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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 have

,
and
.

Example 1.15.  Show that    in two ways.

Solution.  (Method 1):  The binomial formula (Exercise 14 of Section 1.2) gives

(Method 2):  Using iIdentity stated above and Example 1.12 yields

Explore Solution 1.15.

Which method would you use if you were asked to compute ?

Example 1.16.  Evaluate  .

Solution.
Explore Solution 1.16.

Extra Example.  Evaluate    in two ways.
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.

Example 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 repeated roots.

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.

To solve   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

for   .

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 given by

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

for   .

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 solving    is 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

for   .

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 if    is any particular solution to the equation , then all solutions can be generated by multiplying by the various roots of unity.  That is, the solution set is

.

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  .

Solution.

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.

Theorem 1.5  (Quadratic 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

,
and
.

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.

DeMoivre's Theorem

Roots of Cubic Equations

Roots of Quartic Equations

Complex Roots of Polynomials

Quaternions

The Next Module is

The Topology of Complex Numbers

(c) 2012 John H. Mathews, Russell W. Howell