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The Functions e^x and e^jθ: Roots of Unity and Related Topics

Module by: Louis Scharf. E-mail the author

Note:

This module is part of the collection, A First Course in Electrical and Computer Engineering. The LaTeX source files for this collection were created using an optical character recognition technology, and because of this process there may be more errors than usual. Please contact us if you discover any errors.

The complex number z=ej2π/Nz=ej2π/N is illustrated in Figure 1. It lies on the unit circle at angle θ=2π/Nθ=2π/N. When this number is raised to the nthnth power, the result is zn=ej2πn/Nzn=ej2πn/N. This number is also illustrated in Figure 1. When one of the complex numbers ej2πn/Nej2πn/N is raised to the NthNth power, the result is

(ej2πn/N)N=ej2πn=1.(ej2πn/N)N=ej2πn=1.
(1)
Figure 1: The Complex Numbers ej2π/Nej2π/N and ej2πn/Nej2πn/N
Figure one is a graph of a circle on a cartesian graph with two line segments from the origin to points on the circle. Above the circle is an expression that reads, (e^j2πn/N)^N = e^(j2πn) = 1. The first line segment begins at the origin and terminates at a point on the circle in the second quadrant of the graph. The point is labeled e^(j2πn/N). The second line segment begins at the origin and terminates at a point on the circle in the first quadrant of the graph. The point is labeled e^(j2π/N).

We say that ej2πn/Nej2πn/N is one of the NthNth roots of unity, meaning that ej2πn/Nej2πn/N is one of the values of z for which

z N - 1 = 0 . z N - 1 = 0 .
(2)

There are N such roots, namely,

e j 2 π n / N , n = 0 , 1 , ... , N - 1 . e j 2 π n / N , n = 0 , 1 , ... , N - 1 .
(3)

As illustrated in Figure 2, the 12th12th roots of unity are uniformly distributed around the unit circle at angles 2πn/122πn/12. The sum of all of the NthNth roots of unity is zero:

S N = n = 0 N - 1 e j 2 π n / N = 0 . S N = n = 0 N - 1 e j 2 π n / N = 0 .
(4)

This property, which is obvious from Figure 2, is illustrated in Figure 3, where the partial sums Sk=n=0k-1ej2πn/NSk=n=0k-1ej2πn/N are plotted for k=1,2,...,Nk=1,2,...,N.

Figure 2: Roots of Unity
Figure two is a circle with 12 line segments from the origin to points on the circle at every incremental 30-degree mark. Starting from the point horizontal and to the right at the 0-degree mark, the labeled points along the curve at the end of the line segments read as follows (in a counter clockwise direction): e^(j2π 0/12), e^(j2π/12), e^(j2π 2/12), e^(j2π 3/12), e^(j2π 4/12), e^(j2π 5/12), e^(j2π 6/12), e^(j2π 7/12), e^(j2π 8/12), e^(j2π 9/12), e^(j2π 10/12), e^(j2π 11/12).

These partial sums will become important to us in our study of phasors and light diffraction in "Phasors" and in our discussion of filters in "Filtering".

Figure 3: Partial Sums of the Roots of Unity
Figure three is a cartesian graph with a dodecagon in the first and second quadrants. The dodecagon is approximately 3.75 units in height. The left vertex of the base of the dodecagon is at the origin, and the base sits along the positive side of the horizontal axis. The angles of the dodecagon are shown to be measured as 2π/12. An arrow from the origin to the fifth vertex of the polygon in the top-right corner, and it is labeled S_5.

Geometric Sum Formula. It is natural to ask whether there is an analytical expression for the partial sums of roots of unity:

S k = n = 0 k - 1 e j 2 π n / N . S k = n = 0 k - 1 e j 2 π n / N .
(5)

We can imbed this question in the more general question, is there an analytical solution for the “geometric sum”

S k = n = 0 k - 1 z n ? S k = n = 0 k - 1 z n ?
(6)

The answer is yes, and here is how we find it. If z=1z=1, the answer is Sk=kSk=k. If z1z1, we can premultiply S k S k by z z and proceed as follows:

z S k = n = 0 k - 1 z n + 1 = m = 1 k z m = m=0k-1zm+zk-1 = S k + z k - 1 . z S k = n = 0 k - 1 z n + 1 = m = 1 k z m = m=0k-1zm+zk-1 = S k + z k - 1 .
(7)

From this formula we solve for the geometric sum:

S k = 1 - z k 1 - z z 1 k , z = 1 . S k = 1 - z k 1 - z z 1 k , z = 1 .
(8)

This basic formula for the geometric sum Sk is used throughout electromagnetic theory and system theory to solve problems in antenna design and spectrum analysis. Never forget it.

Exercise 1

Find formulas for Sk=n=0k-1ejnθSk=n=0k-1ejnθ and for Sk=n=0k-1ej2π/Nn.Sk=n=0k-1ej2π/Nn.

Exercise 2

Prove n=0N-1ej2πn/N=0n=0N-1ej2πn/N=0.

Exercise 3

Find formulas for the magnitude and phase of the partial sum Sk=n=0k-1ej2πn/NSk=n=0k-1ej2πn/N.

Exercise 4

(MATLAB) Write a MATLAB program to compute and plot the partial sum Sk=n=0k-1ej2πn/NSk=n=0k-1ej2πn/N for k=1,2,...,Nk=1,2,...,N. You should observe Figure 3.

Exercise 5

Solve the equation (z+1)3=z3(z+1)3=z3.

Exercise 6

Find all roots of the equation z3+z2+3z-15=0z3+z2+3z-15=0.

Exercise 7

Find c c so that (1+j)(1+j) is a root of the equation z17+2z15-c=0z17+2z15-c=0.

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