For n in N,left(frac{1+cos theta+i sin theta} | vedclass

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(C) Given $n \in N$,let $z = \left(\frac{1+\cos 	heta+i \sin 	heta}{1+\cos 	heta-i \sin 	heta}ight)^n$.Using the identities $1+\cos 	heta = 2 \

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given $n \in N$,let $z = \left(\frac{1+\cos 	heta+i \sin 	heta}{1+\cos 	heta-i \sin 	heta}ight)^n$.Using the identities $1+\cos 	heta = 2 \cos^2 \frac{	heta}{2}$ and $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$,we get:$z = \left(\frac{2 \cos^2 \frac{	heta}{2} + 2i \sin \frac{	heta}{2} \cos \frac{	heta}{2}}{2 \cos^2 \frac{	heta}{2} - 2i \sin \frac{	heta}{2} \cos \frac{	heta}{2}}ight)^n$$z = \left(\frac{2 \cos \frac{	heta}{2} (\cos \frac{	heta}{2} + i \sin \frac{	heta}{2})}{2 \cos \frac{	heta}{2} (\cos \frac{	heta}{2} - i \sin \frac{	heta}{2})}ight)^n$$z = \left(\frac{\cos \frac{	heta}{2} + i \sin \frac{	heta}{2}}{\cos \frac{	heta}{2} - i \sin \frac{	heta}{2}}ight)^n$Using the exponential form $e^{i\phi} = \cos \phi + i \sin \phi$,we have:$z = \left(\frac{e^{i	heta/2}}{e^{-i	heta/2}}ight)^n = (e^{i	heta})^n = e^{in	heta}$$z = \cos (n 	heta) + i \sin (n 	heta)$.

List-$I$	List-$II$
$P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j = 1$	$1.$ True
$Q.$ There exists a $k \in \{1, 2, \ldots, 9\}$ such that $z_1 \cdot z = z_k$ has no solution $z$ in the set of complex numbers.	$2.$ False
$R.$ $\frac{\|1-z_1\|\|1-z_2\| \ldots \|1-z_9\|}{10}$ equals	$3.$ $1$
$S.$ $1 - \sum_{k=1}^9 \cos \left(\frac{2k\pi}{10}\right)$ equals	$4.$ $2$

For $n \in N$,$\left(\frac{1+\cos \theta+i \sin \theta}{1+\cos \theta-i \sin \theta}\right)^n=$

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