Imaginary^Imaginary

Express $\begin{array}{rl} (ik)^{ik} &= i^{ik} \cdot k^{ik}\\ &= \left(i^i\right)^k \cdot k^{ik}\\ &= e^{-\frac{\pi}{2}k} \cdot k^{ik} \end{array}$ The first factor is a real number, so it remains to focus on the second factor $k^{ik}$ , which leads to the possible solutions. Observe that $\begin{array}{rl} k^{ik} &= e^{ik\ln(k)}\\ &= \cos\left(k\ln(k)\right) + i\sin\left(k\ln(k)\right) \end{array}$ For $k^{ik} \in \mathbb{R}$ , we need $\sin\left(k\ln(k)\right) = 0$ . Then, $k\ln(k) = n\pi$ for integer $n \in \mathbb{Z}$ . Thus, the solutions are $\boxed{k\ln(k) = n\pi,\, n\in\mathbb{N}}$ (as given), which also makes $e^{-\frac{\pi}{2}k}$ real.

Being $k \in \mathbb{R}^{+}$ , $\ln (ik) = \{\ln |ik| + i\cdot \alpha; \space \alpha \in \text{arg(ik)}\}$ . Taking the main branch of the multivaluated function arg, (i.e, $\text{ Arg} (\cdot)$ and sorry, because I'm not giving the domain and codomain of $\text{ Arg} (\cdot)$ ), we can say predetermined $\ln (ik) = \ln |k| + i\cdot \frac{\pi}{2} \Rightarrow$ (remembering $k \in \mathbb{R}^{+}$ ) $\large (ik)^{ik} = e^{ik \cdot (\ln k + i\cdot \frac{\pi}{2})} = e^{(-\frac{k\pi}{2} + ik \cdot \ln k)} = e^{-\frac{k\pi}{2}} \cdot e^{ik \cdot \ln k}$ and hence, $(ik)^{ik} \in \mathbb{R} \iff k \cdot \ln k = n\pi; \text{ such that } n \in \mathbb{Z}$