Solving a system of two cubic equations

Question

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!

Answer 1

$$\displaystyle\left\{\begin{matrix}x^3-3xy^2+4y=0\\-y^3+3x^2y+4x=0\end{matrix}\right.$$ 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.

Answer 2

$$z^3=-4i\bar{z}\Rightarrow (-i)z^4=4|z|^2\Rightarrow |z|^4=4|z|^2$$ 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 $$x=\pm\sqrt{2+\sqrt{2}}$$