If i is the root of the equation x^2+1=0,then | vedclass

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

Question

(B) Let $z_1 = 1+\sqrt{3}i = 2(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 2(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}) = 2e^{i\pi/3}$.Similarly,$z_2 = 1-\sqr

Vedclass · Accepted Answer

$2^{2023}$

Vedclass · Answer

(B) Let $z_1 = 1+\sqrt{3}i = 2(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 2(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}) = 2e^{i\pi/3}$.Similarly,$z_2 = 1-\sqrt{3}i = 2(\cos \frac{\pi}{3} - i\sin \frac{\pi}{3}) = 2e^{-i\pi/3}$.Then,$(1+\sqrt{3}i)^{2023} + (1-\sqrt{3}i)^{2023} = (2e^{i\pi/3})^{2023} + (2e^{-i\pi/3})^{2023}$.$= 2^{2023} (e^{i2023\pi/3} + e^{-i2023\pi/3})$.Since $2023 = 3 	imes 674 + 1$,we have $\frac{2023\pi}{3} = 674\pi + \frac{\pi}{3}$.Thus,$e^{i2023\pi/3} = e^{i(674\pi + \pi/3)} = e^{i\pi/3} = \cos \frac{\pi}{3} + i\sin \frac{\pi}{3} = \frac{1}{2} + i\frac{\sqrt{3}}{2}$.Similarly,$e^{-i2023\pi/3} = \frac{1}{2} - i\frac{\sqrt{3}}{2}$.Sum $= 2^{2023} (\frac{1}{2} + i\frac{\sqrt{3}}{2} + \frac{1}{2} - i\frac{\sqrt{3}}{2}) = 2^{2023}(1) = 2^{2023}$.

If $i$ is the root of the equation $x^2+1=0$,then $(1+\sqrt{3}i)^{2023}+(1-\sqrt{3}i)^{2023}=$

Similar Questions

If $z^2+z+1=0$,where $z$ is a complex number,then $\left(z+\frac{1}{z}\right)^3+\left(z^4+\frac{1}{z^4}\right)^3$ is equal to

$\frac{(\sin \frac{\pi}{8} + i \cos \frac{\pi}{8})^8}{(\sin \frac{\pi}{8} - i \cos \frac{\pi}{8})^8}$ is equal to

If $\omega$ is a complex cube root of unity,then $\sin \left\{\left(\omega^{10}+\omega^{23}\right) \pi-\frac{\pi}{4}\right\}=$

If $1, \omega, \omega^2$ are the cube roots of unity,then the roots of the equation $8z^3 - 12z^2 + 6z - 28 = 0$ are

Let $\omega (\neq 1)$ be a cubic root of unity. Then the minimum value of the set $\{|a+b\omega+c\omega^2|^2 : a, b, c \text{ are distinct non-zero integers}\}$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module