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

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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)$.

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

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