If z = frac{1}{1 - cos theta + i sin theta} a | vedclass

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(A) Given $z = \frac{1}{1 - \cos 	heta + i \sin 	heta}$.Using trigonometric identities $1 - \cos 	heta = 2 \sin^2 \frac{	heta}{2}$ and $\sin 	het

Vedclass · Accepted Answer

$(\frac{1}{2} \operatorname{cosec} \frac{	heta}{2}, -(\frac{\pi}{2} - \frac{	heta}{2}))$

Vedclass · Answer

(A) Given $z = \frac{1}{1 - \cos 	heta + i \sin 	heta}$.Using trigonometric identities $1 - \cos 	heta = 2 \sin^2 \frac{	heta}{2}$ and $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$,we get:$z = \frac{1}{2 \sin^2 \frac{	heta}{2} + i (2 \sin \frac{	heta}{2} \cos \frac{	heta}{2})}$$z = \frac{1}{2 \sin \frac{	heta}{2} (\sin \frac{	heta}{2} + i \cos \frac{	heta}{2})}$Since $\sin \frac{	heta}{2} + i \cos \frac{	heta}{2} = i (\cos \frac{	heta}{2} - i \sin \frac{	heta}{2}) = i e^{-i 	heta/2}$,we have:$z = \frac{1}{2 \sin \frac{	heta}{2} \cdot i e^{-i 	heta/2}} = \frac{1}{2 \sin \frac{	heta}{2}} \cdot \frac{1}{i} e^{i 	heta/2}$Since $\frac{1}{i} = -i = e^{-i \pi/2}$,we get:$z = \frac{1}{2 \sin \frac{	heta}{2}} e^{-i \pi/2} e^{i 	heta/2} = \frac{1}{2} \operatorname{cosec} \frac{	heta}{2} e^{i (	heta/2 - \pi/2)}$Thus,the modulus is $\frac{1}{2} \operatorname{cosec} \frac{	heta}{2}$ and the amplitude is $-(\frac{\pi}{2} - \frac{	heta}{2})$.

If $z = \frac{1}{1 - \cos \theta + i \sin \theta}$ and $\theta$ is acute,then the modulus and amplitude of $z$ respectively are:

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