Find all solution in $\mathbb{R}$ for the following system of equations:

\begin{cases} x + \frac{3x-y}{x^2+y^2} = 3 \\ y – \frac{x+3y}{x^2+y^2} = 0 \end{cases}

I've tried few method, but none bring a success.

First I tried to substitute, and tried to express $y$ in terms of $x$, but I wasn't able to do that. I tried to use $y = \frac{x+3y}{x^2+y^2}$, but as you guess that substitution made the things even more complicated.

Next I tried to multiply the second equation with $\frac{3x-y}{x+3y}$, assuming it's not zero. And I end up with:

\begin{cases} x + \frac{3x-y}{x^2+y^2} = 3 \\ y\left(\frac{3x-y}{x+3y}\right) – \frac{x+3y}{x^2+y^2} = 0 \end{cases}

Now I add the together and I end up with:

$$x + y\left(\frac{3x-y}{x+3y}\right) = 3$$

But solution to this equation aren't solution to the system, because there are infinite amount of them.

Next I tried just to add them and I end up with:

$$x+y \frac{2x-3y}{x^2+y^2} = 3$$

But again I faced the same problem, this equation has an infinite amount of solution, unlike the system of equations.

So how can I solve this system of equations?


1 Answer 1


Complexify. Let $z = x+iy$. Then for $x^2+y^2 \neq 0$:

$$\begin{align} &&\left(x + \frac{3x-y}{x^2+y^2}\right) + i\left(y - \frac{x+3y}{x^2+y^2}\right) &= (3+i0)\\ &\iff& (x+iy) + \frac{(3-i)(x-iy)}{x^2+y^2} &= 3\\ &\iff& z + \frac{3-i}{z} &= 3\\ &\iff& z^2 - 3z + (3-i) &= 0\\ &\iff& \left(z-\frac32\right)^2 &= i - \frac34\\ &\iff& z &= \frac{3 \pm \sqrt{4i-3}}{2}. \end{align}$$

Now, $(1+2i)^2 = -3+4i$, so the solutions are

$$z = \frac{3 \pm (1+2i)}{2},$$

either $z = 2+i$ or $z = 1-i$, which, in real form, yields $(x,y) = (2,1)$ or $(x,y) = (1,-1)$.

  • 1
    $\begingroup$ Great solution and answer. I didn't know that complexifying will provide a real solution. Thanks again. $\endgroup$
    – Stefan4024
    Sep 17, 2013 at 1:12
  • 3
    $\begingroup$ Heh, there is a saying (I forgot by whom), that the shortest path between two real points passes through the complex domain. Fits here. $\endgroup$ Sep 17, 2013 at 1:13
  • 1
    $\begingroup$ @DanielFischer Hadamard. $\endgroup$
    – Pedro
    Oct 22, 2013 at 2:25
  • $\begingroup$ Quite neat. And simple. $\endgroup$ Oct 22, 2013 at 3:30

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