Question
यदि $z = r{e^{i\theta }},$तो $|{e^{iz}}|$बराबर है
$iz = ir(\cos \theta + i\sin \theta ) = - r\sin \theta + ir\cos \theta $
या ${e^{iz}} = {e^{( - r\sin \theta + ir\cos \theta )}} = {e^{ - \sin \theta }}{e^{ri\cos \theta }}$
या $|{e^{iz}}| = |{e^{ - r\sin \theta }}||{e^{ri\cos \theta }}|$$ = {e^{ - r\sin \theta }}|{e^{ir\,\cos \theta }}|$
$ = {e^{ - r\sin \theta }}{[\{ {\cos ^2}(r\cos \theta ) + {\sin ^2}(r\cos \theta )\} ]^{1/2}} = {e^{ - r\sin \theta }}$
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