**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.

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