left(frac{1+cos (3 theta)+i sin (3 theta)}{1+ | vedclass

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

Question

(A) Let $z = \frac{1+\cos (3 	heta)+i \sin (3 	heta)}{1+\cos (3 	heta)-i \sin (3 	heta)}$.Using the identities $1+\cos (2A) = 2\cos^2 A$ and $\sin

Vedclass · Accepted Answer

$\cos (60 	heta)+i \sin (60 	heta)$

Vedclass · Answer

(A) Let $z = \frac{1+\cos (3 	heta)+i \sin (3 	heta)}{1+\cos (3 	heta)-i \sin (3 	heta)}$.Using the identities $1+\cos (2A) = 2\cos^2 A$ and $\sin (2A) = 2\sin A \cos A$,we have:$z = \frac{2\cos^2(\frac{3	heta}{2}) + i 2\sin(\frac{3	heta}{2})\cos(\frac{3	heta}{2})}{2\cos^2(\frac{3	heta}{2}) - i 2\sin(\frac{3	heta}{2})\cos(\frac{3	heta}{2})}$$z = \frac{2\cos(\frac{3	heta}{2}) [\cos(\frac{3	heta}{2}) + i\sin(\frac{3	heta}{2})]}{2\cos(\frac{3	heta}{2}) [\cos(\frac{3	heta}{2}) - i\sin(\frac{3	heta}{2})]}$$z = \frac{\cos(\frac{3	heta}{2}) + i\sin(\frac{3	heta}{2})}{\cos(\frac{3	heta}{2}) - i\sin(\frac{3	heta}{2})} = \frac{e^{i(3	heta/2)}}{e^{-i(3	heta/2)}} = e^{i(3	heta/2 + 3	heta/2)} = e^{i(3	heta)}$.Therefore,$z^{20} = (e^{i(3	heta)})^{20} = e^{i(60	heta)} = \cos(60	heta) + i\sin(60	heta)$.

$\left(\frac{1+\cos (3 \theta)+i \sin (3 \theta)}{1+\cos (3 \theta)-i \sin (3 \theta)}\right)^{20} = ?$

Similar Questions

If $\theta = \frac{\pi}{6}$,then the $10^{th}$ term of the series $1 + (\cos \theta + i \sin \theta) + (\cos \theta + i \sin \theta)^2 + (\cos \theta + i \sin \theta)^3 + \ldots$ is equal to:

The product $(1 - \omega + {\omega ^2})(1 - {\omega ^2} + {\omega ^4})(1 - {\omega ^4} + {\omega ^8}) \dots$ up to $2n$ factors is:

If $\omega$ is a complex cube root of unity,then $(1 - \omega )(1 - {\omega ^2})(1 - {\omega ^4})(1 - {\omega ^8}) = $

If the cube roots of unity are $1, \omega, \omega^2$,then the roots of the equation $(x - 1)^3 + 8 = 0$ are

If the roots of the equation $(z-4)^3=8 i$ are $a-2 i, b+i$,and $c+i$,then $\sqrt{a b c}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module