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for
1.4 The Geometry of Complex Numbers, Continued
In Secion
1.3 we saw that a complex number
could
be viewed as a vector in the xy-plane whose tail is at the origin and
whose head is at the point (x,y). A
vector can be uniquely specified by giving its magnitude (i.e., its
length) and direction (i.e., the angle it makes with the positive
x-axis). In this section, we focus on these two geometric aspects of
complex numbers.
Let r
be the modulus of z (i.e.,
r = |z|), and
let
be
the angle that the line from the origin to the complex number
z makes with the positive
x-axis. (Note: The
number
is
undefined if z=0). We make
the following definition.
Definition 1.9, (Polar
Representation). The identity
(1-27)
is known as a polar representation of z,
and the values r
and
are
called polar coordinates of z, and is
illustrated in Figure 1.11 (a).
![[Graphics:Images/ComplexGeometryContinuedMod_gr_6.gif]](complexgeometry2/ComplexGeometryContinuedMod/Images/ComplexGeometryContinuedMod_gr_6.gif)
Figure 1.11 Polar representation of complex numbers.
As Figure 1.11(b) shows, the angle
can
be any value for which the identities
hold
true. Given
, the
collection of all values of
for which
is
denoted by
. Formally,
we have the following definitions.
Example
1.7. If
, then
and
is
a polar representation of z. The
polar coordinates in this case are
,
and
.
Explore
Solution 1.7.
Definition 1.10, (Argument,
arg z). If
, we
denote
by
(1-28)
.
If
, we
say that
is an argument of z.
Notice that we write
as
opposed to
. This
is because
is
a set, and the designation
indicates that
belongs to that set. Notice also that,
if
, then
there exists some integer n such
that
(1-29)
.
Example 1.8. Since
,
we have
.
Explore
Solution 1.8.
Mathematicians have agreed to single out a
special choice of
. It
is that value of
for which
as the following definition indicates.
Definition 1.11, (The principal value of
argument, Arg z). Let
, be
a complex number. Then
(1-30)
, provided
and
.
If
, we
call
the argument of z.
Example
1.9.
.
Explore
Solution 1.9.
Remark. Clearly if
,
where
,
then
.
where
. Note
that, as with arg, arctan
is a set (as opposed to Arctan, which
is a number). We specifically identify
as a proper subset of
because
has period
,
whereas cos and sin
have period
. In
selecting the proper values for
, we must be careful in specifying the choices of
so that the poin z associated with
r and
lies in the appropriate quadrant.
Example
1.10. If
, then
and
.
It would be a mistake to use
as an acceptable value for
,
as the point z associated with
and
is in the first quadrant, whereas
is in the third quadrant. A correct choice for
is
. Thus
, and
,
where n is any integer. In
this case,
, and
.
Remark. Note
that
is
indeed a proper subset of
.
If
is
on the y-axis, it would be a mistake
to attempt to find
by
looking at
, as
, so
is undefined. We emphasize these special cases:
If
, then
, and
If
, then
.
Example
1.11. If
,
it would be a mistake to attempt to find
by
looking at
, as
, so
is undefined. In this case
, and
.
As we shall see in Section
2.4,
is a discontinuous function of z because it "jumps'' by an amount of
as z crosses the negative real
axis.
In Section
5.1 we define
for any complex number z. You
will see that this complex exponential has all the properties of real
exponentials that you studied in earlier mathematics
courses. That is,
, and
so forth. You will also see, amazingly, that if
,
then
(1-31)
.
We will establish this result rigorously
in Chapter 5, but there
is a plausible explanation we can give now.
If
has
the normal properties of an exponential, it must be
that
. Now,
recall from Calculus the values of three infinite series:
,
, and
.
Substituting
for
in the infinite series for
gives
. At
this point, our argument loses rigor because we have not talked about
infinite series of complex numbers, let alone whether such series
converge. Nevertheless, if we merely take the last series
as a formal expression and split it into two series according to
whether the index k is even
(k=2n) or odd (k=2n+1),
we get
![[Graphics:Images/ComplexGeometryContinuedMod_gr_119.gif]](complexgeometry2/ComplexGeometryContinuedMod/Images/ComplexGeometryContinuedMod_gr_119.gif)
Thus, it seems the only possible value for
is
. We
will use this result freely from now on, and, as stated, supply a
rigorous proof in Chapter
5.
If we set x=0 and let
take the role of y in the above
equation, we get a famous result known as Euler
Formula:
(1-32)
.
If
is a real number,
will be located somewhere on the circle with radius 1
centered at the origin. This assertion is easy to verify
because
(1-33)
.
Figure 1.12 illustrates the location of the points
for various values of
.
![[Graphics:Images/ComplexGeometryContinuedMod_gr_130.gif]](complexgeometry2/ComplexGeometryContinuedMod/Images/ComplexGeometryContinuedMod_gr_130.gif)
Figure
1.12 The location of
for various values of
.
Notice that, when
,
we get
, so
(1-34)
.
Euler
was the first to discover this relationship; it is
referred to as Euler's identity. It has been labeled by
many mathematicians as the most amazing relation in analysis---and
with good reason. Symbols with a rich history are
miraculously woven together---the constant used by
Hippocrates as early as Hippocrates
as early as 400 B.C.;
the base of the natural logarithms; the basic concepts of
addition (+) and equality (=); the foundational whole
numbers 0 and 1; and
,
the number that is the central focus of this book.
Euler's formula is of tremendous use in
establishing important algebraic and geometric properties of complex
numbers. You will see shortly that it enables you to
multiply complex numbers with great ease. It also allows you to
express a polar form of the complex number z in a more compact
way. Recall that if
and
, then
. Using
Euler's formula we can now write z in
its exponential
form:
(1-35)
.
Example 1.12. In
example 1.10 we investigated the polar form of
, and
saw that
and
. Now
we have
.
Explore
Solution 1.12.
Together with the rules for exponentiation
that we will verify in Chapter
5, the exponential form has interesting
applications. If
and
,
then
(1-36)
Figure 1.13 illustrates the geometric significance of this
equation.
![[Graphics:Images/ComplexGeometryContinuedMod_gr_152.gif]](complexgeometry2/ComplexGeometryContinuedMod/Images/ComplexGeometryContinuedMod_gr_152.gif)
Figure
1.13 The product of two complex numbers
.
We have already seen that the modulus of
the product is the product of the moduli; that is,
. The
above identity establishes that an argument of
is
an argument of
plus an argument of
. It
also answers the question posed at the end of Section
1.3 regarding why the product
was in a different quadrant than either
or
. It
further offers an interesting explanation as to why the product of
two negative real numbers is a positive real number. The
negative numbers, each of which has an angular displacement of
radians, combine to produce a product that is rotated to a point with
an argument of
radians,
coinciding with the positive real axis.
Using exponential form, if
,
we can write
, a
little more compactly as
(1-37)
.
Doing so enables us to see a nice relationship between the sets
,
,
and
.
Theorem
1.3. If
and
, then
as sets,
(1-38)
.
Before proceeding with the proof, we
recall two important facts about sets. First, to establish
the equality of two sets, we must show that each is a subset of the
other. Second, the sum of two sets is the sum of all
combinations of elements from the first and second sets,
respectively. In this case,
.
Proof Let
. Because
, it
follows that
. Hence,
there is some integer n such
that
. Further,
as
we
have
. Likewise,
gives
. But
if
, then
. This
result shows that
. Thus,
. The
proof that
is
left as an exercise.
Using the exponential
form,
, we
see that
. In
other words,
Recalling that
and
,
we also have
and
If z is in the first quadrant, the
positions of the numbers
are
as shown in Figure 1.14 when
. Figure
1.15 depicts the situation when
.
![[Graphics:Images/ComplexGeometryContinuedMod_gr_196.gif]](complexgeometry2/ComplexGeometryContinuedMod/Images/ComplexGeometryContinuedMod_gr_196.gif)
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Example
1.13. If
, then
and
. Therefore
![[Graphics:Images/ComplexGeometryContinuedMod_gr_201.gif]](complexgeometry2/ComplexGeometryContinuedMod/Images/ComplexGeometryContinuedMod_gr_201.gif)
the modulus is
, and
the argument is
.
Explore
Solution 1.13.
Example
1.14. Given
, compute
using
polar computations.
If
and
, then
representative polar forms for these numbers
are
and
. Hence
the modulus is
, and
the argument is
.
Explore
Solution 1.14.
Caveat. The
formula
does not hold for all complex numbers
. Whereas,
the formula
does
hold for all complex numbers
,
and is left for the reader to verify.
Exercises for Section 1.4. The Geometry of Complex Numbers, Continued
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(c) 2006 John H. Mathews, Russell W. Howell