Real-imaginary plane ```Name: yendor Status: N/A Age: N/A Location: N/A Country: N/A Date: Around 1995 ``` Question: 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? Replies: 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? jluWhen 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. rcwinther Click here to return to the Mathematics Archives

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