## Monday, February 13, 2006

### Complex Stuff! Topic: Complex Numbers/Geometry. Level: AIME.

Introduction: First, I'm going to go over a cool property of complex numbers that helps us immensely with the following AIME problem. Given any complex number $z = a+bi = re^{i\theta}$ (make sure you understand all these forms; for more information, go here), we can rotate $z$ about the origin by an angle $\phi$ by multiplying by $e^{i\phi}$.

The easiest way to show this is by saying $ze^{i\phi} = (re^{i\theta})e^{i\phi} = re^{i(\theta+\phi)}$ from which we easily see that the angle of $z$ simply increases by $\phi$.

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Problem: (1994 AIME - #8) The points $(0,0)$, $(a,11)$, and $(b,37)$ are the vertices of an equilateral triangle. Find the value of $ab$.

Solution: At first glance, we're like AIME problem + analytic geometry = brute force! But then that doesn't turn out so nicely, so we remember that complex numbers are cool.

We know that in an equilateral triangle the angles are all $\frac{\pi}{6}$. And since the origin is one of our vertices, by rotating the first point (call it $A$) by $\frac{\pi}{6}$ counterclockwise should result in the second point (call it $B$). Note that it isn't the other way around by looking at a simple picture.

But by our little useful fact above, we rewrite $A = a+11i$ and $B = b+37i$ and thus by rotation $Ae^{i\frac{\pi}{6}} = (a+11i)\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right) = \frac{a-11\sqrt{3}}{2}+\frac{11+a\sqrt{3}}{2}i$ must equal $B = b+37i$.

Equating the real and imaginary parts, we have

$\frac{a-11\sqrt{3}}{2} = b$
$\frac{11+a\sqrt{3}}{2} = 37$

The second equation gives us $a = 21\sqrt{3}$ and plugging back into the first we have $b = 5\sqrt{3}$, giving us the answer $ab = 315$. QED.

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Comment: If you truly want to experience the elegance of the complex number solution, try using the distance formula to solve it; it's definitely not fun. In any case, the use of complex numbers can be generalized to many geometry problems - a lot of the time you can just put stuff on the complex plane and see what happens.

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Practice Problem: Using the complex plane, show that the sum of the $n$th roots of unity is zero.