The value of left(frac{-1+i sqrt{3}}{1-i}righ | vedclass

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

Question

(C) Let $z = \frac{-1+i \sqrt{3}}{1-i}$.First,express the numerator in polar form: $-1+i \sqrt{3} = 2 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{

Vedclass · Accepted Answer

$-2^{15} i$

Vedclass · Answer

(C) Let $z = \frac{-1+i \sqrt{3}}{1-i}$.First,express the numerator in polar form: $-1+i \sqrt{3} = 2 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right) = 2e^{i 2\pi/3}$.Next,express the denominator: $1-i = \sqrt{2} \left( \cos \frac{-\pi}{4} + i \sin \frac{-\pi}{4} \right) = \sqrt{2}e^{-i \pi/4}$.Then,$z = \frac{2e^{i 2\pi/3}}{\sqrt{2}e^{-i \pi/4}} = \sqrt{2} e^{i (2\pi/3 + \pi/4)} = \sqrt{2} e^{i 11\pi/12}$.Now,$z^{30} = (\sqrt{2})^{30} e^{i (11\pi/12) \cdot 30} = 2^{15} e^{i 55\pi/2}$.Since $e^{i 55\pi/2} = e^{i (26\pi + 3\pi/2)} = e^{i 3\pi/2} = -i$.Therefore,$z^{30} = 2^{15} \cdot (-i) = -2^{15} i$.

The value of $\left(\frac{-1+i \sqrt{3}}{1-i}\right)^{30}$ is

Similar Questions

Let $S = \{z \in \mathbb{C} : z^2 + 4z + 16 = 0\}$. Then $\sum_{z \in S} |z + \sqrt{3}i|^2$ is equal to:

If $z = x + iy$ and $x^2 + y^2 = 1$,then $\frac{1 + x + iy}{1 + x - iy} = $

If $\sqrt{-3-4 i}=re^{i \theta}$,then $r^2 \tan \theta=$

If $\frac{3-2 i \sin \theta}{1+2 i \sin \theta}$ is a purely imaginary number,then $\theta=$

The biquadratic equation,two of whose roots are $1+i$ and $1-\sqrt{2}$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module