vedclass · Answer

(B) The given complex number is $z = \frac{1}{2} + i\frac{\sqrt{3}}{2}$.In polar form,this is $z = \cos\left(\frac{\pi}{3}ight) + i\sin\left(\frac{\pi}{3}ight) = cis\left(\frac{\pi}{3}ight)$.By De Moivre's Theorem,the $n$-th roots of $cis(	heta)$ are given by $cis\left(\frac{	heta + 2k\pi}{n}ight)$ for $k = 0, 1, 2, 3$.For the fourth root $(n=4)$ and $k=0$,we get $cis\left(\frac{\pi/3}{4}ight) = cis\left(\frac{\pi}{12}ight)$.Thus,$cis\left(\frac{\pi}{12}ight)$ is a fourth root of the given number.

Which of the following is a fourth root of $\frac{1}{2} + \frac{i\sqrt{3}}{2}$?

Similar Questions

$(-1 + i\sqrt{3})^{20}$ is equal to

Find the value of $(1 + 2\omega + \omega^2)^{3n} - (1 + \omega + 2\omega^2)^{3n}$.

If $\omega$ is a complex cube root of unity,then the value of $\omega^{99} + \omega^{100} + \omega^{101}$ is

$n$ is a positive integer and not a multiple of $3$. If $\omega$ is a non-real cube root of unity,then $\omega^n + \omega^{2n}$ is equal to

If $\omega$ is a complex cube root of unity, then the value of the expression $2(1 + \frac{1}{\omega})(1 + \frac{1}{\omega^2}) + 3(2 + \frac{1}{\omega})(2 + \frac{1}{\omega^2}) + ... + (n + 1)(n + \frac{1}{\omega})(n + \frac{1}{\omega^2})$ is...

Vedclass Test Series

Exam Paper Generator

Online Exam Module