Find all ordered pairs $(a, b)$ of complex numbers with $a^2+b^2\neq 0,$ and
$\displaystyle a+\frac{10b}{a^2+b^2} = 5\;\;,b+\frac{10a}{a^2+b^2}=4$
$\bf{My\; Solution}::$ Using Complex number..
Given $\displaystyle a+\frac{10b}{a^2+b^2}=5..............(1)$ and $\displaystyle b+\frac{10a}{a^2+b^2}=4...................(2)\times i$
Now Add $\bf{(1)}$ and $\bf{(2)\;,}$ We get
$\displaystyle (a+ib)+\frac{10}{(a+ib)\cdot (a-ib)}\cdot (b+ia) = 5+4i$
So $\displaystyle (a+ib)+\frac{10i\cdot (a-ib)}{(a+ib)\cdot (a-ib)} = 5+4i\Rightarrow (a+ib)+\frac{10i}{(a+ib)}=5+4i$
Now Let $(a+ib)=z\;,$ Then equation is $\displaystyle z+\frac{10i}{z}=5+4i\Rightarrow z^2-(5+4i)z+10i=0$
Now Solving the above equation we get $\displaystyle z = \frac{(5+4i)\pm \sqrt{(5+4i)^2-40i}}{2}=\frac{(5+4i)\pm \sqrt{25-16}}{2}=\frac{(5+4i)\pm 3}{2}=4+2i\;,1+2i$
So $z=x+iy=4+2i\Rightarrow (x,y)=(4,2)$ and $z=x+iy=1+2i\Rightarrow (x,y)=(1,2)$
My Question is , can we solve it without Using Complex number
If yes, The please explain here
Thanks