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

Write $z=r{e}^{i\theta }$ with $-\pi <\theta \le \pi$. Its principal sixth root is $r^{1/6}{e}^{i\theta /6}$, whose complex argument lies in $(-\pi /6,\pi /6]$. Equality with $2e^{i\pi /6}$ forces $r=64$ and $\theta =\pi$, so $z=-64$.