Algebra of Complex Numbers, Revisited
1.5 The Algebra of Complex Numbers, Revisited
The real numbers are deficient in the sense that not all algebraic operations on them produce real numbers. Thus, for to make sense, we must consider the domain of complex numbers. Do complex numbers have this same deficiency? That is, if we are to make sense of expressions such as , must we appeal to yet another new number system? The answer to this question is no. In other words, any reasonable algebraic operation performed on complex numbers gives complex numbers. Later we show how to evaluate intriguing expressions such as . For now we only look at integral powers and roots of complex numbers.
The important players in this regard are
the exponential and polar forms of a non-zero complex
number . By
the laws of exponents (which, you recall, we have promised to prove
in Section 5.1) we
Example 1.15. Show
Solution. (Method 1): The binomial formula (Exercise 14 of Section 1.2) gives
(Method 2): Using iIdentity stated above and Example
Explore Solution 1.15.
Which method would you use if you were asked to compute ?
1.16. Evaluate .
Explore Solution 1.16.
Example. Evaluate in
Explore Extra Solution.
An interesting application of the laws of
exponents comes from putting the equation
into its polar form. Doing so gives
which is known as De Moivre's Formula, in honor of the French mathematician Abraham de Moivre (1667--1754).
Example 1.17. Use
DeMoivre's formula to show that .
Solution. If we let n=5 and use the binomial formula to expand the left side of De Moivre's formula, we obtain
The real part of right side of this expression is . Equating it to the real part of establishes the desired result. Furthermore, it can be shown that .
Explore Solution 1.17.
A key aid in determining roots of complex numbers is a corollary to the Fundamental Theorem of Algebra. We prove this theorem in Section 6.6. Our proofs must be independent of the conclusions we derive here because we are going to make use of the corollary now.
Theorem 1.4 (Fundamental Theorem of Algebra). If P(z) is a polynomial of degree n () with complex coefficients, then the equation has precisely n (not necessarily distinct) solutions.
Proof. Refer to Section 6.6.
1.18. Let and
find its zeros. This polynomial of degree 3 can be written
as . Hence
the equation has
solutions and . Thus,
in accordance with the Fundamental Theorem of Algebra, we have three
solutions, with being
Graphs of Arg[P(z)] and |P(z)| for .
Explore Solution 1.18.
The corollary implies that if we can find n distinct solutions to the equation (or ) , we will have found all the solutions. We begin our search for these solutions by looking at the simpler equation . Solving this equation will enable us to handle the more general one quite easily.
we first note an important condition that determines when two nonzero
complex numbers are equal. If we let and , then
(i.e., ) iff and ,
where k is an integer. That is, two complex numbers are equal iff their moduli agree and an argument of one equals an argument of the other to within an integral multiple of .
We now find all solutions to in two stages, with each stage corresponding to one direction in the iff part of the above relation. First, we show that if we have a solution to , then the solution must have a certain form. Second, we show that any quantity with that form is indeed a solution.
For the first stage, suppose that is a solution to . Putting the latter equation in exponential form gives , so we must have and . In other words, and , where k is an integer.
So, if is a solution to , then and must be true. This observation completes the first stage of our solution strategy. For the second stage, we note that if , and , then is indeed a solution to because . For example, if n=7 and k=3, then is a solution to because .
Furthermore, it is easy to verify that we
get n distinct solutions to
(and, therefore, all solutions) by
setting k=0,1,2,...,n-1. The solutions
for k=n,n+1,... merely repeat those
for k=0,1,..., because the arguments so
generated agree to within an integral multiple of . As
we stated in Section 1.1, the
n solutions can be expressed as
They are called the roots of unity.
When k=0 in the above equation, we get , which is a rather trivial result. The first interesting root of unity occurs when k=1, giving . This particular value shows up so often that mathematicians have given it a special symbol.
Definition 1.9 (Primitive
nth Root of Unity).
For any natural number n, the value
is called the primitive root of unity.
By De Moivre's formula, the
roots of unity can be expressed as
Geometrically, the roots of unity are equally spaced points that lie on the unit circle and form the vertices of a regular polygon with n sides.
Example 1.19. The
solutions to the equation
are given by the eight values
In Cartesian form, these solutions are . The primitive root of unity is
Figure 1.18 illustrates this result.
Figure 1.18 The eight eighth roots of unity.
Explore Solution 1.19.
The procedure for
easy to generalize in solving for
any nonzero complex number c. If and , then iff . But
this last equation is satisfied iff
, and , where k is an integer.
As before, we get n distinct solutions given by
Each of the above solutions can be considered an root of c. Geometrically, the roots of c are equally spaced points that lie on the circle and form the vertices of a regular polygon with n sides. Figure 1.19 illustrates the case for n=5.
Figure 1.19 The five solutions to the equation .
It is interesting to note that
any particular solution to the equation ,
then all solutions can be generated by multiplying by the
roots of unity. That is, the solution set
The reason for this is that if , then for any we have
and that multiplying a number by increases an argument of that number by , so that contain n distinct values.
Example 1.20. Find
all the cube roots of , i.e.
find all solutions to the equation .
for . The Cartesian forms of the solutions (shown in Figure 1.20) are
Figure 1.20 The point and its three cube roots .
Explore Solution 1.20.
Is the quadratic formula valid in the complex domain? The answer is yes, provided we are careful with our terms.
Formula). If , then
the solution set for z is
where by we mean all distinct square roots of the number inside the parenthesis.
Proof . The proof is left as an exercise for the reader.
Example 1.21. Find
all solutions to the equation .
Solution. The quadratic formula gives . As , we compute for and . In Cartesian form, this expression reduces to
Thus, our solution set is .
Explore Solution 1.21.
In Exercise 5 of Section 1.2 we asked you to show that a polynomial with nonreal coefficients must have some roots that do not occur in complex conjugate pairs. This last example gives an illustration of such a phenomenon.
Exercises for Section 1.5. The Algebra of Complex Numbers, Revisited
Roots of Cubic Equations
Roots of Quartic Equations
Complex Roots of Polynomials
The Next Module is
The Topology of Complex Numbers
Return to the Complex Analysis Modules
Return to the Complex Analysis Project
This material is coordinated with our book Complex Analysis for Mathematics and Engineering.
(c) 2012 John H. Mathews, Russell W. Howell