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Question-75851




Question Number 75851 by aliesam last updated on 18/Dec/19
Answered by MJS last updated on 18/Dec/19
10((8+6i)/(3−i))=18+26i=(3+i)^3   (x+yi)^3 =x^3 −3xy^2 +(3x^2 y−y^3 )i  ⇒  (1)  x^3 −3xy^2 =18 ⇒ y=±((√(x^3 −18))/( (√(3x))))  (2)  (3x^2 −y^2 )y=26           ±(3x^2 −((x^3 −18)/(3x)))((√(x^3 −18))/( (√(3x))))=26           (3x^2 −((x^3 −18)/(3x)))^2 ((x^3 −18)/(3x))=676           x^9 −((27)/2)x^6 −((2889)/8)x^3 −((729)/8)=0           trying factors of ((729)/8) ⇒ x=3 ⇒ y=1  z_1 =3+i  z_2 =z_1 ω  z_3 =z_1 ω^2
$$\mathrm{10}\frac{\mathrm{8}+\mathrm{6i}}{\mathrm{3}−\mathrm{i}}=\mathrm{18}+\mathrm{26i}=\left(\mathrm{3}+\mathrm{i}\right)^{\mathrm{3}} \\ $$$$\left({x}+{y}\mathrm{i}\right)^{\mathrm{3}} ={x}^{\mathrm{3}} −\mathrm{3}{xy}^{\mathrm{2}} +\left(\mathrm{3}{x}^{\mathrm{2}} {y}−{y}^{\mathrm{3}} \right)\mathrm{i} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{1}\right)\:\:{x}^{\mathrm{3}} −\mathrm{3}{xy}^{\mathrm{2}} =\mathrm{18}\:\Rightarrow\:{y}=\pm\frac{\sqrt{{x}^{\mathrm{3}} −\mathrm{18}}}{\:\sqrt{\mathrm{3}{x}}} \\ $$$$\left(\mathrm{2}\right)\:\:\left(\mathrm{3}{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){y}=\mathrm{26} \\ $$$$\:\:\:\:\:\:\:\:\:\pm\left(\mathrm{3}{x}^{\mathrm{2}} −\frac{{x}^{\mathrm{3}} −\mathrm{18}}{\mathrm{3}{x}}\right)\frac{\sqrt{{x}^{\mathrm{3}} −\mathrm{18}}}{\:\sqrt{\mathrm{3}{x}}}=\mathrm{26} \\ $$$$\:\:\:\:\:\:\:\:\:\left(\mathrm{3}{x}^{\mathrm{2}} −\frac{{x}^{\mathrm{3}} −\mathrm{18}}{\mathrm{3}{x}}\right)^{\mathrm{2}} \frac{{x}^{\mathrm{3}} −\mathrm{18}}{\mathrm{3}{x}}=\mathrm{676} \\ $$$$\:\:\:\:\:\:\:\:\:{x}^{\mathrm{9}} −\frac{\mathrm{27}}{\mathrm{2}}{x}^{\mathrm{6}} −\frac{\mathrm{2889}}{\mathrm{8}}{x}^{\mathrm{3}} −\frac{\mathrm{729}}{\mathrm{8}}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{trying}\:\mathrm{factors}\:\mathrm{of}\:\frac{\mathrm{729}}{\mathrm{8}}\:\Rightarrow\:{x}=\mathrm{3}\:\Rightarrow\:{y}=\mathrm{1} \\ $$$${z}_{\mathrm{1}} =\mathrm{3}+\mathrm{i} \\ $$$${z}_{\mathrm{2}} ={z}_{\mathrm{1}} \omega \\ $$$${z}_{\mathrm{3}} ={z}_{\mathrm{1}} \omega^{\mathrm{2}} \\ $$
Commented by aliesam last updated on 18/Dec/19
god bless you sir
$${god}\:{bless}\:{you}\:{sir} \\ $$

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