6
$\begingroup$

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?

$\endgroup$

1 Answer 1

14
$\begingroup$

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 $(x,y) = (1,-1)$.

$\endgroup$
4
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.