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Question Number 223203 by ajfour last updated on 17/Jul/25
Find x and y ix+y=ix^3 −y^3 and xy(y−ix)=c
$${Find}\:{x}\:{and}\:{y} \\ $$$$\:\:\:\:{ix}+{y}={ix}^{\mathrm{3}} −{y}^{\mathrm{3}} \\ $$$$\:\:\:{and}\: \\ $$$$\:\:\:\:{xy}\left({y}−{ix}\right)={c} \\ $$
Answered by mr W last updated on 17/Jul/25
let p=ix py(y−p)=ic ⇒y−p=((ic)/(py)) p+y=−p^3 −y^3 =−(p+y)(p^2 +y^2 −py) 1) p+y=0 ⇒y=−p 2p^3 =ic ⇒p=−i(√(c/2)) ⇒x=−((c/2))^(1/) ⇒y=i(√(c/2)) 2) p+y≠0 p^2 +y^2 −py=−1 (y−p)^2 +py=−1 (((ic)/(py)))^2 +py=−1 −(c^2 /((py)^2 ))+py=−1 (1/((py)^3 ))−(1/(c^2 (py)))−(1/c^2 )=0 ⇒(1/(py))=(1/c)((((√((1/4)−(1/(27c^2 ))))+(1/2)))^(1/3) −(((√((1/4)−(1/(27c^2 ))))−(1/2)))^(1/3) ) py=(c/( (((√((1/4)−(1/(27c^2 ))))+(1/2)))^(1/3) −(((√((1/4)−(1/(27c^2 ))))−(1/2)))^(1/3) ))=h −p+y=i((((√((1/4)−(1/(27c^2 ))))+(1/2)))^(1/3) −(((√((1/4)−(1/(27c^2 ))))−(1/2)))^(1/3) )=ik −p, y are roots of z^2 −ikz+h=0 −ix, y=((ik±(√(−k^2 +4h)))/2) ⇒x=((−k∓(√(k^2 +4h)))/2) ⇒y=(((k±(√(k^2 +4h)))i)/2)
$${let}\:{p}={ix} \\ $$$${py}\left({y}−{p}\right)={ic}\:\Rightarrow{y}−{p}=\frac{{ic}}{{py}} \\ $$$${p}+{y}=−{p}^{\mathrm{3}} −{y}^{\mathrm{3}} =−\left({p}+{y}\right)\left({p}^{\mathrm{2}} +{y}^{\mathrm{2}} −{py}\right) \\ $$$$\left.\mathrm{1}\right) \\ $$$${p}+{y}=\mathrm{0}\:\Rightarrow{y}=−{p} \\ $$$$\mathrm{2}{p}^{\mathrm{3}} ={ic} \\ $$$$\Rightarrow{p}=−{i}\sqrt{\frac{{c}}{\mathrm{2}}}\:\Rightarrow{x}=−\sqrt[{}]{\frac{{c}}{\mathrm{2}}} \\ $$$$\Rightarrow{y}={i}\sqrt{\frac{{c}}{\mathrm{2}}} \\ $$$$\left.\mathrm{2}\right) \\ $$$${p}+{y}\neq\mathrm{0} \\ $$$${p}^{\mathrm{2}} +{y}^{\mathrm{2}} −{py}=−\mathrm{1} \\ $$$$\left({y}−{p}\right)^{\mathrm{2}} +{py}=−\mathrm{1} \\ $$$$\left(\frac{{ic}}{{py}}\right)^{\mathrm{2}} +{py}=−\mathrm{1} \\ $$$$−\frac{{c}^{\mathrm{2}} }{\left({py}\right)^{\mathrm{2}} }+{py}=−\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\left({py}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{{c}^{\mathrm{2}} \left({py}\right)}−\frac{\mathrm{1}}{{c}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{py}}=\frac{\mathrm{1}}{{c}}\left(\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{27}{c}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{27}{c}^{\mathrm{2}} }}−\frac{\mathrm{1}}{\mathrm{2}}}\right) \\ $$$${py}=\frac{{c}}{\:\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{27}{c}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{27}{c}^{\mathrm{2}} }}−\frac{\mathrm{1}}{\mathrm{2}}}}={h} \\ $$$$−{p}+{y}={i}\left(\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{27}{c}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{27}{c}^{\mathrm{2}} }}−\frac{\mathrm{1}}{\mathrm{2}}}\right)={ik} \\ $$$$−{p},\:{y}\:{are}\:{roots}\:{of} \\ $$$${z}^{\mathrm{2}} −{ikz}+{h}=\mathrm{0} \\ $$$$−{ix},\:{y}=\frac{{ik}\pm\sqrt{−{k}^{\mathrm{2}} +\mathrm{4}{h}}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{−{k}\mp\sqrt{{k}^{\mathrm{2}} +\mathrm{4}{h}}}{\mathrm{2}} \\ $$$$\Rightarrow{y}=\frac{\left({k}\pm\sqrt{{k}^{\mathrm{2}} +\mathrm{4}{h}}\right){i}}{\mathrm{2}} \\ $$
Answered by Frix last updated on 17/Jul/25
ix+y=ix^3 −y^3 ix+y=(ix+y)(x^2 +ixy−y^2 ) ix+y=0∨x^2 +ixy−y^2 =1 y=−ix inserting in (2) ⇒ x^3 =−(c/2) x^2 +ixy−y^2 =1 y=±((√(3x^2 −4))/2)+i(x/2) inserting in (2) ⇒ x^3 −x−c=0 ⇒ if c∈R: x=−ω^k ((c/2))^(1/3) ∧y=ix; ω=−(1/2)+((√3)/2)i; k=0, 1, 2 ∨ depending on the value of c: (a) x_1 =(((c/2)−(√((c^2 /4)−(1/(27))))))^(1/3) +(((c/2)+(√((c^2 /4)−(1/(27))))))^(1/3) x_2 =ω(((c/2)−(√((c^2 /4)−(1/(27))))))^(1/3) +ω^2 (((c/2)+(√((c^2 /4)−(1/(27))))))^(1/3) x_3 =ω^2 (((c/2)−(√((c^2 /4)−(1/(27))))))^(1/3) +ω(((c/2)+(√((c^2 /4)−(1/(27))))))^(1/3) y=±((√(3x^2 −4))/2)+i(x/2) (b) x=−((2(√3))/3)sin ((−2kπ+sin^(−1) ((3(√3)c)/2))/3) ; k=0, 1, 2 y=±((√(3x^2 −4))/2)+i(x/2)
$$\mathrm{i}{x}+{y}=\mathrm{i}{x}^{\mathrm{3}} −{y}^{\mathrm{3}} \\ $$$$\mathrm{i}{x}+{y}=\left(\mathrm{i}{x}+{y}\right)\left({x}^{\mathrm{2}} +\mathrm{i}{xy}−{y}^{\mathrm{2}} \right) \\ $$$$\mathrm{i}{x}+{y}=\mathrm{0}\vee{x}^{\mathrm{2}} +\mathrm{i}{xy}−{y}^{\mathrm{2}} =\mathrm{1} \\ $$$$ \\ $$$${y}=−\mathrm{i}{x} \\ $$$$\mathrm{inserting}\:\mathrm{in}\:\left(\mathrm{2}\right)\:\Rightarrow\:{x}^{\mathrm{3}} =−\frac{{c}}{\mathrm{2}} \\ $$$$ \\ $$$${x}^{\mathrm{2}} +\mathrm{i}{xy}−{y}^{\mathrm{2}} =\mathrm{1} \\ $$$${y}=\pm\frac{\sqrt{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}+\mathrm{i}\frac{{x}}{\mathrm{2}} \\ $$$$\mathrm{inserting}\:\mathrm{in}\:\left(\mathrm{2}\right)\:\Rightarrow\:{x}^{\mathrm{3}} −{x}−{c}=\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow \\ $$$$\mathrm{if}\:{c}\in\mathbb{R}: \\ $$$${x}=−\omega^{{k}} \sqrt[{\mathrm{3}}]{\frac{{c}}{\mathrm{2}}}\wedge{y}=\mathrm{i}{x};\:\omega=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i};\:{k}=\mathrm{0},\:\mathrm{1},\:\mathrm{2} \\ $$$$\vee \\ $$$$\mathrm{depending}\:\mathrm{on}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{c}: \\ $$$$\left({a}\right) \\ $$$${x}_{\mathrm{1}} =\sqrt[{\mathrm{3}}]{\frac{{c}}{\mathrm{2}}−\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{27}}}}+\sqrt[{\mathrm{3}}]{\frac{{c}}{\mathrm{2}}+\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{27}}}} \\ $$$${x}_{\mathrm{2}} =\omega\sqrt[{\mathrm{3}}]{\frac{{c}}{\mathrm{2}}−\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{27}}}}+\omega^{\mathrm{2}} \sqrt[{\mathrm{3}}]{\frac{{c}}{\mathrm{2}}+\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{27}}}} \\ $$$${x}_{\mathrm{3}} =\omega^{\mathrm{2}} \sqrt[{\mathrm{3}}]{\frac{{c}}{\mathrm{2}}−\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{27}}}}+\omega\sqrt[{\mathrm{3}}]{\frac{{c}}{\mathrm{2}}+\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{27}}}} \\ $$$${y}=\pm\frac{\sqrt{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}+\mathrm{i}\frac{{x}}{\mathrm{2}} \\ $$$$\left({b}\right) \\ $$$${x}=−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{sin}\:\frac{−\mathrm{2}{k}\pi+\mathrm{sin}^{−\mathrm{1}} \:\frac{\mathrm{3}\sqrt{\mathrm{3}}{c}}{\mathrm{2}}}{\mathrm{3}}\:;\:{k}=\mathrm{0},\:\mathrm{1},\:\mathrm{2} \\ $$$${y}=\pm\frac{\sqrt{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}+\mathrm{i}\frac{{x}}{\mathrm{2}} \\ $$

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