College Algebra with Modeling and Visualizaitions

Report
Chapter 3
Section 3.4
The Fundamental Theorem
of Algebra
The Fundamental Theorem

Complex Numbers and the Imaginary i

Definition:
The number x such that x2 = –1 is defined to be i

Note:
Since x =   –1 then i =  –1

and –i = – –1
iy
Standard Form
a + bi , b = 0
a + bi
Real Numbers
Complex
Numbers
a + bi , b ≠ 0
Imaginary
Numbers
● a + bi
x
Sometimes
b ≠ 0 and a = 0
The Complex
Plane
Note: a + bi is also written a + ib for real a and b
… especially when b is a radical or other functional form
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The Fundamental Theorem

Radical Expressions and Arithmetic
 The expression  –a
 –a =  –1  a = i  a

The sum and difference of complex numbers
(a + bi) ± (c + di) = (a ± c) ± (b ± d)i

Examples:
(3 + 4i) – (2 – 5i) = (3 – 2) + (4 + 5)i = 1 + 9i
(7 – 3i) + (2 – 5i) = (7 + 2) – (3 + 5)i = 9 – 8i

The product of complex numbers
(a + bi)(c + di) = ac + bdi2 + (ad + bc)i = (ac – bd) + (ad + bc)i

Example:
(3 + 4i)(2 – 5i) = 6 – 20i2 + (8i – 15i) = 26 – 7i
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The Fundamental Theorem

Complex Conjugates

Definition: a + bi and a – bi are a complex conjugate pair

Example: 7 + 3i and 7 – 3i are complex conjugates

Fact: The product of complex conjugates is always real
(a + bi )  (a – bi) = a2 + b2


Example: (7 + 3i)  (7 – 3i) = 72 + 32 = 49 + 9 = 58
The quotient of complex numbers
(ac + bd) + (bc – ad)i
a + bi
a + bi c – di
=
=
c + di
c2 + d2
c + di c – di
ac + bd
ac + bd (bc – ad)i
bc – ad
= 2
= 2
i
+
+ 2
c + d2
c + d2
c2 + d2
c + d2
(
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The Fundamental Theorem

Quotient Examples

1.
(15 + 65i) (1 – 2i)
15 + 130 + (65 – 30)i
15 + 65i
= (1 + 2i)(1 – 2i) =
12 + 22
1 + 2i
= 29 + 7i


2.
3.
3
3 ( –i )
=
= –3i
i
( i )( –i )
(–2 + i) (–2i)
–2 + i
(–2 + i) (1 – i )2
=
=
(1 + i )2 (1 + i )2 (1 – i )2 ((1 + i )(1 – i ))2
2 + 4i
=
( 2 )2
=
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1
+i
2
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The Fundamental Theorem

Quadratic Formula Examples

1. Solve: x2 – 2x + 1 = 0
x=
–b
±  b2 – 4ac
2a
=
Solution set: { 1 }

+2
± (–2)2 – 4(1)(1)
2(1)
=
2± 0
2
=1
Question: Is there and easier way ?
2. Solve: x2 – 4x – 5 = 0
x=
–b
±  b2 – 4ac
2a
=
+4
± (–4)2 – 4(1)(–5)
2(1)
=
4 ±  36
=
2
–1
5
Solution set: { –1, 5 }

3. Solve: x2 – 2x + 5 = 0
x=
–b
±  b2 – 4ac
2a
=
+2
± (–2)2 – 4(1)(5)
2(1)
=
2 ±  –16
= 1± 2i
2
Solution set: { 1 – 2i , 1 + 2i }
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The Fundamental Theorem

Complex Solutions of Quadratic Equations

Complex solutions occur in pairs

Each has a complex conjugate solution

Depends on discriminant:
x =
–b
±  b2 – 4ac
2a
WHY ?
The Quadratic Formula
b2 – 4ac = 0
One real solution
b2 – 4ac > 0
Two real solutions
b2 – 4ac < 0
Two complex (non-real) solutions
Note the  radical
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The Fundamental Theorem

The Fundamental Theorem of Algebra
A polynomial of degree n ≥ 1 has at least one complex zero

Consequence:
Every polynomial has a complete factorization

WHY ?
For polynomial f(x) there is some zero k1 so that
f(x) = (x – k1)Q1(x)

Then
Q1(x) = (x – k2)Q2(x)
so that
f(x) = (x – k1)(x – k2)Q2(x)

Applying the Fundamental Theorem n times
f(x) = (x – k1)(x – k2)...(x – kn)Cn
... where Cn is a non-zero constant
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The Fundamental Theorem

The Number of Zeros Theorem
A polynomial of degree n has at most n distinct zeros

Follows from the Fundamental Theorem factorization
f(x) = (x – k1)(x – k2)...(x – kn)Cn

If all the ki are distinct there are n distinct zeros
... otherwise one or more zeros are repeated

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Examples

1. f(x) = 5x(x + 3)(x – 1)(x – 4)

2. f(x) = 4x2(x – 2)3(x – 7)
How many distinct zeros ?
How many distinct zeros ?
Total zeros ?
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The Fundamental Theorem

The Conjugate Zeros Theorem

If f(x) is a polynomial with only real coefficients then its
complex zeros occur in conjugate pairs

Examples

1. f(x) = x4 + 5x2 – 36
Let u = x2
= u2 + 5u – 36
= (u – 4)(u + 9)
= (x2 – 4)(x2 + 9) = (x + 2)(x – 2)(x + 3i)(x – 3i)

Zeros ?
2. f(x) = x3 + x
= x(x2 + 1)
= x(x + i)(x – i)
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Zeros ?
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The Fundamental Theorem

Solving Equations Using Zeros




Factor polynomial
Set completely factored form equal to 0
Apply zero product property
Examples

1. f(x) = x4 + 5x2 – 36 = (x + 2)(x – 2)(x + 3i)(x – 3i) = 0
(x + 2) = 0 OR (x – 2) OR (x + 3i) = 0 OR (x – 3i) = 0
Solution set is { –2, 2, –3i, 3i }

2. f(x) = x3 + x = x(x + i)(x – i) = 0
x=0
OR
x+i=0
OR
x–i=0
Solution set is { 0, – i , i }
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Think about it !
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