{left[ {frac{{1 + cos (pi /8) + isin (pi /8)} | vedclass

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

Question

(A) Let $z = \frac{1 + \cos(\pi/8) + i\sin(\pi/8)}{1 + \cos(\pi/8) - i\sin(\pi/8)}$.Using the identities $1 + \cos 	heta = 2\cos^2(	heta/2)$ and $\s

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(A) Let $z = \frac{1 + \cos(\pi/8) + i\sin(\pi/8)}{1 + \cos(\pi/8) - i\sin(\pi/8)}$.Using the identities $1 + \cos 	heta = 2\cos^2(	heta/2)$ and $\sin 	heta = 2\sin(	heta/2)\cos(	heta/2)$:$z = \frac{2\cos^2(\pi/16) + 2i\sin(\pi/16)\cos(\pi/16)}{2\cos^2(\pi/16) - 2i\sin(\pi/16)\cos(\pi/16)}$$z = \frac{2\cos(\pi/16) [\cos(\pi/16) + i\sin(\pi/16)]}{2\cos(\pi/16) [\cos(\pi/16) - i\sin(\pi/16)]}$$z = \frac{\cos(\pi/16) + i\sin(\pi/16)}{\cos(\pi/16) - i\sin(\pi/16)}$Using Euler's formula $e^{i	heta} = \cos 	heta + i\sin 	heta$,we have $z = \frac{e^{i\pi/16}}{e^{-i\pi/16}} = e^{i\pi/8}$.Then $z^8 = (e^{i\pi/8})^8 = e^{i\pi} = \cos \pi + i\sin \pi = -1$.

${\left[ {\frac{{1 + \cos (\pi /8) + i\sin (\pi /8)}}{{1 + \cos (\pi /8) - i\sin (\pi /8)}}} \right]^8}$ is equal to

Similar Questions

If $\omega$ is the complex cube root of unity and $\left(\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}\right)^k+\left(\frac{a+b \omega+c \omega^2}{b+a \omega^2+c \omega}\right)^l=2$,then $2k+l$ is always

If $z = \frac{\sqrt{3} + i}{-2}$,then $z^{69}$ is equal to

If $\alpha$ is an imaginary cube root of unity,then for $n \in N$,the value of $\alpha^{3n + 1} + \alpha^{3n + 3} + \alpha^{3n + 5}$ is

If the cube roots of unity are $1, \omega, \omega^2$,then the roots of the equation $(x - 2)^3 + 27 = 0$ are

$\left[\frac{1+\cos \left(\frac{\pi}{12}\right)+i \sin \left(\frac{\pi}{12}\right)}{1+\cos \left(\frac{\pi}{12}\right)-i \sin \left(\frac{\pi}{12}\right)}\right]^{72}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module