1© 2011 Pearson Education, Inc. All rights reserved 1© 2010 Pearson Education, Inc. All rights reserved

© 2011 Pearson Education, Inc. All rights reserved

Chapter 7

Applications of Trigonometric

Functions

OBJECTIVES

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Polar Form of Complex Numbers; DeMoivre’s TheoremSECTION 7.8

1

2Represent complex numbers geometrically.Find the absolute value of a complex number.Write a complex number in polar form.Find products and quotients of complex numbers in polar form.Use DeMoivre’s Theorem to find powers of a complex number.Use DeMoivre’s Theorem to find the nth roots of a complex number.

3

4

5

6

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GEOMETRIC REPRESENTATION OFCOMPLEX NUMBERS

The geometric representation of the complex number a + bi is the point P(a, b) in a rectangular coordinate system.

When a rectangular coordinate system is used to represent complex numbers, the plane is called the complex plane.

The x-axis is also called the real axis, because the real part of a complex number is plotted along the x-axis. Similarly, the y-axis is also called the imaginary axis.

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GEOMETRIC REPRESENTATION OFCOMPLEX NUMBERS

A complex number a + bi may be viewed as a position vector with initial point (0, 0) and terminal point (a, b).

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EXAMPLE 1 Plotting Complex Numbers

Plot each number in the complex plane.1 + 3i, –2 + 2i, –3, –2i, 3 – i

Solution

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ABSOLUTE VALUE OFA COMPLEX NUMBER

The absolute value (or magnitude or modulus) of a complex number z = a + bi is

.22 babiaz

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POLAR FORM OF A COMPLEX NUMBER

cos sin ,z r i

The complex number z = a + bi can be written in polar form

where a = r cos θ, b = r sin θ, and

When a nonzero complex number is written in polar form, the positive number r is the modulus or absolute value of z; the angle θ is called the argument of z (written θ = arg z).

2 2 ,r a b

.tanab

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EXAMPLE 4Writing a Complex Number in Rectangular Form

Write the complex number in rectangular form.Solution

The rectangular form of z is

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PRODUCT AND QUOTIENT RULES FOR TWO COMPLEX NUMBERS IN POLAR FORM

Let z1 = r1(cos + i sin ) andz2 = r2(cos θ2 + isin θ2) be two complex numbers in polar form. Then

and

21212121 sincos irrzz

.0 ,sincos 221212

1

2

1 zirr

zz

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EXAMPLE 5 Finding the Product and Quotient of Two Complex Numbers

Leave the answers in polar form.

Let z1 3 cos65ºi sin 65º and

z2 4 cos15º i sin15º . Find z1z2 and z1

z2

.

Solution

80sin80cos121565sin1565cos43

15sin15cos465sin65cos321

ii

iizz

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EXAMPLE 5 Finding the Product and Quotient of Two Complex Numbers

Solution continued

50sin50cos43

1565sin1565cos43

15sin15cos465sin65cos3

2

1

i

i

ii

zz

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DEMOIVRE’S THEOREM

Let z = r(cos + i sin) be a complex number in polar form. Then for any integer n,

cos sin .n nz r n i n

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EXAMPLE 7 Finding the Power of a Complex Number

Let z = 1 + i. Use DeMoivre’s Theorem to find each power of z. Write answers in rectangular form.

a. z16 b. z 10

SolutionConvert z to polar form. Find r and .

r a2 b2 12 12 2

tan ba

11

1 so 4

z 1 i 2 cos4

i sin4

,

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EXAMPLE 7 Finding the Power of a Complex Number

a. z16Solution continued

256012564sin4cos2

416sin

416cos2

4sin

4cos2

4sin

4cos2

8

1616

1616

ii

iz

iz

iz

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EXAMPLE 7 Finding the Power of a Complex Number

b. z 10Solution continued

ii

i

iz

iz

iz

32110

321

25sin

25cos

321

410sin

410cos2

4sin

4cos2

4sin

4cos2

1010

1010

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DEMOIVRE’S nth ROOTS THEOREM

The nth roots of a complex number w = r(cos + i sin ), where r > 0 and is in degrees, are given by

If is in radians, replace 360º with 2π in zk.

zk r1 n cos 360º k

n

i sin

360º kn

,

for k = 0, 1, 2, …, n – 1.

Let z and w be two complex numbers and let n be a positive integer. The complex number z is called an nth root of w if zn = w.

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EXAMPLE 8 Finding the Roots of a Complex Number

Find the three cube roots of 1 + i in polar form, with the argument in degrees.SolutionIn the previous example, we showed that

1 i 2 cos4

i sin4

1 i 2 cos 45º i sin 45º

.2 ,1 ,0 ,336045sin

336045cos2

3/1

kkikzk

Use DeMoivre’s Theorem with n = 3.

and

.

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Solution continued

EXAMPLE 8 Finding the Roots of a Complex Number

15sin15cos2

3036045sin

3036045cos2

6/1

6/10

i

iz

135sin135cos2

3136045sin

3136045cos2

6/1

6/11

i

iz

Substitute k = 0, 1, and 2 in the expression for zk and simplify to find the three cube roots.

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Solution continued

EXAMPLE 8 Finding the Roots of a Complex Number

1/61 2 cos135º sin135ºz i

1/60 2 cos15º sin15ºz i

1/62 2 cos255º sin 255ºz i

The three cube roots of 1 + i are as follows:

255sin255cos2

3236045sin

3236045cos2

6/1

6/12

i

iz