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Question Number 63803 by aliesam last updated on 09/Jul/19

Commented by Prithwish sen last updated on 09/Jul/19

$${(\mathrm{x}+\mathrm{iy})}^3=\frac{{(1+\sqrt{7}\mathrm{i})}^2}{1-\sqrt{7}\mathrm{i}}=\frac{{(1+\sqrt{7}\mathrm{i})}^3}{8}$$$$\therefore \mathrm{x}=\frac{1}{2}\mathrm{and}\mathrm{y}=\frac{\sqrt{7}}{2}$$

Commented by mathmax by abdo last updated on 09/Jul/19

$$letZ=x+iyweget{Z}^3=\frac{-6+2\sqrt{7}i}{1-\sqrt{7}i}letsolvethis\left(e\right)$$$$\mid -6+2\sqrt{7}i\mid =\sqrt{36+28}=\sqrt{64}=8$$$$\mid 1-\sqrt{7}i\mid =\sqrt{1+7}=\sqrt{8}\Rightarrow \mid \frac{-6+2\sqrt{i7}}{1-\sqrt{7}i}\mid =\frac{8}{\sqrt{8}}=2\sqrt{2}$$$$-6+2\sqrt{7}i=8(\frac{-6}{8}+\frac{2\sqrt{7}i}{8})=8(-\frac{3}{4}+\frac{\sqrt{7}}{4}i)=8e^{iarctan(-\frac{\sqrt{7}}{3})}$$$$1-\sqrt{7}i=2\sqrt{2}(\frac{1}{2\sqrt{2}}-\frac{\sqrt{7}}{2\sqrt{2}}i)=2\sqrt{2}{e}^{iarctan(-\sqrt{7})}\Rightarrow$$$$\frac{-6+2\sqrt{7}i}{1-\sqrt{7}i}=2\sqrt{2}{e}^{i(arctan\left(\frac{-\sqrt{7}}{3}\right)+arctan\left(\sqrt{7}\right))}=2\sqrt{2}{e}^{i(arctan\left(\sqrt{7}\right)-arctan\left(\frac{\sqrt{7}}{3}\right))}$$$$letZ=\rho e^{i\theta }\left(e\right)\Rightarrow {\rho }^3{e}^{i3\theta }=2\sqrt{2}{e}^{i(arctan\sqrt{7}-arctan\left(\frac{\sqrt{7}}{3}\right))}\Rightarrow$$$$\rho ={\left(2\sqrt{2}\right)}^{\frac{1}{3}}and3\theta =arctan\left(\sqrt{7}\right)-arctan\left(\frac{\sqrt{7}}{3}\right)+2k\pi \Rightarrow$$$$\Rightarrow {\theta }_k=\frac{arctan\left(\sqrt{7}\right)-arctan\left(\frac{\sqrt{7}}{3}\right)}{3}+\frac{2k\pi }{3}andk\in \left[[0,2]\right]$$$$therootsare{Z}_k=e^{i{\theta }_k}=x_k+{iy}_k$$$$Z_0=e^{i{\theta }_0}=x_0+{iy}_0$$$$Z_1=e^{i{\theta }_1}=x_1+{iy}_1$$$$Z_2=e^{i{\theta }_2}=x_2+{iy}_2...$$

Commented by mathmax by abdo last updated on 09/Jul/19

$$sorry{Z}_k=\rho e^{i{\theta }_k}\Rightarrow$$$$Z_0=\rho e^{i{\theta }_0}=x_0+{iy}_0$$$$Z_1=\rho e^{i{\theta }_1}=x_1+{iy}_1$$

Commented by mathmax by abdo last updated on 09/Jul/19

$$Z_2=\rho e^{i{\theta }_2}=x_2+{iy}_2$$$$thereels\left(x_i\right)and\left(y_i\right)aredetremined.$$

Answered by MJS last updated on 09/Jul/19

$$\frac{-6+2\sqrt{7}\mathrm{i}}{1-\sqrt{7}\mathrm{i}}=-\frac{5}{2}-\frac{\sqrt{7}}{2}\mathrm{i}$$$$\mathrm{in}\mathrm{this}\mathrm{case}\mathrm{we}\mathrm{have}\mathrm{good}\mathrm{luck}\mathrm{because}$$$${(\frac{1}{2}+\frac{\sqrt{7}}{2}\mathrm{i})}^3=-\frac{5}{2}-\frac{\sqrt{7}}{2}\mathrm{i}$$$$\Rightarrow z_1=\frac{1}{2}+\frac{\sqrt{7}}{2}\mathrm{i}$$$$z^3+\frac{5}{2}+\frac{\sqrt{7}}{2}\mathrm{i}=0$$$$(z-\frac{1}{2}-\frac{\sqrt{7}}{2}\mathrm{i})(z^2+(\frac{1}{2}+\frac{\sqrt{7}}{2}\mathrm{i})z-\frac{3}{2}+\frac{\sqrt{7}}{2}\mathrm{i})=0$$$$\Rightarrow z_{2,3}=\pm \frac{1+\sqrt{21}}{4}-\frac{\sqrt{7}-\sqrt{3}}{4}\mathrm{i}$$

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