Higher Degree Functions and Complex Numbers rn this chapter you will study Functions in which the indepen­ dent variable. x, is cubed, or raised to a higher power. The '1lost significant mathematical ';oncept concerns values of x that 7lake y = O. To make sense out ')f this concept. you will make '.l.se of imaginary numbers. to .vhich you were introduced in ".:hapters 1 and 5. You will use '1igher degree functions as math­ 'matical models for such things 1S the bending of beams and the Jayload carried by airplanes. Deflection of diving board

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/

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/ Distance

514

Chapter to

~

Higher Degree Functions and Complex Numbers

_,_ _I

10-1

INTRODUCTION TO HIGHER DEGREE FUNCTIONS

You recall that a polynomial function is a function with a general equation of the form

y

a polynomial in x.

You studied linear (first degree) and quadratic (second degree) functions in Chapters 3 and 5. In this section you will explore the graph of a cubic function.

Objective: Discover by pointwise plotting what the graph of a cubic function looks like.

The exercise which follows is designed to allow you to accomplish this ob­ jective.

EXERCISE 10-1

1.

Plot a graph of the cubic function

f(x)

=

- 4x 2

3x + 2

in the domain -3 ~ x ~ 6. Do this by calculating the value off(x) for each integer value of x in the domain. For example,f(6) = 56. Then plot the resulting points, and connect them with a smooth curve. You may wish to use different scales for the two axes so that the graph fits conveniently onto the graph paper. Be careful with the

[0-2

Complex :--':umlwr Hc\'j('\\

arithmetic, since even one point out of place can make the graph look completely different! 2. From your graph, you should be able to answer the following ques­ tions: a. How many x-intercepts does a cubic function seem to have? b. How many vertices does a cubic function graph seem to have? c. A quadratic function graph (a parabola) has two x-intercepts and one vertex. How many x-intercepts and vertices would you ex­ pect in the graph of a quartic (fourth degree) function? 3. Plot graphs of

= x3

-

4x 2

-

3x + 18,

h(x) = x 3

-

4x 2

-

3x + 34.

g(x)

and

This is easy if you observe that g(x) = f(x) + 16, and h(x) = f(x) + 32, wheref(x) is defined in Problem 1, above. a. How do the graphs of g and h compare with the graph off? b. How many x-intercepts do the graphs of g and h have? c. What conclusion can you reach about the number of x-intercepts a cubic function can have?

10-2

CO:\IPLEX .'\L':\IBER REVIE\V

In the last section you explored the functions

f(x)

x3

4x 2

g(x)

x

3

2

h(x)

x3

4x -

4x 2

3x + 2, -

3x + 18,

and

3x + 34.

Each is a cubic function, and each equation differs only in the constant term. The graphs, as drawn by the program PLOT CUBIC on the accom­ panying disk, are shown in Figure 1O-2a. Each has the same shape, but is displaced 16 units from each other in the y-direction. This fact is reason­ able because

g(x) h(x)

+ 16, and g(x) + 16. f(x)

The graph of function f has three distinct x-intercepts. The graph of func­ tion g is raised just enough so that the two intercepts on the right merge into one. When the graph is raised still higher, as for function h, these two intercepts vanish! You will find that they have disappeared into the set of complex numbers!

)\ )

316

Chapter 10

Higher Degree Functions ami Complex ;\.'umtJers

x

'1/'-, :~ /

)

-20 f(x)=x 3

4x 2 -3x+2

g(X)=X3

4x 2 -3x+18

f(x)=x 3 -4x 2

3x+34

Figure 10-2a _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ In this section you will refresh your memory about complex numbers, to which you were exposed in Chapter 5. After you explore complex solu­ tions of quadratic equations in the next section, you will be ready to find out some general conclusions about the x-intercepts of cubic and higher de­ gree functions.

Objective: Be able to add, subtract. multiply, and divide complex numbers. An imaginary number is defined to be a square root of a negative real number. For instance,

vC3,

and

v=;;:

are imaginary numbers. To make these numbers fit into what you already know, it is customary to give a name to -v=1". The name is i (for "imaginary"). With this name, any imaginary number can be expressed in terms of i.

vCi7 = Y(-l)(l7) = v=Iw = iW = 4.1231

... i

Similarly,

Y-100

lOi,

-

v=3

- rV3,

and

v=; = i v:;;..

\0·2

Complex :'\1I1111)('r Ik\ie\\"

Since the symbol "Vn" means, "a number you can square and get n for the answer," the symbol i means "a number you can square and get - 1 for the answer." The number i is not a real number. If you square any real number, the answer is non-negative. However, i is just as "real" as any other number. Like fractions and negative numbers, it was invented by people to give answers to certain kinds of problem. The name "imaginary" was picked a long time ago when some people started using these num­ bers, because other people found it hard to imagine such numbers. The above discussion leads to the definitions that were first mentioned in Chapter 5.

DEFINITIONS

IMAGINARY AND COMPLEX NUMBERS 1. Unit Imaginary Number i is a number whose square is -1. That is,

or equivalently, i

=

v'=l

2. Imaginary Numbers in terms of i If x is a non-negative real number, then

3. Complex Number A complex number is a number of the form a + bi, where the real number a is called the real part of a + bi, the real number b is called the imaginary part of a + bi, and i is v=T. 4. Complex Conjugates The complex numbers a + bi and a - bi are called complex con­ jugates of each other.

Complex numbers can be plotted on a complex number plane. Figure 10­ 2b shows the complex number 4 + 3i. The real part, 4, is plotted as the abscissa. The imaginary part, 3 (a real number!), is plotted as the ordi­ nate.

,"')17

,:)18

Chapter iO

Iligher Degree Functions (lnd Complex :'\umbcrs

Imaginaries

5i 4i

3i

-

- - - - ' t 4 + 3;

\

2i

-5

-4 -3

-2

-1

I I

0

2

-i

3

4

5 Reals

-2i -3i -4i

-5i

Fi~ure

1O-2b _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __

Since the real-number line and the imaginary-number line both lie in the complex plane, you can conclude that the set of real numbers and the set of imaginary numbers are both subsets of the set of complex numbers. Since 0 lies on both the real-number line and on the imaginary-number line, you can also conclude that 0 is both a real number and an imaginary number!! With these definitions and conclusions in mind, you are ready to accom­ plish the objective of this section. 1. Powers of j:

By definition, j2 -1. To evaluate other integer powers of j, it is neces­ sary only to use the definition of exponentation with integer exponents:

;3 = ;2. i = -I . j =

-j

1)(- 1)

is = i· i4

=

i· 1 = i

· . j3 = j4

· . is

-i. 1.

=

i.

6

i = -1.

· . j1

= -i.

i8

1.

1()·2

Complcx :\umlwr I{('\'jcw

Note that any time the exponent is a multiple of 4, the power equals 1. Thus

because 76 is a multiple of 4. Suppose that you wish to evaluate i 39. Since 39 multiple of 4,

36

+

3, and 36 is a

A faster technique is to divide the exponent by 4, getting 9 with a remain­ der of 3. Throwaway the quotient, 9, and keep the remainder, 3. So i 39 j3, which equals - i. EXAMPLE I Evaluate i '066.

Solution: Divide 1066 by 4. The quotient is 266 (which is irrelevant!) and the re­ mainder is 2. So l.

•

2. Sums and Differences of Complex Numbers Complex numbers behave just like linear binomials when they are added or subtracted . EXAMPLE 2

If z, = 7 - 8i and a. z, + Z2. b. Z, Z2.

Z2

=2 +

Iii, find

Solution:

a.

ZI

+ Z2

= (7

b.

=7

- 8i

= 9

+

3i

z, -

Zz

= (7

+ (2 + 11i) + 2 + Iii

8i)

8i) - (2

= 7 - 8i - 2

5 - 19i

+

11 i)

IIi •

:-; I~)

520

Chapter 10

I Hgller Degree Functions ane! Complex :--';umlJers

3. Products of Complex Numbers Complex numbers also behave like linear binomials when you multiply them. The only extra thing to remember is that i 2 = 1. EXAMPLE 3

Multiply; (7 - 8i)(2

+ 11 i)

Solution:

8i)(2 + IIi)

(7

14 = 14

+ 6li

88i 2

+ 6li +

88

102

+

Multiply each term in one binomial by each term in the other.

•

61i

EXAMPLE 4 Multiply: (4

+ 90(4 - 90

Solution: (4

+

90(4

16 - 81i 2 16

9i) Middle term is zero

+ 81

•

= 97

Note that the complex numbers 4 + 9i and 4 9i are complex conjugates of each other. Their product is a real number. This fact has application in dividing by a complex number. CONCLUSION

PRODUCT OF COMPLEX CONJUGATES The product of two complex conjugates is a real number.

I

(a

+ bi)(a - bi)

= a2

+ b2

1

4. Quotients of Complex Numbers Dividing one complex number by another is a special case of transforming to simple radical form. You rationalize the denominator by multiplying by 1 in the form (conjugate of denominator) I (conjugate of denominator).

10-2

EXA~lPLE

complex .';umber Hl'Yi