The amplitude of {e^{{e^{ - itheta }}}} is eq | vedclass

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Question

(B) Let $z = {e^{{e^{ - i	heta }}}} = {e^{\cos 	heta - i\sin 	heta }} = {e^{\cos 	heta }}{e^{ - i\sin 	heta }}$.$z = {e^{\cos 	heta }}[\cos (\si

Vedclass · Accepted Answer

$ - \sin 	heta $

Vedclass · Answer

(B) Let $z = {e^{{e^{ - i	heta }}}} = {e^{\cos 	heta - i\sin 	heta }} = {e^{\cos 	heta }}{e^{ - i\sin 	heta }}$.$z = {e^{\cos 	heta }}[\cos (\sin 	heta ) - i\sin (\sin 	heta )]$.$z = {e^{\cos 	heta }}\cos (\sin 	heta ) - i{e^{\cos 	heta }}\sin (\sin 	heta )$.$amp(z) = {	an ^{ - 1}}\left[ { - \frac{{{e^{\cos 	heta }}\sin (\sin 	heta )}}{{{e^{\cos 	heta }}\cos (\sin 	heta )}}} ight]$.$amp(z) = {	an ^{ - 1}}[	an ( - \sin 	heta )] = - \sin 	heta $.

The amplitude of ${e^{{e^{ - i\theta }}}}$ is equal to

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