*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.

That's ok Daniel, you'll do it ;)

ReplyDeleteThat has happened to me before but not even close to this level of mathematics O_o