On the method of finding roots of unity.

Remember that method of root-finding I worked out last night? Turns out it was an important part of an attempt to prove Fermat's Last Theorem, one of the most famous theorems in the history of mathematics. In other words, probably every mathematician worth their salt already knows about it. (Fermat's Last Theorm says that if you have the equation \(x^n + y^n = z^n\),  there are no integer values for x, y, and z (other than 0) that will solve the equation for \(n\ge3\). It was proposed in 1637, but not actually proven until 1995. Yes, that recently.) I found this out because it got brought up in a talk by our Mathematics department chair Dr. Lee that I attended this afternoon, complete with illustration on the board of the method I was using. I learned a few other useful tidbits of information about it while listening, so I'm going to set down a formal notation of the process here, before I forget it.

Start with your equation, \(z^n=\pm1\) Depending on whether 1 is positive or negative, the formulas for the roots take different forms. Since the roots will share several features in common, it is helpful to denote them by a common symbol. I'm going to use omega, \(\omega\), for variety (and 'cause Dr. Lee was using it).

In both cases the roots are given by \[\omega = e^{(i\pi k/n)}\] however, if \(z^n=+1\), the number k is given by \(k = 0, 2, 4, ... , (2n-2)\), whereas if \(z^n=-1\), k is given by \(k = 1, 3, 5,  ... , (2n-1)\).

I had originally thought that the evenness or oddness of n would effect the formulas, but upon looking over the examples I worked out, I see that it doesn't. The only thing that changes is the range of the number k. Note that one interesting thing happens if n is even and 1 is negative: all the roots will be complex numbers, which makes them very difficult to find without using an explicit method like this.

 Well, once again, it looks like I have managed to independently discover something that was already worked out long before me. Perhaps, someday, I will actually be able to produce something original.