The least positive integral value of n such t | vedclass

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

Question

(C) Let $z = \frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}$.Using $\sin 	heta = \cos(\frac{\pi}

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(C) Let $z = \frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}$.Using $\sin 	heta = \cos(\frac{\pi}{2} - 	heta)$ and $\cos 	heta = \sin(\frac{\pi}{2} - 	heta)$,we have $\frac{\pi}{2} - \frac{2 \pi}{9} = \frac{5 \pi}{18}$.So,$z = \frac{1+\cos \frac{5 \pi}{18}+i \sin \frac{5 \pi}{18}}{1+\cos \frac{5 \pi}{18}-i \sin \frac{5 \pi}{18}}$.Using half-angle formulas $1+\cos 	heta = 2 \cos^2 \frac{	heta}{2}$ and $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$,we get:$z = \frac{2 \cos^2 \frac{5 \pi}{36} + i 2 \sin \frac{5 \pi}{36} \cos \frac{5 \pi}{36}}{2 \cos^2 \frac{5 \pi}{36} - i 2 \sin \frac{5 \pi}{36} \cos \frac{5 \pi}{36}} = \frac{\cos \frac{5 \pi}{36} + i \sin \frac{5 \pi}{36}}{\cos \frac{5 \pi}{36} - i \sin \frac{5 \pi}{36}} = \frac{e^{i 5 \pi / 36}}{e^{-i 5 \pi / 36}} = e^{i 5 \pi / 18}$.Given $z^n = 1$,we have $(e^{i 5 \pi / 18})^n = e^{i 5 n \pi / 18} = 1$.This implies $\frac{5 n \pi}{18} = 2 k \pi$ for some integer $k$.$n = \frac{36 k}{5}$.For $n$ to be the least positive integer,we set $k=5$,which gives $n = 36$.

The least positive integral value of $n$ such that $\left[\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right]^n=1$ is

Similar Questions

The sum of the least positive arguments of the distinct cube roots of the complex number $(1-i \sqrt{3})$ is

If $\alpha, \beta$ are the roots of the equation $x^2-2x+4=0$ and for any $n \in N, \alpha^n+\beta^n=k \cos \frac{n \pi}{3}$, then $k=$

Let ${z_1}$ and ${z_2}$ be $n^{th}$ roots of unity which are ends of a line segment that subtend a right angle at the origin. Then $n$ must be of the form

The product of the four values of the complex number $(1+i)^{3/4}$ is

If $\alpha, \beta$ are the roots of the equation $x^2 - 2x + 4 = 0$,then the value of $\alpha^n + \beta^n$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module