Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

$${god}\:{bless}\:{you}\:{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com