Complex Numbers (AMC 12A - #23)

OMFG COMPLEX NUMBERS!!

The solutions of the equation are the vertices of a convex polygon in the complex plane. What is the area of the polygon?


Wow, at first look, that looks really complicated. It's not. Things you should know:

A good understand of polar form
How to find roots of complex numbers

Ok, one thing should pop out immediately, and thats the 1-4-6-4-1 form of the equation. I hate solutions to problems where they said "it should be apparent that...". Well if it was so fucking apparent, then I would've gotten it right. Unfortunately, that's what this comes down to. The reason 1-4-6-4-1 is important is because those are the general coefficients of the binomial expansion .

Seeing the various negative signs and i coefficients, you should think of . Again, I hate saying that "you should think of", but once again, that's really what it comes down to. Actually expanding this is:





A little different from our actual equation, but we can easily alter to to fit it:







Now we can use De Moivre's Theorem to find the fourth roots of i + 1

First convert i + 1 to polar form:



Now find the four fourth roots:










So now what? We have a bunch of angles with which to attach a radius value to, AND once we convert all of them to rectangular, we have to subtract an i. Anybody can tell that the values are going to get very, very ugly. Luckily, we can take a shortcut. We only have to find the area of the polygon formed by the four solutions. That means:

The area will be unchanged if we translate it.
The area will be unchanged if we rotate it.

So how can we make our points easier to look at? Simple. First off, forget about subtracting i from each one. All that would do is move all of the points down one, which is translation, so it doesn't matter if we don't do it, we'll get the same area.

Second, we can subtract 11.25 from each of the angles. All this will do is rotate each of the points around the center of the polygon, keeping the area the same.

So now, the points of our polygon are:



Turn them into rectangular:



So now we have a square with its vertices on the axes. Using elementary properties of a square, we can easily solve for one of the sides of it, which is .

Squaring this, the area of our polygon is: