A) \[\omega \]
B) \[-{{\omega }^{2}}\]
C) \[\pi /2\]
D) none of these
Correct Answer: D
Solution :
Let \[A={{i}^{i}}\Rightarrow \log A=i\log i\] \[\Rightarrow \]\[\log A=i\log (0+i)=i[\log 1+i{{\tan }^{-1}}\infty ]\] \[\Rightarrow \]\[\log A=i\,[0+i\pi /2]=-\pi /2\] \[\Rightarrow \]\[A={{e}^{-\pi /2}}\]You need to login to perform this action.
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