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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 $ab$.

I don't understand how I can express this in the form $az + b \overline{z} = 38$. I know that the equation of a line in the complex plane is $z = u + t(v-u)$ where t is any real number.

David
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Forget complex numbers for a moment, and write down an equation in terms of $x,y$: the usual thing with point and slope. Then plug $$x=\frac12 (z+\bar z),\quad y=\frac{1}{2i}(z-\bar z)$$ to convert the equation into $z$ and $\bar z$.

Finally, multiply the equation by a constant to make the right hand side $38$.

  • So I plug those values into y= -1/6(x-7) + 2 ? – David Sep 06 '14 at 23:49
  • @David Yes. To make it easier, multiply by 12 first: $12y -2(x-7) = 24$. –  Sep 07 '14 at 00:00
  • How do I combine 6iz - 6ioverline(z) - z - overline(z) + 14 = 24? – David Sep 07 '14 at 00:08
  • Combine like terms as usual; $6iz-z = (6i-1)z$, and so on. Also, use $\bar z$ for $\bar z$. See math notation guide. –  Sep 07 '14 at 00:16
  • I got (z - $\bar z$)(6i-1) = 10. So technically a = b = 114/5i - 19/5. Is this correct? – David Sep 07 '14 at 00:25
  • No, does not look right. Let's backtrack: equation of line could be written in terms of $x,y$ as $x+6y=19$. Plug the complex form in: $(z+\bar z)/2 + 6 (z-\bar z)/(2i) = 19$. Multiply by $2$: $z+\bar z -6iz + 6i\bar z = 38$. Combine: $(1-6i)z + (1+6i)\bar z = 38$. –  Sep 07 '14 at 00:32
  • Oh wait I didn't realize that when u said to multiply by 12 first that took care of the 38 part. Anyways so a = b= (1+6i) right? – David Sep 07 '14 at 00:36
  • No. This is not what I got in my previous comment. Also, if you have $a=b$ or $a=-b$, that means either $x$ or $y$ cancel out, which is not the case. –  Sep 07 '14 at 00:38
  • Ohhhhhhh wait I see where I messed up anyways I got the answer to be 37 :) – David Sep 07 '14 at 00:41