left(frac{cos theta+i sin theta}{sin theta+i | vedclass

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

Question

(B) Let $z_1 = \frac{\cos 	heta+i \sin 	heta}{\sin 	heta+i \cos 	heta} = \frac{\cos 	heta+i \sin 	heta}{i(\cos 	heta-i \sin 	heta)} = \frac{1}

Vedclass · Accepted Answer

$2 \cos 16 	heta$

Vedclass · Answer

(B) Let $z_1 = \frac{\cos 	heta+i \sin 	heta}{\sin 	heta+i \cos 	heta} = \frac{\cos 	heta+i \sin 	heta}{i(\cos 	heta-i \sin 	heta)} = \frac{1}{i} \cdot \frac{e^{i 	heta}}{e^{-i 	heta}} = -i e^{i 2 	heta}$.Then $z_1^8 = (-i)^8 (e^{i 2 	heta})^8 = 1 \cdot e^{i 16 	heta} = \cos 16 	heta + i \sin 16 	heta$.Let $z_2 = \frac{1+\cos 	heta-i \sin 	heta}{1+\cos 	heta+i \sin 	heta} = \frac{2 \cos^2 \frac{	heta}{2} - i 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}}{2 \cos^2 \frac{	heta}{2} + i 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}} = \frac{\cos \frac{	heta}{2} - i \sin \frac{	heta}{2}}{\cos \frac{	heta}{2} + i \sin \frac{	heta}{2}} = \frac{e^{-i 	heta/2}}{e^{i 	heta/2}} = e^{-i 	heta}$.Then $z_2^{16} = (e^{-i 	heta})^{16} = e^{-i 16 	heta} = \cos 16 	heta - i \sin 16 	heta$.Adding the two terms: $z_1^8 + z_2^{16} = (\cos 16 	heta + i \sin 16 	heta) + (\cos 16 	heta - i \sin 16 	heta) = 2 \cos 16 	heta$.

$\left(\frac{\cos \theta+i \sin \theta}{\sin \theta+i \cos \theta}\right)^8+\left(\frac{1+\cos \theta-i \sin \theta}{1+\cos \theta+i \sin \theta}\right)^{16}=$

Similar Questions

If $1+x^2=\sqrt{3} x$,then $\sum_{n=1}^{24}\left(x^n-\frac{1}{x^n}\right)^2$ is equal to

If $Z_1, Z_2, Z_3$ are three complex numbers with unit modulus such that $|Z_1-Z_2|^2+|Z_1-Z_3|^2=4$,then $Z_1 \overline{Z_2}+\overline{Z_1} Z_2+Z_1 \overline{Z_3}+\overline{Z_1} Z_3=$

If $\left|z-\frac{2}{z}\right|=2$,then the greatest value of $|z|$ is

$(r, \theta)$ denotes $r(\cos \theta + i \sin \theta)$. If $x = (1, \alpha)$,$y = (1, \beta)$,$z = (1, \gamma)$ and $x + y + z = 0$,then $\sum \cos (2\alpha - \beta - \gamma) = $

If $a \pm ib$ and $b \pm ai$ are the roots of $x^4-10x^3+50x^2-130x+169=0$,then $\frac{a}{b}+\frac{b}{a}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module