2

There exist two complex numbers $c$, say $c_1$ and $c_2$, so that $2 + 2i$, $5 + i$, and $c$ form the vertices of an equilateral triangle. Find the product $c_1 c_2$.enter image description here

So far, I have used the distance formula to get $(a−2)^2+(b−2)^2=(a−5)^2+(b−1)^2$ for $c_1(a,b)$. I expanded and simolified this to get $b=3a-9$. What do I do next?

Thanks

Rohini Sharma
  • 449
  • Following your approach, you know how long the sides of each equilateral triangle should be by looking at the distance between the vertices at 2+2i and 5+i – Winston Jul 07 '15 at 15:08
  • If any of the answers below were useful to you, then you should upvote (if you can) all answers you find useful and accept the one that was most useful to you. It is a way to show that you have found the answer to your question and it shows your appreciation. Now it seems like you still need help. If answers are not useful to you, then it helps if you say why not. This helps others to help you. For more information read this. – gebruiker Nov 10 '17 at 15:47

1 Answers1

3

Take the short path, instead. In the complex plane, a $\pm 60^\circ$ rotation is encoded by the multiplication by $\omega^{\pm 1}=\exp\left(\pm \frac{\pi i}{3}\right)$. That gives:

$$ c_1 = (2+2i)+\omega ((5+i)-(2+2i)), $$ $$ c_2 = (2+2i)+\omega^{-1} ((5+i)-(2+2i)),$$

so: $$ c_1 c_2 = (2+2i)^2 + ((5+i)-(2+2i))^2 + (2+2i)((5+i)-(2+2i)) $$ since $\omega+\omega^{-1}=1$. Simplifying, $c_1 c_2 = \color{red}{16+6i}.$

Jack D'Aurizio
  • 353,855
  • 1
    Interestingly, that expression for $c_1 c_2$ can be rearranged further into the form $$c_1 c_2 = (2+2i)^2-(2+2i)(5+i)+(5+i)^2,$$ which treats the two vertices symmetrically. – Semiclassical Jul 07 '15 at 16:07