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Question Number 129809 by mohammad17 last updated on 19/Jan/21

$${(x+iy)}^3=\frac{-6+2\sqrt{7}i}{1-\sqrt{7}i}findx,y$$

Answered by Olaf last updated on 20/Jan/21

$${(x+iy)}^3=\frac{-6+2\sqrt{7}i}{1-\sqrt{7}i}$$$$\mathrm{Let}x+iy=\rho e^{i\theta }$$$$\mid \frac{-6+2\sqrt{7}i}{1-\sqrt{7}i}\mid =\frac{\sqrt{36+28}}{\sqrt{1+7}}=2\sqrt{2}$$$$\arg \left(\frac{-6+2\sqrt{7}i}{1-\sqrt{7}i}\right)=\arg (-6+2\sqrt{7})-\arg (1-\sqrt{7})+2k\pi$$$$=\arctan (-\frac{\sqrt{7}}{3})-\arctan (-\sqrt{7})+2k\pi$$$$=\arctan \left(\sqrt{7}\right)-\arctan \left(\frac{\sqrt{7}}{3}\right)+2k\pi$$$$=\arctan \left(\frac{\sqrt{7}-\frac{\sqrt{7}}{3}}{1+\sqrt{7}\times \frac{\sqrt{7}}{3}}\right)+2k\pi$$$$=\arctan \left(\frac{\sqrt{7}}{5}\right)+2k\pi$$$${\rho }^3{e}^{i3\theta }=2\sqrt{2}{e}^{i\arctan \left(\frac{\sqrt{7}}{5}\right)+2k\pi }$$$$\rho =\sqrt[3]{2}\mathrm{and}\theta =\frac{1}{3}\arctan \left(\frac{\sqrt{7}}{5}\right)+\frac{2}{3}k\pi$$$$x=\rho \cos \theta \mathrm{and}y=\rho \sin \theta$$

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