Question Number 23996 by klok9917@gmail last updated on 10/Nov/17
$$\mathrm{z}^{−\mathrm{4}_{=\mathrm{1}/\mathrm{3}\left(\mathrm{1}−\sqrt{\left.\mathrm{3i}\right)}\right.} } \\ $$
Answered by FilupES last updated on 11/Nov/17
$${z}^{−\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}−\sqrt{\mathrm{3}{i}}\right) \\ $$$${z}^{−\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}−\sqrt{\mathrm{3}}×\left(−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \right) \\ $$$${z}^{−\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}−\sqrt{\mathrm{3}}×\left({e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\right) \\ $$$${z}^{−\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}−\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\left(\mathrm{1}+{i}\right)\right) \\ $$$${z}^{−\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{3}}−\sqrt{\frac{\mathrm{1}}{\mathrm{6}}}\left(\mathrm{1}+{i}\right) \\ $$$${z}^{−\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}\left(\mathrm{1}+{i}\right) \\ $$$$\therefore\:{z}\:=\:\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}\left(\mathrm{1}+{i}\right)\right)^{\mathrm{4}} \\ $$