HMMT February

Algebra Difficulty 5.1 AIME, harder Find the answer

Find all ordered pairs $(a, b)$ of complex numbers with $a^{2}+b^{2} \neq 0, a+\frac{10b}{a^{2}+b^{2}}=5$ , and $b+\frac{10a}{a^{2}+b^{2}}=4$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

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.

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