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I'm trying to solve a system of two cubic equations with two variables x and y.

The original problem was to solve the equation $z^3=-4i \overline{z}$. I know how to solve it using polar form.

Now I want to solve it using Cartesian form, say $z=x+yi$.

Doing the algebra and simplifying I got the next system of equations: $$\displaystyle\left\{\begin{matrix}x^3-3xy^2+4y=0\\-y^3+3x^2y+4x=0\end{matrix}\right.$$

It is trivial that $\displaystyle (0,0)$ is a solution, but I couldn't find the other four.

The best I got is $(3x^2-y^2)(x^2-3y^2)=16$, but I don't how to continue.

Please help, thanks!

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  • $\begingroup$ I'm not clear how you got from $z = -4i\overline{z}$ to a system of two cubic equations. The latter is an interesting topic in itself, but in general position two cubic curves will have nine points of intersection (not the five you seem to have expected). $\endgroup$ – hardmath Oct 15 '14 at 11:41
  • $\begingroup$ I had a typo, supposed to be $z^3=-4i \overline{z}$, I fixed it. If $z=x+yi$ where $x,y \in \mathbb{R}$ the equation has exactly five solutions (and I found them all using polar form). $\endgroup$ – Galc127 Oct 15 '14 at 11:47
  • $\begingroup$ I'm afraid the best way to solve that system is to recognize it as the Cartesian form of the original problem and then solve that using polar form, and then translate the answer back to Cartesian. They invented complex numbers for a reason, you know. $\endgroup$ – Gerry Myerson Oct 15 '14 at 12:28
  • $\begingroup$ @GerryMyerson, as mentioned, I solved it using polar form and I'm aware that it is the proper way and the easier one. I'm interested to know if there is a way to solve the set of equations that I got using Cartesian form. $\endgroup$ – Galc127 Oct 15 '14 at 12:31
  • $\begingroup$ I understand, and what I'm telling you is that the best way to solve the equations you got is to go back to the equation you started with. Any other way will just be a disguised way of the polar way. $\endgroup$ – Gerry Myerson Oct 15 '14 at 22:14
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$$\displaystyle\left\{\begin{matrix}x^3-3xy^2+4y=0\\-y^3+3x^2y+4x=0\end{matrix}\right.$$ To solve this system of equations first multiply the first equation with $x$ and second with $y$. Then subtract the second equation from the first.

We get $$x^4+y^4-6x^2y^2=0$$

which is same as $$(x^2-2 x y-y^2) (x^2+2 x y-y^2)=0$$ So now we have two cases: $x^2-2xy-y^2=0$ or $x^2+2xy-y^2=0$

From here it should be pretty easy.

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  • $\begingroup$ Thanks a lot! your answer is very helpful. $\endgroup$ – Galc127 Oct 15 '14 at 13:26
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$$z^3=-4i\bar{z}\Rightarrow (-i)z^4=4|z|^2\Rightarrow |z|^4=4|z|^2$$ A trivial solution would be $|z|=0\Leftrightarrow z=0$. Assume $|z|\neq 0$ then $$|z|^2=4 \Rightarrow x^2+y^2=4$$ Expressing the main equation in cartesian coordinates: $$(x+iy)^3=-4i(x-iy)\Rightarrow x^3+3ix^2y-3xy^2-iy^3=-4ix-4y$$ So one gets after equalizing the real and imaginary parts $$x^3-3xy^2+4y=0$$ and $$3x^2y-y^3+4x=0$$ Adding the last two equations together yields $$x^3-y^3+3xy(x-y)+4(x+y)=0\Rightarrow (x-y)(x^2+xy+y^2)+3xy(x-y)+4(x+y)=0\Rightarrow(x-y)(x^2+4xy+y^2)+4(x+y)=0$$ Using $x^2+y^2=4$ we get $$(x-y)(1+xy)+(x+y)=0\Rightarrow x+x^2y-y-xy^2+x+y=0\Rightarrow x(2+xy-y^2)=0$$ So $x=0\Rightarrow y=0$ or $2+xy-y^2=0\Rightarrow 2+x\sqrt{4-x^2}-4+x^2=0$ The last result is equivalent $$(2-x^2)^2=x^2(4-x^2)\Leftrightarrow 4-4x^2+x^4=4x^2-x^4\Leftrightarrow x^4-4x^2+2=0$$ This equation has four roots $$x=\pm\sqrt{2-\sqrt{2}}$$ and $$x=\pm\sqrt{2+\sqrt{2}}$$

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  • $\begingroup$ He's question was to solve the system of equation. $\endgroup$ – Harto Saarinen Oct 15 '14 at 12:16
  • $\begingroup$ @Harto: I am not even sure if that system of equation is true. I have not checked it yet. $\endgroup$ – Arian Oct 15 '14 at 12:20
  • $\begingroup$ There are 4 specific solutions such that $|z|=2$, not every $z \in \mathbb{C}$ is a solution. $\endgroup$ – Galc127 Oct 15 '14 at 12:23
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    $\begingroup$ Actually, there are infinite, but they have a form. The form is $z=2 cis(67.5+90k)$ as $k \in \mathbb{Z}$. We may find that there are 4 solutions such that $0<\theta<360$. If for example you'll take $z=2cis(30)$ then you'll find that it is not a proper solution of the equation. $\endgroup$ – Galc127 Oct 15 '14 at 12:28
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    $\begingroup$ Your edit is correct. $\endgroup$ – Arian Oct 15 '14 at 14:19

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