If |x+iy|=sqrt{x^2+y^2},then |(1-sqrt{3}i)^9+ | vedclass

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

Question

(D) Let $z_1 = 1-\sqrt{3}i$ and $z_2 = \sqrt{3}+i$. Converting to polar form: $z_1 = 2(\cos(-\frac{\pi}{3}) + i\sin(-\frac{\pi}{3})) = 2e^{-i\pi/3}$.

Vedclass · Accepted Answer

$2^{\frac{19}{2}}$

Vedclass · Answer

(D) Let $z_1 = 1-\sqrt{3}i$ and $z_2 = \sqrt{3}+i$. Converting to polar form: $z_1 = 2(\cos(-\frac{\pi}{3}) + i\sin(-\frac{\pi}{3})) = 2e^{-i\pi/3}$. $z_2 = 2(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6})) = 2e^{i\pi/6}$. Now,$z_1^9 = 2^9 e^{-i3\pi} = 2^9(\cos(-3\pi) + i\sin(-3\pi)) = 2^9(-1) = -2^9$. $z_2^9 = 2^9 e^{i3\pi/2} = 2^9(\cos(\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2})) = 2^9(0 - i) = -i2^9$. Then,$|z_1^9 + z_2^9| = |-2^9 - i2^9| = |2^9(-1-i)| = 2^9|-1-i|$. $|z_1^9 + z_2^9| = 2^9 \sqrt{(-1)^2 + (-1)^2} = 2^9 \sqrt{2} = 2^9 \cdot 2^{1/2} = 2^{19/2}$. Thus,the correct option is $D$.

If $|x+iy|=\sqrt{x^2+y^2}$,then $|(1-\sqrt{3}i)^9+(\sqrt{3}+i)^9|=$

Similar Questions

The roots of the equation $(x-1)^5=32(x+1)^5$ are

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

$\sum_{k=1}^6 \left[ \sin \frac{2 k \pi}{7} - i \cos \frac{2 k \pi}{7} \right]$ is equal to

If $\omega$ is a complex cube root of unity,then $\cos \left[\left(\omega^{1234}+\omega^{2021}\right) \pi-\frac{\pi}{4}\right]$ is equal to

What is the product of all the values of $(1-i)^{\frac{2}{5}}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module