If |Z|=2, Z_1=frac{Z}{2} e^{i alpha} and thet | vedclass

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(C) Given $|Z|=2$ and $\operatorname{amp}(Z)=	heta$, we can write $Z=2e^{i	heta}$.Then $Z_1 = \frac{2e^{i	heta}}{2} e^{i\alpha} = e^{i(	heta+\alph

Vedclass · Accepted Answer

$i 	an (n 	heta+n \alpha)$

Vedclass · Answer

(C) Given $|Z|=2$ and $\operatorname{amp}(Z)=	heta$, we can write $Z=2e^{i	heta}$.Then $Z_1 = \frac{2e^{i	heta}}{2} e^{i\alpha} = e^{i(	heta+\alpha)}$.Now, $Z_1^n = (e^{i(	heta+\alpha)})^n = e^{in(	heta+\alpha)} = \cos(n(	heta+\alpha)) + i\sin(n(	heta+\alpha))$.Similarly, $Z_1^{-n} = e^{-in(	heta+\alpha)} = \cos(n(	heta+\alpha)) - i\sin(n(	heta+\alpha))$.Substituting these into the expression:$\frac{Z_1^n-Z_1^{-n}}{Z_1^n+Z_1^{-n}} = \frac{(\cos(n(	heta+\alpha)) + i\sin(n(	heta+\alpha))) - (\cos(n(	heta+\alpha)) - i\sin(n(	heta+\alpha)))}{(\cos(n(	heta+\alpha)) + i\sin(n(	heta+\alpha))) + (\cos(n(	heta+\alpha)) - i\sin(n(	heta+\alpha)))}$$= \frac{2i\sin(n(	heta+\alpha))}{2\cos(n(	heta+\alpha))} = i	an(n(	heta+\alpha))$.Thus, the correct option is $C$.

If $|Z|=2$, $Z_1=\frac{Z}{2} e^{i \alpha}$ and $\theta$ is the $\operatorname{amp}(Z)$, then $\frac{Z_1^n-Z_1^{-n}}{Z_1^n+Z_1^{-n}}=$

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