Think Outside the Box

The equations given are the real and imaginary parts of $x-yi + \frac{3+i}{x-yi} = 3.$ Let $z = x-yi$ ; then $z^2-3z + (3+i) = 0$ , which factors as $(z-(2-i))(z-(1+i)) = 0$ . So $z = 2-i$ or $z = 1+i$ , which leads to $(2,1)$ and $(1,-1)$ as the two solutions. The sum of the coordinates is $\fbox{3}$ .

The fractions are in the way, how do we get rid of them?

We will attempt to multiply the first equation by $y$ and the second equation by $x$ . Adding the new equations, we have $2xy+\frac{-(y^2+x^2)}{x^2 + y^2} = 3y$

Or $2xy-1=3y$ . Solving for $x$ , we have $x=\dfrac{3y+1}{2y}$

Substituting $x=\frac{3y+1}{2y}$ to our second equation and simplifying, we have $4y^4-3y^2-1=0$ $\implies (y^2-1)(4y^2+1)=0$ $\implies y=\pm1$

If $y=1$ , we have $x=2$ . If $y=-1$ , we have $x=1$ . So the solutions are $(2,1)$ and $(1,-1)$ for $x$ and $y$ , respectively.

Therefore $\displaystyle \sum_{i=1}^{n} x_i+y_i = 2+1+1-1=\boxed{3}$