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how to prove i^i is a real number?
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Use ln(-1)
\[i^i\] first write \[i=e^{\frac{\pi}{2}i}\] and then raise it to the ith power
arnab
you get \[e^{-\frac{\pi}{2}}\]
beautiful!
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thnx and only one cup of coffee
cant get the probability question though
Let i^i = e^x -> i ln i = x i ln (sqrt(-1)) = x i/2 ln -1 = x but since ln -1 = ipi x = -pi/2
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