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If \(x=\sqrt{-1}\), then the value of \(x^x\) is

1. \(e^{-\pi/2}\)
2. \(e^{\pi/2}\)
3. \(x\)
4. 1

Correct answer: \(e^{-\pi/2}\)

Solution

Taking the principal value, \(i=e^{i\pi/2}\), so \(\log i=i\pi/2\). Then \(i^i=e^{i\log i}=e^{i(i\pi/2)}=e^{-\pi/2}\).

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