For $z=x+iy$,
$$
i^z=e^{z\operatorname{Log}i}=e^{i\pi z/2}=e^{-\pi y/2}e^{i\pi x/2}.
$$
Matching its <modulus> and <complex argument> with $2e^{i\pi/6}$ gives
$$
\boxed{z=\frac13+4k-\frac{2i\log2}{\pi},\qquad k\in\mathbb Z.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 1 / 1C / iii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ia/paper-1/1c/iii/solution