(-1 + isqrt{3})^{20} is equal to | vedclass

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

Question

(D) Let $z = -1 + i\sqrt{3}$. The modulus $r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$. We can write $z = 2 \left( -\frac{1}{2} + i\frac{\sqr

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Let $z = -1 + i\sqrt{3}$. The modulus $r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$. We can write $z = 2 \left( -\frac{1}{2} + i\frac{\sqrt{3}}{2} \right) = 2 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right)$. Using De Moivre's Theorem,$z^{20} = 2^{20} \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right)^{20} = 2^{20} \left( \cos \frac{40\pi}{3} + i \sin \frac{40\pi}{3} \right)$. Since $\frac{40\pi}{3} = 13\pi + \frac{\pi}{3} = 12\pi + \pi + \frac{\pi}{3} = 6(2\pi) + \frac{4\pi}{3}$,we have $\cos \frac{40\pi}{3} = \cos \frac{4\pi}{3} = -\frac{1}{2}$ and $\sin \frac{40\pi}{3} = \sin \frac{4\pi}{3} = -\frac{\sqrt{3}}{2}$. Thus,$z^{20} = 2^{20} \left( -\frac{1}{2} - i\frac{\sqrt{3}}{2} \right) = 2^{19}(-1 - i\sqrt{3})$. None of the given options match this result.

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

Similar Questions

Two values of $(-8-8 \sqrt{3} i)^{1/4}$ are

$(1+\sqrt{5}+i \sqrt{10-2 \sqrt{5}})^5=$

If ${z_1}, {z_2}, {z_3}, ......, {z_n}$ are the $n^{th}$ roots of unity,then for $k = 1, 2, ....., n-1$:

If $\omega$ is a complex cube root of unity,then $(1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) \dots$ to $2n$ factors is equal to:

If $x+\frac{1}{x}=2 \cos \theta$, then for any integer $n$, $x^n+\frac{1}{x^n}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module