left(frac{sqrt{6}-sqrt{2}}{4}+frac{sqrt{6}+sq | vedclass

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

Question

(D) Given expression: $z = \left(\frac{\sqrt{6}-\sqrt{2}}{4}+i \frac{\sqrt{6}+\sqrt{2}}{4}\right)^{2020}$ We know that $\cos \frac{5 \pi}{12} = \frac{

Vedclass · Accepted Answer

$\frac{1}{2}-\frac{\sqrt{3}}{2} i$

Vedclass · Answer

(D) Given expression: $z = \left(\frac{\sqrt{6}-\sqrt{2}}{4}+i \frac{\sqrt{6}+\sqrt{2}}{4}ight)^{2020}$ We know that $\cos \frac{5 \pi}{12} = \frac{\sqrt{6}-\sqrt{2}}{4}$ and $\sin \frac{5 \pi}{12} = \frac{\sqrt{6}+\sqrt{2}}{4}$. So,$z = \left(\cos \frac{5 \pi}{12} + i \sin \frac{5 \pi}{12}ight)^{2020}$. Using De Moivre's theorem,$(\cos 	heta + i \sin 	heta)^n = \cos(n 	heta) + i \sin(n 	heta)$: $z = \cos \left(2020 	imes \frac{5 \pi}{12}ight) + i \sin \left(2020 	imes \frac{5 \pi}{12}ight)$ $z = \cos \left(\frac{505 	imes 5 \pi}{3}ight) + i \sin \left(\frac{505 	imes 5 \pi}{3}ight)$ $z = \cos \left(\frac{2525 \pi}{3}ight) + i \sin \left(\frac{2525 \pi}{3}ight)$ Since $\frac{2525 \pi}{3} = 841 \pi + \frac{2 \pi}{3} = 842 \pi - \frac{\pi}{3}$,$z = \cos \left(-\frac{\pi}{3}ight) + i \sin \left(-\frac{\pi}{3}ight)$ $z = \cos \frac{\pi}{3} - i \sin \frac{\pi}{3} = \frac{1}{2} - \frac{\sqrt{3}}{2} i$.

$\left(\frac{\sqrt{6}-\sqrt{2}}{4}+\frac{\sqrt{6}+\sqrt{2}}{4} i\right)^{2020} =$

Similar Questions

$(\sqrt{3}+i)^7+(\sqrt{3}-i)^7$ is equal to (in $\sqrt{3}$)

If $1, \alpha_1, \alpha_2, \alpha_3, \alpha_4$ are the roots of $z^5-1=0$ and $\omega$ is a cube root of unity,then $(\omega-1)(\omega-\alpha_1)(\omega-\alpha_2)(\omega-\alpha_3)(\omega-\alpha_4)+\omega$ is equal to

If $\alpha$ is an imaginary cube root of unity,then for $n \in N$,the value of $\alpha^{3n + 1} + \alpha^{3n + 3} + \alpha^{3n + 5}$ is

If $z$ is a complex number such that $z^2+z+1=0$,then $\left(z+\frac{1}{z}\right)^3+\left(z^2+\frac{1}{z^2}\right)^3+\left(z^3+\frac{1}{z^3}\right)^3+\ldots+\left(z^{2020}+\frac{1}{z^{2020}}\right)^3=$

If $x = a + b$,$y = a\omega + b\omega^2$,and $z = a\omega^2 + b\omega$,then the value of $x^3 + y^3 + z^3$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module