Wednesday, January 20, 2010

A novel method for finding nth roots.

By dint of hard work, I have tonight come across a previously unknown (to me at least) method by which one can find higher order roots of equations of the form x to the nth power equals a constant. Since in several hours of Internet searching for just such a solution I did not come across anything comparable, I'm going to set it down here so I don't forget it later on.

The problem arose while I was working on my Complex Analysis homework. I needed to find the roots to the equation z to the 6th power equals 1. In other words, I needed to find 6 roots of 1, each of which, when raised to the 6th power, would equal 1. Now, I know that I learned procedures to do just that back in algebra, but I could not for the life of me remember them tonight. After some fruitless searching on the Internet, I came across a Wikipedia article on roots of one. It too did not contain what I was looking for, but it did contain some very useful pieces of information, among them that the n roots of one were always evenly spaced on the unit circle in the complex plane, and that they formed a regular n-sided polygon within the unit circle.

Now, if you remember complex numbers from algebra, you know that they can be written in the form \(a+bi\), where a and b are real numbers. However, if you do some playing around with very complicated math like infinite series, you can derive Euler's formula, which says that e to the \(i\times x\) power equals the cosine of x plus i times the sine of x, or, in symbols, \(e^{ix} = \cos(x) + i \sin(x)\). This allows complex numbers to be written in the equivalent polar form \(re^{ix}\), where r is the absolute distance (positive) from the origin to the point in question, and x is the angle from the real axis in the complex plane.

Since the vertices of the regular n-sided polygon formed by the roots of the equation you're looking at are evenly spaced, it's a trivial matter to quickly write down all the roots of the equation in polar form just by reading the values of the angles for the polar form off a well-drawn picture. You can then undertake the almost-as-simple process of converting them back into the more familiar rectangular form of \(a+bi\).

As an example, take my original problem of \(z^6 = 1\). The picture, hopefully, will make things a little clearer. There, I've marked the polygon that the 6 roots make inscribed on the unit circle. In the complex plane, the x-axis is referred as the "real" axis and the y-axis as the "imaginary" axis. With a number of the form \( a+bi\), a is the real part, b the imaginary part. When working the problem, start off by checking if 1 is a solution. In this case it is, so one vertex of the polygon must be at the point (1,0), and the coordinates of the rest of the vertices can be determined from there. With the polar form, only the angle is necessary, and I've marked the angles of all the vertices for you. Once you have the polar form, you can convert it to a rectangular form using the equations
Real part = \(r \cos(x)\)
Imaginary part = \(r \sin(x)\)
where \(r = 1\) since we're using a unit circle.

Anyway, to cut a long process short, since it's getting late, I will simply give you the 6 roots of the equation given first in their polar forms, derived simply by reading off the diagram, then in their more familiar rectangular forms, arrived at using the definitions just given.
\[z=1,e^{i\pi/3},e^{i2\pi/3},-1,e^{i4\pi/3},e^{i5\pi/3}\\
z=1,\frac{1+\sqrt{3}i}{2},\frac{-1+\sqrt{3}i}{2},-1,\frac{-1-\sqrt{3}i}{2},\frac{1-\sqrt{3}i}{2}\]When I have a bit more time, I'll work out a formal notation for the process, so it can be a bit more rigorous and useful to people encountering it for the first time. Until then, a hui hou!

Final musing: what do you think the chances are that this has probably been worked out hundreds of years ago already by much greater mathematicians than I? I'll ask my professor tomorrow...

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