(sqrt{3}+i)^{10}+(sqrt{3}-i)^{10}= | vedclass

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

Question

(B) Let $z = \sqrt{3} + i$. Converting to polar form,$r = \sqrt{(\sqrt{3})^2 + 1^2} = 2$. The argument $	heta = 	an^{-1}(\frac{1}{\sqrt{3}}) = \frac

Vedclass · Accepted Answer

$1024$

Vedclass · Answer

(B) Let $z = \sqrt{3} + i$. Converting to polar form,$r = \sqrt{(\sqrt{3})^2 + 1^2} = 2$. The argument $	heta = 	an^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$.So,$z = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$.By De Moivre's Theorem,$z^{10} = 2^{10}(\cos \frac{10\pi}{6} + i \sin \frac{10\pi}{6}) = 1024(\cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3})$.Since $\cos \frac{5\pi}{3} = \frac{1}{2}$ and $\sin \frac{5\pi}{3} = -\frac{\sqrt{3}}{2}$,we have $z^{10} = 1024(\frac{1}{2} - i \frac{\sqrt{3}}{2}) = 512 - 512i\sqrt{3}$.Similarly,for $\bar{z} = \sqrt{3} - i$,$\bar{z}^{10} = 1024(\cos(-\frac{5\pi}{3}) + i \sin(-\frac{5\pi}{3})) = 1024(\frac{1}{2} + i \frac{\sqrt{3}}{2}) = 512 + 512i\sqrt{3}$.Adding these,$(\sqrt{3}+i)^{10} + (\sqrt{3}-i)^{10} = (512 - 512i\sqrt{3}) + (512 + 512i\sqrt{3}) = 1024$.

$(\sqrt{3}+i)^{10}+(\sqrt{3}-i)^{10}=$

Similar Questions

The product of the distinct $(2n)^{\text{th}}$ roots of $1+i\sqrt{3}$ is equal to:

If $z = \frac{1 + i\sqrt{3}}{\sqrt{3} + i}$,then $(\bar{z})^{100}$ lies in

$\alpha, \beta$ are the roots of the equation $x^2+2x+4=0$. If the point representing $\alpha$ in the Argand diagram lies in the $2^{nd}$ quadrant and $\alpha^{2024}-\beta^{2024}=ik, (i=\sqrt{-1})$,then $k=$

If $\omega$ is a complex cube root of unity,then $(1 + \omega - \omega^2)(1 - \omega + \omega^2) = $

If $\alpha_1, \alpha_2, \ldots, \alpha_{23}$ are the $23^{rd}$ roots of unity,then $\alpha_1^{47} + \alpha_2^{47} + \ldots + \alpha_{23}^{47} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module