**for**

**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

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 .

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 .

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.

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 .

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 .

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**** **

**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