The imaginary part of (sqrt{3}-i)^{2016}+(-sq | vedclass

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

Question

(C) Let $z_1 = \sqrt{3}-i = 2(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6}))$. Then $z_1^{2016} = 2^{2016}(\cos(-\frac{2016\pi}{6}) + i\sin(-\frac{2016

Vedclass · Accepted Answer

$-2^{2019}$

Vedclass · Answer

(C) Let $z_1 = \sqrt{3}-i = 2(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6}))$. Then $z_1^{2016} = 2^{2016}(\cos(-\frac{2016\pi}{6}) + i\sin(-\frac{2016\pi}{6})) = 2^{2016}(\cos(-336\pi) + i\sin(-336\pi)) = 2^{2016}(1 + 0i) = 2^{2016}$.Let $z_2 = -\sqrt{3}-i = 2(\cos(-\frac{5\pi}{6}) + i\sin(-\frac{5\pi}{6}))$. Then $z_2^{2019} = 2^{2019}(\cos(-\frac{2019 	imes 5\pi}{6}) + i\sin(-\frac{2019 	imes 5\pi}{6})) = 2^{2019}(\cos(-\frac{3365\pi}{2}) + i\sin(-\frac{3365\pi}{2}))$.Since $-\frac{3365\pi}{2} = -1682\pi - \frac{\pi}{2}$,we have $\cos(-\frac{3365\pi}{2}) = \cos(-\frac{\pi}{2}) = 0$ and $\sin(-\frac{3365\pi}{2}) = \sin(-\frac{\pi}{2}) = -1$.Thus,$z_2^{2019} = 2^{2019}(0 - i) = -i 2^{2019}$.The sum is $2^{2016} - i 2^{2019}$.The imaginary part is $-2^{2019}$.

The imaginary part of $(\sqrt{3}-i)^{2016}+(-\sqrt{3}-i)^{2019}$ is

Similar Questions

If $\cos \alpha + \cos \beta + \cos \gamma = 0$ and $\sin \alpha + \sin \beta + \sin \gamma = 0$,then $\cos 3\alpha + \cos 3\beta + \cos 3\gamma$ equals to

If $\alpha, \beta$ are the roots of the equation $x^2+x+1=0$,then $(\alpha+\beta)^2+(\alpha^2+\beta^2)^2+(\alpha^3+\beta^3)^2+\ldots+(\alpha^{12}+\beta^{12})^2=$

If $z = \sin \theta - i \cos \theta,$ then for any integer $n$

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module