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Name: yendor
Status: N/A
Age: N/A
Location: N/A
Country: N/A
Date: Around 1995

I heard that the roots to the equation x^5 - 1 = 0 can be represented graphically on a real and imaginary cartesian plane as five equally spaced points on a circle located at the origin of this plane and with a radius of

1. I was wondering how this is so, and why exactly?

It is hard to answer your question without knowing how much you know about the complex plane. But let us start with the equation x^2 = 1, meaning x * x = 1. Can you represent the solutions as two equally spaced points in the complex plane?


When talking about the complex plane, we measure the real part of a complex number z along one axis (say, the x axis) which we call the real axis, and the imaginary part [i.e., the coefficient of i = sqrt(-1)] along the other axis, which we call the imaginary axis; and we write z = x + i*y. Just as we can designate points in the plane by rectangular coordinates (x,y), we can also use polar coordinates (r, theta). Recall the relationship between these sets of coordinates: x = r*cos(theta), y = r*sin(theta). Substituting these into the expression for z, we get
     z = r*cos(theta) + i*r*sin(theta) = r*[cos(theta) + i*sin(theta)]. 

There is a result due to Euler which says
     e^(i*theta) = cos(theta) + i*sin(theta).  Thus, z = r*e^(i*theta). 

This is often called the polar form of the complex number z. Without loss of generality, we can require that r be non-negative. Note that the modulus of e^(i*theta) = sqrt[cos^2(theta) + sin^2(theta)] which equals 1 regardless of what theta is. Thus, the modulus of z is r. Now to your question. Note that z^5 = r^5 * e^(i * 5 * theta). {The point of going to the polar form is to exploit here the fact that
     [e^(i * theta)]^5 = e^(i * 5 * theta).}  We want to solve z^5 = 1, or
     r^5 * e^(i * 5 * theta) = 1 + 0*i.

First, take modulus of both sides. This gives r^5 = 1. Since r is real and non-negative, we must have r = 1. Putting that in, we now must solve e^(i * 5 * theta) = 1 + 0*i. Converting back to sines and cosines and matching up real and imaginary parts, we must solve cos(5 * theta) = 1 and sin(5 * theta) = 0. It suffices to solve only the cosine equation, since any solution to it will also solve the sine equation (but not conversely). Thus, we get 5 * theta = 2 * k * pi, where k is any integer. The five choices k = 0, 1, 2, 3, 4 give distinct solutions; thus, a complete solution set (5 points) is r = 1 and theta = 0, 2*pi/5, 4*pi/5, 6*pi/5, or 8*pi/5. Each of these points is a unit distance from the origin, and they form a regular pentagon. In general, the solution set of z^n = 1, where n = 3, 4, . . . , consists of n points, each of unit distance from the origin, forming a regular n-gon with z = 1 + 0*i being one of the points.


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