Additional Topics
in Trigonometry
Copyright © Cengage Learning. All rights reserved.
6
6.5
TRIGONOMETRIC FORM OF A COMPLEX NUMBER
Copyright © Cengage Learning. All rights reserved.
What You Should Learn
• Plot complex numbers in the complex plane and
find absolute values of complex numbers.
• Write the trigonometric forms of complex
numbers.
• Multiply and divide complex numbers written in
trigonometric form.
• Use DeMoivre’s Theorem to find powers of
complex numbers.
• Find nth roots of complex numbers.
3
The Complex Plane
4
The Complex Plane
Just as real numbers can be represented by points on the
real number line, you can represent a complex number
z = a + bi
as the point (a, b) in a coordinate plane (the complex
plane).
5
The Complex Plane
The horizontal axis is called the real axis and the vertical
axis is called the imaginary axis, as shown in Figure 6.44.
Figure 6.44
6
The Complex Plane
The absolute value of the complex number a + bi is
defined as the distance between the origin (0, 0) and the
point (a, b).
If the complex number a + bi is a real number (that is,
if b = 0), then this definition agrees with that given for the
absolute value of a real number
| a + 0i| =
= |a|.
7
Example 1 – Finding the Absolute Value of a Complex Number
Plot z = –2 + 5i and find its absolute value.
Solution:
The number is plotted in Figure 6.45.
Figure 6.45
It has an absolute value of
8
Trigonometric Form of a
Complex Number
9
Trigonometric Form of a Complex Number
We have learned how to add, subtract, multiply, and divide
complex numbers.
To work effectively with powers
and roots of complex numbers,
it is helpful to write complex
numbers in trigonometric form.
In Figure 6.46, consider the
nonzero complex number
a + bi.
Figure 6.46
10
Trigonometric Form of a Complex Number
By letting θ be the angle from the positive real axis
(measured counterclockwise) to the line segment
connecting the origin and the point (a, b), you can write
a = r cos θ
where
and
b = r sin θ
.
Consequently, you have
a + bi = (r cos θ) + (r sin θ)i
from which you can obtain the trigonometric form of a
complex number.
11
Trigonometric Form of a Complex Number
The trigonometric form of a complex number is also called
the polar form. Because there are infinitely many choices
for θ, the trigonometric form of a complex number is not
unique.
Normally, θ is restricted to the interval 0 ≤ θ < 2π, although
on occasion it is convenient to use θ < 0.
12
Example 2 – Writing a Complex Number in Trigonometric Form
Write the complex number z = –2 – 2
form.
i in trigonometric
Solution:
The absolute value of z is
and the reference angle θ ′ is given by
13
Example 2 – Solution
cont’d
Because tan(π /3) =
and because z = –2 – 2
i lies in
Quadrant III, you choose θ to be θ = π + π /3 = 4π /3.
So, the trigonometric form is
z = r (cos θ + i sin θ )
See Figure 6.47.
Figure 6.47
14
Multiplication and Division of
Complex Numbers
15
Multiplication and Division of Complex Numbers
The trigonometric form adapts nicely to multiplication and
division of complex numbers.
Suppose you are given two complex numbers
z1 = r1(cos θ1 + i sin θ1) and z2 = r2(cos θ2 + i sin θ2).
The product of z1 and z2 is given by
z1z2 = r1r2(cos θ1 + i sin θ1)(cos θ2 + i sin θ2)
= r1r2[(cos θ1 cos θ2 – sin θ1 sin θ2)
+ i(sin θ1 cos θ2 + cos θ1 sin θ2)].
16
Multiplication and Division of Complex Numbers
Using the sum and difference formulas for cosine and sine,
you can rewrite this equation as
z1z2 = r1r2[cos(θ1 + θ2) + i sin(θ1 + θ2)].
This establishes the first part of the following rule.
17
Multiplication and Division of Complex Numbers
Note that this rule says that to multiply two complex
numbers you multiply moduli and add arguments, whereas
to divide two complex numbers you divide moduli and
subtract arguments.
18
Example 4 – Multiplying Complex Numbers
Find the product z1z2 of the complex numbers.
Solution:
Multiply moduli
and add
arguments.
19
Example 4 – Solution
cont’d
and
are coterminal.
20
Example 4 – Solution
cont’d
You can check this result by first converting the complex
numbers to the standard forms z1 = –1 +
i and
z2 = 4
– 4i and then multiplying algebraically.
z1z2 = (–1 +
= –4
i)(4
– 4i)
+ 4i + 12i + 4
= 16i
21
Powers of Complex Numbers
22
Powers of Complex Numbers
The trigonometric form of a complex number is used to
raise a complex number to a power.
To accomplish this, consider repeated use of the
multiplication rule.
z = r(cos θ + i sin θ )
z2 = r(cos θ + i sin θ )r(cos θ + i sin θ )
= r 2(cos 2θ + i sin 2θ )
z3 = r 2(cos 2θ + i sin 2θ )r(cos θ + i sin θ )
= r 3(cos 3θ + i sin 3θ )
23
Powers of Complex Numbers
z4 = r 4(cos 4θ + i sin 4θ )
...
z5 = r 5(cos 5θ + i sin 5θ )
This pattern leads to DeMoivre’s Theorem, which is named
after the French mathematician Abraham DeMoivre.
24
Example 6 – Finding Powers of a Complex Number
Use DeMoivre’s Theorem to find (–1 +
)12.
Solution:
First convert the complex number to trigonometric form
using
and
So, the trigonometric form is
25
Example 6 – Solution
cont’d
Then, by DeMoivre’s Theorem, you have
26
Roots of Complex Numbers
27
Roots of Complex Numbers
Recall that a consequence of the Fundamental Theorem of
Algebra is that a polynomial equation of degree n has n
solutions in the complex number system.
So, the equation x6 = 1 has six solutions, and in this
particular case you can find the six solutions by factoring
and using the Quadratic Formula.
x6 – 1 = (x3 – 1)(x3 + 1)
= (x – 1)(x2 + x + 1)(x + 1)(x2 – x + 1)
=0
28
Roots of Complex Numbers
Consequently, the solutions are
and
Each of these numbers is a sixth root of 1. In general, an
nth root of a complex number is defined as follows.
29
Roots of Complex Numbers
To find a formula for an nth root of a complex number, let u
be an nth root of z, where
u = s(cos β + i sin β )
and
z = r(cos θ + i sin θ )
By DeMoivre’s Theorem and the fact that un = z, you have
sn(cos nβ + i sin nβ ) = r(cos θ + i sin θ ).
30
Roots of Complex Numbers
Taking the absolute value of each side of this equation, it
follows that sn = r. Substituting back into the previous
equation and dividing by r, you get
cos nβ + i sin nβ = cos θ + i sin θ.
So, it follows that
cos nβ = cos θ and sin nβ = sin θ.
Because both sine and cosine have a period of 2π, these
last two equations have solutions if and only if the angles
differ by a multiple of 2π.
31
Roots of Complex Numbers
Consequently, there must exist an integer k such that
By substituting this value of β into the trigonometric form
of u, you get the result stated below.
32
Roots of Complex Numbers
When k exceeds n – 1, the roots begin to repeat.
For instance, if k = n, the angle
is coterminal with θ /n, which is
also obtained when k = 0.
The formula for the nth roots
of a complex number z has
a nice geometrical interpretation,
as shown in Figure 6.48.
Figure 6.48
33
Roots of Complex Numbers
Note that because the nth roots of z all have the same
magnitude
, they all lie on a circle of radius
with
center at the origin.
Furthermore, because successive nth roots have
arguments that differ by 2π /n, the n roots are equally
spaced around the circle.
You have already found the sixth roots of 1 by factoring
and by using the Quadratic Formula. The n distinct nth
roots of 1 are called the nth roots of unity.
34
Example 8 – Finding the nth Roots of a Complex Number
Find the three cube roots of z = –2 + 2i.
Solution:
Because z lies in Quadrant II, the trigonometric form of z is
z = –2 + 2i
=
(cos 135° + i sin 135°).
By the formula for nth roots, the cube roots have the form
35
Example 8 – Solution
cont’d
Finally, for k = 0, 1, and 2, you obtain the roots
36
Example 8 – Solution
cont’d
See Figure 6.50.
Figure 6.50
37