

Realimaginary 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?
jlu
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 nonnegative. 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
nonnegative, 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 ngon with z = 1 + 0*i being one of the
points.
rcwinther
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Update: June 2012

