Complex Pair Sum
Let $ (a_1,b_1), $ $ (a_2,b_2), $ $ \dots, $ $ (a_n,b_n) $ be all the ordered pairs $ (a,b) $ of complex numbers with $ a^2+b^2\neq 0, $
\[a+\frac{10b}{a^2+b^2}=5, \quad \text{and} \quad b+\frac{10a}{a^2+b^2}=4.\]Find $ a_1 + b_1 + a_2 + b_2 + \dots + a_n + b_n $.
- 1
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- 5
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- $\frac{a}{b}$
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- 0
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- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
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- $\sin{}$
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- $($
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- $\cap$
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- $\infty$
Solution
If $ a = 0, $ then $ \frac{10}{b} = 5, $ so $ b = 2, $ which does not satisfy the second equation. If $ b = 0, $ then $ \frac{10}{a} = 4, $ so $ a = \frac{5}{2}, $ which does not satisfy the first equation. So, we can assume that both $ a $ and $ b $ are nonzero.
Then
\[\frac{5 - a}{b} = \frac{4 - b}{a} = \frac{10}{a^2 + b^2}.\]Hence,
\[\frac{5b - ab}{b^2} = \frac{4a - ab}{a^2} = \frac{10}{a^2 + b^2},\]so
\[\frac{4a + 5b - 2ab}{a^2 + b^2} = \frac{10}{a^2 + b^2},\]so $ 4a + 5b - 2ab = 10 $. Then $ 2ab - 4a - 5b + 10 = 0, $ which factors as $ (2a - 5)(b - 2) = 0 $. Hence, $ a = \frac{5}{2} $ or $ b = 2 $.
If $ a = \frac{5}{2}, $ then
\[\frac{5/2}{b} = \frac{10}{\frac{25}{4} + b^2}.\]This simplifies to $ 4b^2 - 16b + 25 = 0 $. By the quadratic formula,
\[b = 2 \pm \frac{3i}{2}.\]If $ b = 2, $ then
\[\frac{2}{a} = \frac{10}{a^2 + 4}.\]This simplifies to $ a^2 - 5a + 4 = 0, $ which factors as $ (a - 1)(a - 4) = 0, $ so $ a = 1 $ or $ a = 4 $.
Hence, the solutions are $ (1,2), $ $ (4,2), $ $ \left( \frac{5}{2}, 2 + \frac{3i}{2} \right), $ $ \left( \frac{5}{2}, 2 - \frac{3i}{2} \right), $ and the final answer is
\[1 + 2 + 4 + 2 + \frac{5}{2} + 2 + \frac{3i}{2} + \frac{5}{2} + 2 - \frac{3i}{2} = \boxed{18}.\]
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