Find all ordered pairs $(a, b)$ of complex numbers with $a^{2}+b^{2} \neq 0, a+\frac{10 b}{a^{2}+b^{2}}=5$, and $b+\frac{10 a}{a^{2}+b^{2}}=4$.
Solution 1. First, it is easy to see that $a b \neq 0$. Thus, we can write \[ \frac{5-a}{b}=\frac{4-b}{a}=\frac{10}{a^{2}+b^{2}} \] Then, we have \[ \frac{10}{a^{2}+b^{2}}=\frac{4 a-a b}{a^{2}}=\frac{5 b-a b}{b^{2}}=\frac{4 a+5 b-2 a b}{a^{2}+b^{2}} \] Therefore, $4 a+5 b-2 a b=10$, so $(2 a-5)(b-2)=0$. Now we just plug back in and get the four solutions: $(1,2),(4,2),\left(\frac{5}{2}, 2 \pm \frac{3}{2} i\right)$. It's not hard to check that they all work. Solution 2. The first equation plus $i$ times the second yields $5+4 i=a+b i+\frac{10(b+a i)}{a^{2}+b^{2}}=a+b i-\frac{10 i}{a+b i}$, which is equivalent to $a+b i=\frac{(5 \pm 3)+4 i}{2}$ by the quadratic formula. Similarly, the second equation plus $i$ times the first yields $4+5 i=b+a i-\frac{10 i}{b+a i}$, which is equivalent to $b+a i=\frac{4+(5 \pm 3) i}{2}$. Letting $\epsilon_{1}, \epsilon_{2} \in\{-1,1\}$ be the signs in $a+b i$ and $b+a i$, we get $(a, b)=\frac{1}{2}(a+b i, b+a i)-\frac{1}{2} i(b+a i, a+b i)=$ $\left(\frac{10+\left(\epsilon_{1}+\epsilon_{2}\right) 3}{4}, \frac{8+\left(\epsilon_{2}-\epsilon_{1}\right) 3 i}{4}\right)$. Comment. Many alternative approaches are possible. For instance, $\frac{5-a}{b}=\frac{4-b}{a} \Longrightarrow \quad b-2=$ $\epsilon \sqrt{(a-1)(a-4)}$ for some $\epsilon \in\{-1,1\}$, and substituting in and expanding gives $0=\left(-2 a^{2}+5 a\right) \epsilon \sqrt{(a-1)(a-4)}$. More symmetrically, we may write $a=\lambda(4-b), b=\lambda(5-a)$ to get $(a, b)=\frac{\lambda}{1-\lambda^{2}}(4-5 \lambda, 5-4 \lambda)$, and then plug into $a^{2}+b^{2}=10 \lambda$ to get $0=10\left(\lambda^{4}+1\right)-41\left(\lambda^{3}+\lambda\right)+60 \lambda^{2}=(\lambda-2)(2 \lambda-1)\left(5 \lambda^{2}-8 \lambda+5\right)$. $\fbox{(1,2),(4,2),\left(\frac{5}{2}, 2 \pm \frac{3}{2} i\right)}$.