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The equation of the line joining the complex numbers $-5 + 4i $ and $7 + 2i$ can be expressed in the form $az + b \overline{z} = 38 $ for some complex numbers a and b. Find the product ab.

Well if you let $-5 + 4i $ and $7 + 2i$ equal to $-5,4$ and $7,2$ we can easily find the line's equation...but how to express in the given form? Thx

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    $\begingroup$ You want $az+b\overline z=38$ to be true for $z=-5+4i$, and you also want it to be true for $z=7+2i$. Each of those gives you an equation in the two unknowns $a$ and $b$. $\endgroup$ – Gerry Myerson Jan 31 '14 at 23:11
  • $\begingroup$ oh thanks!!!!!!!!!!!!!!!!!!!!! $\endgroup$ – Math Dude Feb 1 '14 at 21:53
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    $\begingroup$ You're welcome. Now, please write up a full solution, and post it here as an answer. That helps clear up the Unanswered Questions list. $\endgroup$ – Gerry Myerson Feb 2 '14 at 12:11
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Substituting $z = -5 + 4i$ and $z = 7 + 2i$ in the given equation, we obtain the system of equations

$(-5 + 4i) a + (-5 - 4i) b =38$

$(7 + 2i) a + (7 - 2i) b = 38$

Subtracting these equations, we get $(12 - 2i) a + (12 + 2i) b = 0$, so $ b = -\frac{6 - i}{6 + i} a.$ Substituting into the first equation, we get $(-5 + 4i) a - (-5 - 4i) \cdot \frac{6 - i}{6 + i} a = 38.$ Solving for a, we find $a = 1 - 6i$, then $b = 1 + 6i$, so$ ab = (1 - 6i)(1 + 6i) = \boxed{37}. $

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    $\begingroup$ Good. Now, you are permitted (encouraged!) to accept your own answer to your question, by clicking in the check mark to the left of the answer. $\endgroup$ – Gerry Myerson Feb 4 '14 at 6:54

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