Discussion
One of the values of \(i^{(1+i)}\) is :
Answer: D
One value of \(i^{(1+i)}\) is \(i e^{-\pi/2}\).
A complex power is defined by \(z^{w} = e^{w \log z}\), with \(\log i = i(\pi/2 + 2k\pi)\). Taking the principal branch \(k = 0\):
\[ \begin{aligned} i^{1+i} &= e^{(1+i)\, i\pi/2} \\ &= e^{i\pi/2 - \pi/2} \\ &= e^{i\pi/2}\, e^{-\pi/2} = i e^{-\pi/2} \end{aligned} \]
Other branches multiply this by \(e^{-2k\pi}\), so every value is \(i\) times a positive real number. \(i e^{\pi/2}\) has the wrong sign in the exponent (it would need \(k = -1/2\)), and \(e^{\pm \pi i/2} = \pm i\) has modulus 1, which none of the values has.
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