First, it is easy to see that $ab \neq 0$. Thus, we can write $\frac{5-a}{b}=\frac{4-b}{a}=\frac{10}{a^{2}+b^{2}}$. Then, we have $\frac{10}{a^{2}+b^{2}}=\frac{4a-ab}{a^{2}}=\frac{5b-ab}{b^{2}}=\frac{4a+5b-2ab}{a^{2}+b^{2}}$. Therefore, $4a+5b-2ab=10$, so $(2a-5)(b-2)=0$. 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.