What is the value of (1-i sqrt{3})^9? | vedclass

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

Question

(B) Given expression: $(1-i \sqrt{3})^9$We can write this as: $2^9 \left(\frac{1-i \sqrt{3}}{2}\right)^9$$= 2^9 \left(\frac{1}{2} - i \frac{\sqrt{3}}{

Vedclass · Accepted Answer

$-2^9$

Vedclass · Answer

(B) Given expression: $(1-i \sqrt{3})^9$We can write this as: $2^9 \left(\frac{1-i \sqrt{3}}{2}ight)^9$$= 2^9 \left(\frac{1}{2} - i \frac{\sqrt{3}}{2}ight)^9$Using polar form: $\frac{1}{2} - i \frac{\sqrt{3}}{2} = \cos\left(-\frac{\pi}{3}ight) + i \sin\left(-\frac{\pi}{3}ight)$So,$2^9 \left[\cos\left(-\frac{\pi}{3}ight) + i \sin\left(-\frac{\pi}{3}ight)ight]^9$Applying De Moivre's theorem,$(\cos 	heta + i \sin 	heta)^n = \cos(n	heta) + i \sin(n	heta)$:$= 2^9 [\cos(-3\pi) + i \sin(-3\pi)]$$= 2^9 [\cos(3\pi) - i \sin(3\pi)]$Since $\cos(3\pi) = -1$ and $\sin(3\pi) = 0$:$= 2^9 [-1 - 0] = -2^9$

What is the value of $(1-i \sqrt{3})^9$?

Similar Questions

If $\alpha$ is a complex number such that $\alpha^{2}-\alpha+1=0$,then $\alpha^{2011}$ is equal to

If $w = \frac{-1 + i \sqrt{3}}{2}$,where $i = \sqrt{-1}$,then the value of $(3 + w + 3 w^2)^4$ is

The $n^{th}$ roots of unity are in

One of the values of $(\sqrt{3}-i)^{\frac{1}{6}}$ is

$\left(\frac{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}\right)^{12} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module