The value of left(frac{1+sin frac{2 pi}{9}+i | vedclass

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

Question

(C) Let $z = \sin \frac{2 \pi}{9} + i \cos \frac{2 \pi}{9}$.Note that $|z|^2 = \sin^2 \frac{2 \pi}{9} + \cos^2 \frac{2 \pi}{9} = 1$,so $\bar{z} = \fra

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $z = \sin \frac{2 \pi}{9} + i \cos \frac{2 \pi}{9}$.Note that $|z|^2 = \sin^2 \frac{2 \pi}{9} + \cos^2 \frac{2 \pi}{9} = 1$,so $\bar{z} = \frac{1}{z}$.The expression becomes $\left(\frac{1+z}{1+\bar{z}}\right)^3 = \left(\frac{1+z}{1+\frac{1}{z}}\right)^3 = \left(\frac{1+z}{\frac{z+1}{z}}\right)^3 = z^3$.Now,$z = i(\cos \frac{2 \pi}{9} - i \sin \frac{2 \pi}{9}) = i e^{-i \frac{2 \pi}{9}}$.Then $z^3 = (i e^{-i \frac{2 \pi}{9}})^3 = i^3 e^{-i \frac{6 \pi}{9}} = -i e^{-i \frac{2 \pi}{3}}$.$z^3 = -i (\cos \frac{2 \pi}{3} - i \sin \frac{2 \pi}{3}) = -i (-\frac{1}{2} - i \frac{\sqrt{3}}{2}) = i \frac{1}{2} - \frac{\sqrt{3}}{2} = -\frac{1}{2}(\sqrt{3}-i)$.

The value of $\left(\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right)^3$ is

Similar Questions

If $z = x + iy$ satisfies $|z|-2=0$ and $|z-i|-|z+5i|=0$,then

If $\theta \in \mathbb{R}$ and $\frac{1-i \cos \theta}{1+2 i \cos \theta}$ is a real number,then $\theta$ will be (where $I$ is the set of integers):

If $z$ is a point on the circle $|z|=1$ with $\operatorname{Arg}(z)=\frac{\pi}{6}$,then $\frac{z^{12}+1-z^6}{z^{12}+i z^6-1}=$

If the set $\{\operatorname{Re}\left(\frac{z-\bar{z}+z \bar{z}}{2-3 z+5 \bar{z}}\right): z \in \mathbb{C}, \operatorname{Re}(z)=3\}$ is equal to the interval $(\alpha, \beta]$,then $24(\beta-\alpha)$ is equal to

If $(x-iy)^{1/3} = a-ib$,then the value of $\frac{x}{a} + \frac{y}{b}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module