Discussion
The principal value of \((-1)^{(-2i/\pi)}\) is
Answer: A
The principal value is \(e^{2}\).
A complex power is defined through the principal logarithm: \(a^{b} = e^{b\,\mathrm{Log}\,a}\). For \(a = -1\), \(\mathrm{Log}(-1) = \ln 1 + i\pi = i\pi\), since the principal argument of \(-1\) is \(\pi\).
\[ \begin{aligned} (-1)^{-2i/\pi} &= e^{(-2i/\pi)(i\pi)} \\ &= e^{-2i^{2}} = e^{2} \end{aligned} \]
The result is real because \(i \cdot i = -1\) cancels the minus sign. \(e^{-2}\) comes from dropping that sign, and \(e^{2i}\) or \(e^{-2i}\) from forgetting to multiply the two factors of \(i\) together.
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