A) \[1\]
B) \[{{e}^{2r\sin \theta }}\]
C) \[{{e}^{r\sin \theta }}\]
D) \[r{{e}^{\sin \theta }}\]
E) \[{{e}^{-r\sin \theta }}\]
Correct Answer: E
Solution :
If\[z=r{{e}^{i\theta }}=r(\cos \theta +i\sin \theta )\] \[\Rightarrow \] \[iz=ir(\cos \theta +i\sin \theta )\] \[=-r\sin \theta +i\,r\,\cos \theta \] \[\therefore \] \[{{e}^{iz}}={{e}^{-r\sin \theta +r\,i\,\cos \theta }}\] \[|{{e}^{iz}}|=|{{e}^{-r\sin \theta }}||{{e}^{ir\cos \theta }}|\] \[={{e}^{-r\sin \theta }}|{{e}^{ir\cos \theta }}|\] \[={{e}^{-r\sin \theta }}[{{\{{{\cos }^{2}}(r\cos \theta )+{{\sin }^{2}}(r\,\cos \theta )\}}^{1/2}}]\] \[\therefore \] \[|{{e}^{iz}}|={{e}^{-r\sin \theta }}\]
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