If {left( {frac{{1 + cos theta + isin theta } | vedclass

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(D) Let $z = \frac{1 + \cos 	heta + i\sin 	heta}{i + \sin 	heta + i\cos 	heta}$.Simplifying the numerator: $1 + \cos 	heta + i\sin 	heta = 2\cos

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$4$

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(D) Let $z = \frac{1 + \cos 	heta + i\sin 	heta}{i + \sin 	heta + i\cos 	heta}$.Simplifying the numerator: $1 + \cos 	heta + i\sin 	heta = 2\cos^2(	heta/2) + 2i\sin(	heta/2)\cos(	heta/2) = 2\cos(	heta/2) [\cos(	heta/2) + i\sin(	heta/2)]$.Simplifying the denominator: $i + \sin 	heta + i\cos 	heta = i(1 + \cos 	heta) + \sin 	heta = i[2\cos^2(	heta/2)] + 2\sin(	heta/2)\cos(	heta/2) = 2\cos(	heta/2) [\sin(	heta/2) + i\cos(	heta/2)]$.Since $\sin(	heta/2) + i\cos(	heta/2) = i(\cos(	heta/2) - i\sin(	heta/2)) = i(\cos(	heta/2) + i\sin(	heta/2))^{-1}$,we have:$z = \frac{2\cos(	heta/2) [\cos(	heta/2) + i\sin(	heta/2)]}{2\cos(	heta/2) i [\cos(	heta/2) - i\sin(	heta/2)]} = \frac{1}{i} \frac{\cos(	heta/2) + i\sin(	heta/2)}{\cos(	heta/2) - i\sin(	heta/2)} = -i [\cos(	heta/2) + i\sin(	heta/2)]^2 = -i(\cos 	heta + i\sin 	heta)$.Thus,$z^4 = (-i)^4 (\cos 	heta + i\sin 	heta)^4 = 1 \cdot (\cos 4	heta + i\sin 4	heta) = \cos 4	heta + i\sin 4	heta$.Comparing this with $\cos n	heta + i\sin n	heta$,we get $n = 4$.

If ${\left( {\frac{{1 + \cos \theta + i\sin \theta }}{{i + \sin \theta + i\cos \theta }}} \right)^4} = \cos n\theta + i\sin n\theta $,then $n$ is equal to

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