A) w
B) w2
C) \[{{e}^{\frac{\pi }{2}}}\]
D) \[{{e}^{\frac{-\pi }{2}}}\]
Correct Answer: D
Solution :
[d] \[\because i=cos\frac{\pi }{2}+isin\frac{\pi }{2}={{e}^{i\pi /2}}\] \[{{(i)}^{i}}=\left( {{e}^{i.\frac{\pi }{2}}} \right)={{e}^{{{i}^{2}}.\frac{\pi }{2}}}={{e}^{-\frac{\pi }{2}}}\,\,\,(\therefore i=1)\]
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