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Question Number 75851 by aliesam last updated on 18/Dec/19

Answered by MJS last updated on 18/Dec/19

$$10\frac{8+6\mathrm{i}}{3-\mathrm{i}}=18+26\mathrm{i}={(3+\mathrm{i})}^3$$$${(x+y\mathrm{i})}^3=x^3-3{xy}^2+(3x^{2}y-y^3)\mathrm{i}$$$$\Rightarrow$$$$\left(1\right)x^3-3{xy}^2=18\Rightarrow y=\pm \frac{\sqrt{x^3-18}}{\sqrt{3x}}$$$$\left(2\right)(3x^2-y^2)y=26$$$$\pm (3x^2-\frac{x^3-18}{3x})\frac{\sqrt{x^3-18}}{\sqrt{3x}}=26$$$${(3x^2-\frac{x^3-18}{3x})}^2\frac{x^3-18}{3x}=676$$$$x^9-\frac{27}{2}{x}^6-\frac{2889}{8}{x}^3-\frac{729}{8}=0$$$$\mathrm{trying}\mathrm{factors}\mathrm{of}\frac{729}{8}\Rightarrow x=3\Rightarrow y=1$$$$z_1=3+\mathrm{i}$$$$z_2=z_1\omega$$$$z_3=z_1{\omega }^2$$

Commented by aliesam last updated on 18/Dec/19

$$godblessyousir$$

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