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6965_CH06_pp455-518.qxd 1/14/10 1:55 PM Page 503
SECTION 6.6 De Moivre’s Theorem and nth Roots 503
6.6 De Moivre’s Theorem
and nth Roots
What you’ll learn about The Complex Plane
• The Complex Plane
• Trigonometric Form of Complex You might be curious as to why we reviewed complex numbers in Section P.6, then
Numbers proceeded to ignore them for the next six chapters. (Indeed, after this section we will
• Multiplication and Division pretty much ignore them again.) The reason is simply because the key to understand-
of Complex Numbers ing calculus is the graphing of functions in the Cartesian plane, which consists of two
• Powers of Complex Numbers perpendicular real (not complex) lines.
• Roots of Complex Numbers
... and why We are not saying that complex numbers are impossible to graph. Just as every real
number is associated with a point of the real number line, every complex number can
This material extends your be associated with a point of the complex plane. This idea evolved through the work
equation-solving technique to of Caspar Wessel (1745–1818), Jean-Robert Argand (1768–1822), and Carl Friedrich
include equations of the form Gauss (1777–1855). Real numbers are placed along the horizontal axis (the real axis)
n = c,
z n an integer and c a and imaginary numbers along the vertical axis (the imaginary axis), thus associating
complex number. the complex number a + bi with the point 1a, b2. In Figure 6.57 we show the graph of
2 + 3i as an example.
Imaginary axis EXAMPLE 1 Plotting Complex Numbers
a + bi Plot u = 1 + 3i, v = 2 - i, and u + v in the complex plane. These three points and
bi the origin determine a quadrilateral. Is it a parallelogram?
SOLUTION First notice that u + v = 11 + 3i2 + 12 + i2 = 3 + 2i. The num-
bers u, v, and u + v are plotted in Figure 6.58a. The quadrilateral is a parallelogram
because the arithmetic is exactly the same as in vector addition (Figure 6.58b).
Real Now try Exercise 1.
a axis
(a)
Imaginary axis y
Imaginary axis u = 1 + 3i u = 1, 3
2 + 3i u + v = 3, 2
3i u + v = 3 + 2i
Real x
Real O axis O
2 axis
v = 2 – i v = 2, –1
(b)
(a) (b)
FIGURE 6.57 Plotting points in the com- FIGURE 6.58 (a) Two numbers and their sum are plotted in the complex plane. (b) The
plex plane.
arithmetic is the same as in vector addition. (Example 1)
Is There a Calculus of Complex Example 1 shows how the complex plane representation of complex number addition is
Functions? virtually the same as the Cartesian plane representation of vector addition. Another
There is a calculus of complex functions. If you similarity between complex numbers and two-dimensional vectors is the definition of
study it someday, it should only be after acquir- absolute value.
ing a pretty firm algebraic and geometric under-
standing of the calculus of real functions.
6965_CH06_pp455-518.qxd 1/14/10 1:55 PM Page 504
504 CHAPTER 6 Applications of Trigonometry
DEFINITION Absolute Value (Modulus) of a Complex Number
The absolute value or modulus of a complex number z = a + bi is
ƒ zƒ = ƒa + biƒ = 2a2 + b2 .
In the complex plane, ƒa + biƒ is the distance of a + bi from the origin.
Trigonometric Form of Complex Numbers
Figure 6.59 shows the graph of z = a + bi in the complex plane. The distance r
from the origin is the modulus of z. If we define a direction angle u for z just as we
did with vectors, we see that a = r cos u and b = r sin u. Substituting these expres-
sions for a and b gives us the trigonometric form (or polar form) of the complex
Polar Form number z.
What’s in a cis?
Trigonometric (or polar) form appears frequently Imaginary axis
enough in scientific texts to have an abbreviated
form. The expression “cos u + i sin u” is often z = a + bi
shortened to “cis u” (pronounced “kiss u”). Thus
z = r cis u.
r b = r sin u
θ Real
a = r cos u axis
FIGURE 6.59 If ris the distance of z = a + bi from the origin and u is the directional an-
gle shown, then z = r1cos u + i sin u2, which is the trigonometric form of z.
DEFINITION Trigonometric Form of a Complex Number
The trigonometric form of the complex number z = a + bi is
z = r1cos u + i sin u2
where a = r cos u, b = r sin u, r = 2a2 + b2, and tan u = b/a. The number
r is the absolute value or modulus of z, and u is an argument of z.
An angle u for the trigonometric form of z can always be chosen so that 0 … u … 2p,
although any angle coterminal with u could be used. Consequently, the angle u and
argument of a complex number z are not unique. It follows that the trigonometric form
of a complex number z is not unique.
EXAMPLE 2 Finding Trigonometric Forms
Use an algebraic method to find the trigonometric form with 0 … u 6 2p for the
complex number. Approximate exact values with a calculator when appropriate.
(a) 1 - 13 i (b) -3 - 4i
SOLUTION
(a) For ,1 - 13i
2 2
r = ƒ1 - 13 iƒ = 2112 + 1132 = 2.
6965_CH06_pp455-518.qxd 1/14/10 1:55 PM Page 505
SECTION 6.6 De Moivre’s Theorem and nth Roots 505
Imaginary axis Because the reference angle u¿ for u is -p/3 (Figure 6.60),
θ u = 2p + a- pb = 5p .
Real 3 3
θ′ axis
Thus,
1 – 3i 1 - 13 i = 2 cos 5p + 2i sin 5p .
3 3
FIGURE 6.60 The complex number for (b) For ,-3 - 4i
Example 2a. 2 2
ƒ -3 - 4iƒ = 21-32 + 1-42 = 5.
The reference angle u¿ for u (Figure 6.61) satisfies the equation
Imaginary axis tan u¿=4 , so
θ 3
Real 4
axis -1
θ′ u¿=tan 3 = 0.927. Á
Because the terminal side of u is in the third quadrant, we conclude that
–3 – 4i u = p + u¿L4.07.
FIGURE 6.61 The complex number for Therefore,
Example 2b. -3 - 4i L 51cos 4.07 + i sin 4.072.
Now try Exercise 5.
Multiplication and Division
of Complex Numbers
The trigonometric form for complex numbers is particularly convenient for multiplying
and dividing complex numbers. The product involves the product of the moduli and the
sum of the arguments. (Moduli is the plural of modulus.) The quotient involves the quo-
tient of the moduli and the difference of the arguments.
Product and Quotient of Complex Numbers
Let . z = r 1cos u + i sin u and z = r 1cos u + i sin u 2 Then
1# 1 1 12 2 2 2 2
1. z z = r r 3cos 1u + u 2 + i sin 1u + u 24.
1 2 1 2 1 2 1 2
2. z1 = r1 3cos 1u - u 2 + i sin 1u - u 24, r Z 0.
z r 1 2 1 2 2
2 2
Proof of the Product Formula
z # z = r 1cos u + i sin u 2 # r 1cos u + i sin u 2
1 2 1 1 1 2 2 2
= r r 31cos u cos u - sin u sin u 2 + i 1sin u cos u + cos u sin u 24
1 2 1 2 1 2 1 2 1 2
= r r 3cos 1u + u 2 + i sin 1u + u 24
1 2 1 2 1 2
You will be asked to prove the quotient formula in Exercise 63.
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506 CHAPTER 6 Applications of Trigonometry
EXAMPLE 3 Multiplying Complex Numbers
Use an algebraic method to express the product of z1 and z2 in standard form.
Approximate exact values with a calculator when appropriate.
z = 2512acos -p + i sin -pb, z = 14acos p + i sin pb
1 4 4 2 3 3
SOLUTION
z # z = 2512acos -p + i sin -pb #14acos p + i sin pb
1 2 4 4 3 3
= 25# 1412 ccos a -p + pb + i sina -p + pbd
4 3 4 3
= 35012 acos p + i sin p b
12 12
L 478.11 + 128.11i Now try Exercise 19.
EXAMPLE 4 Dividing Complex Numbers
Use an algebraic method to express the product z /z in standard form. Approximate
1 2
exact values with a calculator when appropriate.
z1 = 2121cos 135° + i sin 135°2, z2 = 61cos 300° + i sin 300°2
SOLUTION 212 1cos 135° + i sin 135°2
z1 =
z2 61cos 300° + i sin 300°2
= 12 3cos 1135° - 300°2 + i sin 1135° - 300°24
3
= 12 3cos 1-165°2 + i sin 1-165°24
3
L-0.46 - 0.12i Now try Exercise 23.
Powers of Complex Numbers
We can use the product formula to raise a complex number to a power. For example, let
z = r1cos u + i sin u2. Then
z2 = z # z
= r 1cos u + i sin u2 # r 1cos u + i sin u2
Imaginary axis = r23cos 1u + u2 + i sin 1u + u24
= r21cos 2u + i sin 2u2
Figure 6.62 gives a geometric interpretation of squaring a complex number: Its argu-
z2 ment is doubled and its distance from the origin is multiplied by a factor of r, increased
if r 7 1 or decreased if r 6 1.
z 3 2
r2 θ We can find z by multiplying z by :z
2 r
θ Real z3 = z # z2
axis = r 1cos u + i sin u2 # r21cos 2u + i sin 2u2
= r33cos 1u + 2u) + i sin 1u + 2u24
= r31cos 3u + i sin 3u2
FIGURE 6.62 A geometric
interpretation of z2.
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