The real part of frac{1}{1 - cos theta + i si | vedclass

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Question

(B) Given expression: $z = \frac{1}{(1 - \cos 	heta) + i \sin 	heta}$.Multiply the numerator and denominator by the conjugate $(1 - \cos 	heta) - i

Vedclass · Accepted Answer

$1/2$

Vedclass · Answer

(B) Given expression: $z = \frac{1}{(1 - \cos 	heta) + i \sin 	heta}$.Multiply the numerator and denominator by the conjugate $(1 - \cos 	heta) - i \sin 	heta$:$z = \frac{(1 - \cos 	heta) - i \sin 	heta}{(1 - \cos 	heta)^2 + \sin^2 	heta}$.Simplify the denominator:$(1 - \cos 	heta)^2 + \sin^2 	heta = 1 - 2 \cos 	heta + \cos^2 	heta + \sin^2 	heta = 1 - 2 \cos 	heta + 1 = 2 - 2 \cos 	heta = 2(1 - \cos 	heta)$.Thus,$z = \frac{1 - \cos 	heta}{2(1 - \cos 	heta)} - i \frac{\sin 	heta}{2(1 - \cos 	heta)}$.The real part is $\frac{1 - \cos 	heta}{2(1 - \cos 	heta)} = \frac{1}{2}$.

The real part of $\frac{1}{1 - \cos \theta + i \sin \theta}$ is equal to

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