Precalculus Polynomial & Rational – Part Two V. J. Motto

PrecalculusPolynomial & Rational – Part Two

V. J. Motto

Introduction

In Section 1-5, we saw that, if the

discriminant of a quadratic equation is

negative, the equation has no real solution.

• For example, the equation

x2 + 4 = 0

has no real solution.

Introduction

If we try to solve this equation, we get:

x2 = –4

So,

• However, this is impossible—since the square of any real number is positive.

• For example, (–2)2 = 4, a positive number.

• Thus, negative numbers don’t have real square roots.

4.x

Complex Number System

To make it possible to solve all quadratic

equations, mathematicians invented

an expanded number system—called

the complex number system.

Complex Number

First, they defined the new number

• This means i 2 = –1.

• A complex number is then a number of the form a + bi, where a and b are real numbers.

1i

Complex Number—Definition

A complex number is an expression

of the form

a + bi

where:• a and b are real numbers.• i 2 = –1.

Real and Imaginary Parts

The real part of this complex number is a.

The imaginary part is b.

• Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.

Real and Imaginary Parts

Note that both the real and

imaginary parts of a complex

number are real numbers.

E.g. 1—Complex Numbers

Here are examples of complex numbers.

3 + 4i Real part 3, imaginary part 4

½ – 2/3i Real part ½, imaginary part -2/3

6i Real part 0, imaginary part 6

–7 Real part –7, imaginary part 0

Pure Imaginary Number

A number such as 6i, which has

real part 0, is called:

• A pure imaginary number.

Complex Numbers

A real number like –7 can be thought

of as:

• A complex number with imaginary part 0.

Complex Numbers

In the complex number system, every

quadratic equation has solutions.

• The numbers 2i and –2i are solutions of x2 = –4 because:

(2i)2 = 22i2 = 4(–1) = –4 and

(–2i)2 = (–2)2i2 = 4(–1) = –4

Complex Numbers

Though we use the term imaginary here,

imaginary numbers should not be thought of

as any less “real”—in the ordinary rather than

the mathematical sense of that word—than

negative numbers or irrational numbers.

• All numbers (except possibly the positive integers) are creations of the human mind—the numbers –1 and as well as the number i.2

Complex Numbers

We study complex numbers as they complete

—in a useful and elegant fashion—our study

of the solutions of equations.

• In fact, imaginary numbers are useful not only in algebra and mathematics, but in the other sciences too.

• To give just one example, in electrical theory, the reactance of a circuit is a quantity whose measure is an imaginary number.

Arithmetic Operations on Complex Numbers

Complex numbers are added, subtracted,

multiplied, and divided just as we would

any number of the form a + b

• The only difference we need to keep in mind is that i2 = –1.

.c

Arithmetic Operations on Complex Numbers

Thus, the following calculations are valid.

(a + bi)(c + di)

= ac + (ad + bc)i + bdi2 (Multiply and collect

all terms)

= ac + (ad + bc)i + bd(–1) (i2 = –1)

= (ac – bd) + (ad + bc)i (Combine real and

imaginary parts)

Arithmetic Operations on Complex Numbers

We therefore define the sum,

difference, and product of complex

numbers as follows.

Adding Complex Numbers

(a + bi) + (c + di)

= (a + c) + (b + d)i

• To add complex numbers, add the real parts and the imaginary parts.

Subtracting Complex Numbers

(a + bi) – (c + di)

= (a – c) + (b – d)i

• To subtract complex numbers, subtract the real parts and the imaginary parts.

Multiplying Complex Numbers

(a + bi) . (c + di)

= (ac – bd) + (ad + bc)i

• Multiply complex numbers like binomials, using i 2 = –1.

E.g. 2—Adding, Subtracting, and Multiplying

Express the following in the form a + bi.

(a) (3 + 5i) + (4 – 2i)

(b) (3 + 5i) – (4 – 2i)

(c) (3 + 5i)(4 – 2i)

(d) i 23

E.g. 2—Adding

According to the definition, we add the real

parts and we add the imaginary parts.

(3 + 5i) + (4 – 2i)

= (3 + 4) + (5 – 2)i

= 7 + 3i

Example (a)

E.g. 2—Subtracting

(3 + 5i) – (4 – 2i)

= (3 – 4) + [5 – (– 2)]i

= –1 + 7i

Example (b)

E.g. 2—Multiplying

(3 + 5i)(4 – 2i)

= [3 . 4 – 5(– 2)] + [3(–2) + 5 . 4]i

= 22 + 14i

Example (c)

E.g. 2—Power

i 23 = i

22 + 1 = (i 2)11i

= (–1)11i

= (–1)i

= –i

Example (d)

Dividing Complex Numbers

Division of complex numbers is much like

rationalizing the denominator of a radical

expression—which we considered in

Section 1-2.

Complex Conjugates

For the complex number z = a + bi,

we define its complex conjugate

to be:

z a bi

Dividing Complex Numbers

Note that:

• So, the product of a complex number and its conjugate is always a nonnegative real number.

• We use this property to divide complex numbers.

2 2z z a bi a bi a b

Dividing Complex Numbers—Formula

To simplify the quotient

multiply the numerator and the denominator

by the complex conjugate of the denominator:

a bi

c di

2 2

a bi a bi c di

c di c di c di

ac bd bc ad i

c d

Dividing Complex Numbers

Rather than memorize this entire formula,

it’s easier to just remember the first step

and then multiply out the numerator and

the denominator as usual.

E.g. 3—Dividing Complex Numbers

Express the following in the form a + bi.

• We multiply both the numerator and denominator by the complex conjugate of the denominator to make the new denominator a real number.

3 5a

1 2

7 3b

4

i

i

i

i

E.g. 3—Dividing Complex Numbers

The complex conjugate of 1 – 2i is:

Example (a)

1 2 1 2i i

3 5 3 5 1 2

1 2 1 2 1 2

7 11 7 11

5 5 5

i i i

i i i

ii

E.g. 3—Dividing Complex Numbers

The complex conjugate of 4i is –4i.

Example (b)

7 3 7 3 4

4 4 4

12 28 3 7

16 4 4

i i i

i i i

ii

Square Roots of

Negative Numbers

Square Roots of Negative Numbers

Just as every positive real number r has two

square roots , every negative

number has two square roots as well.

• If -r is a negative number, then its square roots are , because:

and

and r r

i r

2 2i r i r r

2

2i r i r r

Square Roots of Negative Numbers

If –r is negative, then the principal

square root of –r is:

• The two square roots of –r are:

• We usually write instead of to avoid confusion with .

r i r

andi r i r

i b bibi

E.g. 4—Square Roots of Negative Numbers

a 1 1

b 16 16 4

c 3 3

i i

i i

i

Square Roots of Negative Numbers

Special care must be taken when

performing calculations involving square

roots of negative numbers.

• Although when a and b are positive, this is not true when both are negative.

a b ab

Square Roots of Negative Numbers

For example,

However,

• Thus,

22 3 2 3 6 6i i i

2 3 6

2 3 2 3

Square Roots of Negative Numbers

When multiplying radicals of negative

numbers, express them first in the form

(where r > 0)

to avoid possible errors of this type.

i r

E.g. 5—Using Square Roots of Negative Numbers

Evaluate

and express in the form a + bi. 12 3 3 4

12 3 3 4

12 3 3 4

2 3 3 3 2

6 3 2 3 2 2 3 3 3

8 3 3

i i

i i

i

i

Complex Roots of

Quadratic Equations

Complex Roots of Quadratic Equations

We have already seen that, if a ≠ 0,

then the solutions of the quadratic

equation ax2 + bx + c = 0

are:

2 4

2

b b acx

a

Complex Roots of Quadratic Equations

If b2 – 4ac < 0, the equation has no real

solution.

However, in the complex number system,

the equation will always have solutions.

• This is because negative numbers have square roots in this expanded setting.

E.g. 6—Quadratic Equations with Complex Solutions

Solve each equation.

(a) x2 + 9 = 0

(b) x2 + 4x + 5 = 0

E.g. 6—Complex Solutions

The equation x2 + 9 = 0 means x2 = –9.

So,

• The solutions are therefore 3i and –3i.

Example (a)

9 9 3x i i

E.g. 6—Complex Solutions

By the quadratic formula, we have:

• So, the solutions are –2 + i and –2 – i.

Example (b)

24 4 4 5 4 4

2 24 2

22 2

22

x

i

ii

E.g. 7—Complex Conjugates as Solutions of a Quadratic

Show that the solutions of the equation

4x2 – 24x + 37 = 0

are complex conjugates of each other.

E.g. 7—Complex Conjugates as Solutions of a Quadratic

We use the quadratic formula to get:

• So, the solutions are 3 + ½i and 3 – ½i.

• These are complex conjugates.

224 24 4 4 37

2 4

24 16 24 4 13

8 8 2

x

ii