chapter 3 section 3.4 the fundamental theorem of algebra
TRANSCRIPT
Chapter 3
Section 3.4
The Fundamental Theorem of Algebra
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Complex Numbers and the Imaginary i Definition:
The number x such that x2 = –1 is defined to be i
Note:
Since
Standard Form
The Fundamental Theorem
–1 then i =
a + biComplex Numbers
a + bi , b = 0Real Numbers
a + bi , b ≠ 0Imaginary Numbers
x
iy
● a + bi
Note: a + bi is also written a + ib for real a and b … especially when b is a radical or other functional form
Sometimes b ≠ 0 and a = 0
The Complex Plane
and –1 –i = – –1 x =
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Radical Expressions and Arithmetic The expression
The sum and difference of complex numbers
(a + bi) ± (c + di) = (a ± c) ± (b ± d)i Examples:
(3 + 4i) – (2 – 5i)
(7 – 3i) + (2 – 5i)
The product of complex numbers
(a + bi)(c + di)
Example:
(3 + 4i)(2 – 5i)
The Fundamental Theorem
–a
–a –1 a = a i=
= (3 – 2) + (4 + 5)i = 1 + 9i
= (7 + 2) – (3 + 5)i = 9 – 8i
= ac + bdi2 + (ad + bc)i = (ac – bd) + (ad + bc)i
= 6 – 20i2 + (8i – 15i) = 26 – 7i
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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
The Fundamental Theorem
a + bi c + di
c – di c – di
=a + bi c + di
=(bc – ad)i +(ac + bd)
c2 + d2
= +ac + bd c2 + d2
(bc – ad)i c2 + d2
= + i ac + bd c2 + d2
( ) bc – ad c2 + d2
( )
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Quotient Examples
1.
2.
3.
The Fundamental Theorem
15 + 65i 1 + 2i
(15 + 65i) (1 + 2i)
(1 – 2i) (1 – 2i) =
15 + 130 + (65 – 30)i 12 + 22 =
= 29 + 7i
3i
=( i )( –i ) 3( –i )
= –3i
(1 + i )2 –2 + i
=(1 – i )2 (1 + i )2 (1 – i )2 (–2 + i)
=(1 – i ))2 ((1 + i )
(–2i) (–2 + i)
=( 2 )2 2 + 4i
=12
i+
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Quadratic Formula Examples 1. Solve: x2 – 2x + 1 = 0
2. Solve: x2 – 4x – 5 = 0
3. Solve: x2 – 2x + 5 = 0
The Fundamental Theorem
x =2ab2 – 4ac ±–b
= 1=2(1)
(–2)2 – 4(1)(1)±+2 =
2
0±2
Solution set: { 1 } Question: Is there and easier way ?
x =2ab2 – 4ac ±–b
=2(1)
(–4)2 – 4(1)(–5)±+4 =
2
36±4
Solution set: { –1, 5 }
= 5
–1
x =2ab2 – 4ac ±–b ± (–2)2 – 4(1)(5)
=2(1)
+2 =
2
–16 ±2 = 1± 2i
Solution set: { 1 – 2i , 1 + 2i }
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Complex Solutions of Quadratic Equations Complex solutions occur in pairs
Each has a complex conjugate solution
Depends on discriminant:
The Fundamental Theorem
WHY ?
x =2ab2 – 4ac ±–b
The Quadratic Formula
=b2 – 4ac 0
0b2 – 4ac >
0b2 – 4ac <
One real solution
Two real solutions
Two complex (non-real) solutions
Note the radical
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The Fundamental Theorem of Algebra
A polynomial of degree n ≥ 1 has at least one complex zero
Consequence:
Every polynomial has a complete factorization
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
The Fundamental Theorem
WHY ?
... where Cn is a non-zero constant
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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
Examples
1. f(x) = 5x(x + 3)(x – 1)(x – 4)
2. f(x) = 4x2(x – 2)3(x – 7)
The Fundamental Theorem
How many distinct zeros ?
How many distinct zeros ?Total zeros ?
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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
= u2 + 5u – 36
= (u – 4)(u + 9)
= (x2 – 4)(x2 + 9)
2. f(x) = x3 + x
= x(x2 + 1)
= x(x + i)(x – i)
The Fundamental Theorem
Let u = x2
= (x + 2)(x – 2)(x + 3i)(x – 3i) Zeros ?
Zeros ?
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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) = 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)
x = 0 OR x + i = 0 OR x – i = 0
Solution set is { 0, – i , i }
The Fundamental Theorem
= (x + 2)(x – 2)(x + 3i)(x – 3i) = 0
= 0
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Think about it !