# Find all solution in <mi mathvariant="double-struck">R for the following system of equations:

Find all solution in $\mathbb{R}$ for the following system of equations:
$\left\{\begin{array}{l}x+\frac{3x-y}{{x}^{2}+{y}^{2}}=3\\ y–\frac{x+3y}{{x}^{2}+{y}^{2}}=0\end{array}$
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tatoas9f
Complexify. Let $z=x+iy$. Then for ${x}^{2}+{y}^{2}\ne 0$:
$\begin{array}{rlrl}& & \left(x+\frac{3x-y}{{x}^{2}+{y}^{2}}\right)+i\left(y-\frac{x+3y}{{x}^{2}+{y}^{2}}\right)& =\left(3+i0\right)\\ & \phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}& \left(x+iy\right)+\frac{\left(3-i\right)\left(x-iy\right)}{{x}^{2}+{y}^{2}}& =3\\ & \phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}& z+\frac{3-i}{z}& =3\\ & \phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}& {z}^{2}-3z+\left(3-i\right)& =0\\ & \phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}& {\left(z-\frac{3}{2}\right)}^{2}& =i-\frac{3}{4}\\ & \phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}& z& =\frac{3±\sqrt{4i-3}}{2}.\end{array}$
Now, $\left(1+2i{\right)}^{2}=-3+4i$, so the solutions are
$z=\frac{3±\left(1+2i\right)}{2},$
either $z=2+i$ or $z=1-i$, which, in real form, yields $\left(x,y\right)=\left(2,1\right)$ or $\left(x,y\right)=\left(1,-1\right)$.
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Aidyn Cox
Great solution and answer. I didn't know that complexifying will provide a real solution. Thanks again.