If z = 1 + cos theta - i sin theta and 0

Question

(D) Given $z = 1 + \cos 	heta - i \sin 	heta$. $z - 1 = \cos 	heta - i \sin 	heta$. $|z - 1| = \sqrt{\cos^2 	heta + \sin^2 	heta} = 1$. $|z|^2 =

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$\sin \left(\frac{	heta}{2}ight)$

Vedclass · Answer

(D) Given $z = 1 + \cos 	heta - i \sin 	heta$. $z - 1 = \cos 	heta - i \sin 	heta$. $|z - 1| = \sqrt{\cos^2 	heta + \sin^2 	heta} = 1$. $|z|^2 = (1 + \cos 	heta)^2 + (-\sin 	heta)^2 = 1 + 2 \cos 	heta + \cos^2 	heta + \sin^2 	heta = 2 + 2 \cos 	heta$. Now,substitute these into the expression: $\left[|z - 1|^2 - \frac{|z|^2}{4}ight]^{1/2} = \left[1^2 - \frac{2 + 2 \cos 	heta}{4}ight]^{1/2} = \left[1 - \frac{1 + \cos 	heta}{2}ight]^{1/2} = \left[\frac{2 - 1 - \cos 	heta}{2}ight]^{1/2} = \left[\frac{1 - \cos 	heta}{2}ight]^{1/2}$. Using the identity $1 - \cos 	heta = 2 \sin^2 \left(\frac{	heta}{2}ight)$,we get: $\left[\frac{2 \sin^2 \left(\frac{	heta}{2}ight)}{2}ight]^{1/2} = \sqrt{\sin^2 \left(\frac{	heta}{2}ight)} = \sin \left(\frac{	heta}{2}ight)$ (since $0  0$).

If $z = 1 + \cos \theta - i \sin \theta$ and $0 < \theta < \pi$,then $\left[|z - 1|^2 - \frac{|z|^2}{4}\right]^{1/2} =$

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