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Question Number 43757 by Rauny last updated on 15/Sep/18
Probably if x^n =Am ((a/b)π), x=e^(((2k+a)/(bn))iπ)   about 0<(k∈N∪{0})<(n∈N) and b≠0.  p.s. Am (0°)=1, Am (90°)=i etc.,  and s°=(π/(180))s rad(ians)=(π/(180))s.
$$\mathrm{Probably}\:\mathrm{if}\:{x}^{{n}} ={Am}\:\left(\frac{{a}}{{b}}\pi\right),\:{x}={e}^{\frac{\mathrm{2}{k}+{a}}{{bn}}{i}\pi} \\ $$$$\mathrm{about}\:\mathrm{0}<\left({k}\in\mathbb{N}\cup\left\{\mathrm{0}\right\}\right)<\left({n}\in\mathbb{N}\right)\:\mathrm{and}\:{b}\neq\mathrm{0}. \\ $$$$\mathrm{p}.\mathrm{s}.\:{Am}\:\left(\mathrm{0}°\right)=\mathrm{1},\:{Am}\:\left(\mathrm{90}°\right)={i}\:\mathrm{etc}., \\ $$$$\mathrm{and}\:{s}°=\frac{\pi}{\mathrm{180}}{s}\:\mathrm{rad}\left(\mathrm{ians}\right)=\frac{\pi}{\mathrm{180}}{s}. \\ $$

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