Real Solutions Sum
Let $ (a_1, b_1), $ $ (a_2, b_2), $ $ \dots, $ $ (a_n, b_n) $ be the real solutions to
\begin{align*}
a + \frac{17a + 6b}{a^2 + b^2} &= 6, \\
b + \frac{6a - 17b}{a^2 + b^2} &= 0.\end{align*}Find $ a_1 + b_1 + a_2 + b_2 + \dots + a_n + b_n $.
Hint: Use complex numbers.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Multiplying the second equation by $ i $ and adding the first equation, we get
\[a + bi + \frac{17a + 6b + 6ai - 17bi}{a^2 + b^2} = 6.\]We can write
\begin{align*}
17a + 6b + 6ai - 17bi &= (17 + 6i)a + (6 - 17i)b \\
&= (17 + 6i)a - (17 + 6i)bi \\
&= (17 + 6i)(a - bi).\end{align*}Also, $ a^2 + b^2 = (a + bi)(a - bi), $ so
\[a + bi + \frac{(17 + 6i)(a - bi)}{(a + bi)(a - bi)} = 6.\]This simplifies to
\[a + bi + \frac{17 + 6i}{a + bi} = 6.\]Let $ z = a + bi, $ so
\[z + \frac{17 + 6i}{z} = 6.\]This becomes $ z^2 - 6z + (17 + 6i) = 0 $. By the quadratic formula,
\[z = \frac{6 \pm \sqrt{36 - 4(17 + 6i)}}{2} = \frac{6 \pm \sqrt{-32 - 24i}}{2} = 3 \pm \sqrt{-8 - 6i}.\]We want to find the square roots of $ -8 - 6i, $ so let
\[-8 - 6i = (u + vi)^2 = u^2 + 2uvi + v^2 i^2 = u^2 + 2uvi - v^2.\]Equating the real and imaginary parts, we get $ u^2 - v^2 = -8 $ and $ 2uv = -6, $ so $ uv = -3 $. Then $ v = -\frac{3}{u} $. Substituting, we get
\[u^2 - \frac{9}{u^2} = -8.\]Then $ u^4 + 8u^2 - 9 = 0, $ which factors as $ (u^2 - 1)(u^2 + 9) = 0 $. Hence, $ u = 1 $ or $ u = -1 $. If $ u = 1, $ then $ v = -3 $. If $ u = -1, $ then $ v = 3 $. Thus, the square roots of $ -8 - 6i $ are $ 1 - 3i $ and $ -1 + 3i $.
For the square root $ 1 - 3i, $
\[z = 3 + 1 - 3i = 4 - 3i.\]This gives the solution $ (a,b) = (4,-3) $.
For the square root $ -1 + 3i, $
\[z = 3 - 1 + 3i = 2 + 3i.\]This gives the solution $ (a,b) = (2,3) $.
The final answer is then $ 4 + (-3) + 2 + 3 = \boxed{6} $.
Freeflows