(-i+sqrt{3})^{300}+(-i-sqrt{3})^{300}= | vedclass

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

Question

(B) Let $z_1 = -i+\sqrt{3}$ and $z_2 = -i-\sqrt{3}$.We can write $z_1 = -i(1+i\sqrt{3})$ and $z_2 = i(1-i\sqrt{3})$.Alternatively,note that $z_1 = -i(

Vedclass · Accepted Answer

$2^{301}$

Vedclass · Answer

(B) Let $z_1 = -i+\sqrt{3}$ and $z_2 = -i-\sqrt{3}$.We can write $z_1 = -i(1+i\sqrt{3})$ and $z_2 = i(1-i\sqrt{3})$.Alternatively,note that $z_1 = -i(1+i\sqrt{3}) = -2i(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = -2i e^{i\pi/3}$.$z_1^{300} = (-2i)^{300} (e^{i\pi/3})^{300} = 2^{300} (i)^{300} e^{i100\pi} = 2^{300} (1) (1) = 2^{300}$.Similarly,$z_2 = -i(1-i\sqrt{3}) = -2i(\frac{1}{2} - i\frac{\sqrt{3}}{2}) = -2i e^{-i\pi/3}$.$z_2^{300} = (-2i)^{300} (e^{-i\pi/3})^{300} = 2^{300} (i)^{300} e^{-i100\pi} = 2^{300} (1) (1) = 2^{300}$.Thus,$z_1^{300} + z_2^{300} = 2^{300} + 2^{300} = 2 	imes 2^{300} = 2^{301}$.Hence,option $(B)$ is correct.

$(-i+\sqrt{3})^{300}+(-i-\sqrt{3})^{300}=$

Similar Questions

The value of $\sum_{k = 1}^{10} \left( \sin \frac{2k\pi}{11} + i\cos \frac{2k\pi}{11} \right)$ is

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

One of the complex roots of the equation $x^{11}-x^6-x^5+1=0$ is

$(\sin \theta + i\cos \theta )^n$ is equal to

If $\omega$ is an imaginary cube root of unity,then the value of $\sin \left[ (\omega^{10} + \omega^{23})\pi - \frac{\pi}{4} \right]$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module