If z = r{e^{itheta }}, then |{e^{iz}}| = | vedclass

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(B) Given $z = r{e^{i	heta }} = r(\cos 	heta + i\sin 	heta )$.Then $iz = ir(\cos 	heta + i\sin 	heta ) = -r\sin 	heta + ir\cos 	heta$.Therefore

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${e^{ - r\sin 	heta }}$

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(B) Given $z = r{e^{i	heta }} = r(\cos 	heta + i\sin 	heta )$.Then $iz = ir(\cos 	heta + i\sin 	heta ) = -r\sin 	heta + ir\cos 	heta$.Therefore,${e^{iz}} = {e^{-r\sin 	heta + ir\cos 	heta}} = {e^{-r\sin 	heta}} \cdot {e^{i(r\cos 	heta)}}$.Taking the modulus on both sides:$|{e^{iz}}| = |{e^{-r\sin 	heta}}| \cdot |{e^{i(r\cos 	heta)}}|$.Since ${e^{-r\sin 	heta}}$ is a real number,$|{e^{-r\sin 	heta}}| = {e^{-r\sin 	heta}}$.Also,$|{e^{i(r\cos 	heta)}}| = |\cos(r\cos 	heta) + i\sin(r\cos 	heta)| = \sqrt{\cos^2(r\cos 	heta) + \sin^2(r\cos 	heta)} = 1$.Thus,$|{e^{iz}}| = {e^{-r\sin 	heta}} \cdot 1 = {e^{-r\sin 	heta}}$.

If $z = r{e^{i\theta }},$ then $|{e^{iz}}| = $

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