Find all values of $i^i$.

Question

For $z,w\in \mathbb{C}$, we define $\log(z) = \log(|z|) + i \arg(z)$ and $z^w = e^{w\log(z)}$. We have \begin{align*} i^i & = e^{i\log(i)}\\ & = e^{i(\log(1) + i(\pi/2 + 2k\pi))}, & k\in \mathbb{Z}\\ & = e^{-(\pi/2 + 2k\pi))}, & k\in \mathbb{Z}\\ & = e^{-\pi(2k + 1/2)}, & k \in \mathbb{Z}. \end{align*}

Is this correct?

Answer 1

Note that $z,w$ shouldn't be $0$

You answer is true, because $log(z)$ is multivalued, however, if we use principal branch of logarithm, the answer will be $e^{\frac{\pi}{2}}$