A) \[i\]
B) \[-i\]
C) \[1\]
D) \[-1\]
Correct Answer: C
Solution :
We know that, \[{{e}^{i\theta }}=\cos \theta +i\sin \theta \] \[\therefore \] \[{{e}^{iA}}\cdot {{e}^{iB}}\cdot {{e}^{iC}}\cdot {{e}^{iD}}=(\cos A+i\sin A)\]`\[(\cos B+i\sin B)\cdot (\cos C+i\sin C)\cdot \]\[(\cos D+i\sin D)\] \[=\cos (A+B+C+D)+i\sin (A+B+C+D)\] \[=\cos {{360}^{o}}+i\sin {{360}^{o}}\] \[=1\]
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