Download Section 5.6 – Complex Zeros: Fundamental Theorem of Algebra

132
Section 5.6 – Complex Zeros: Fundamental Theorem of
Algebra
Objective #0:
The Fundamental Theorem of Algebra.
We saw that in section 5.5, every non-constant polynomial function with
real coefficients can be uniquely factored into a product of linear and/or
irreducible quadratic factors in the real numbers. In this section, we will see
that every non-constant polynomial can be factored into a product of linear
factors in the complex numbers. We will also find all the complex zeros of a
polynomial. Before we do that, let's do a quick review of complex numbers:
A complex number is a number that contains a real number component and
an imaginary number component. If the imaginary component is zero, then
the complex number is just a real number. If the real component is zero
and the imaginary component is not zero, then the complex number is just
an imaginary number. If both the real component and the imaginary
component are not zero, then the complex number is neither a pure
imaginary nor a pure real number.
Definition of a Complex Number
A complex number is a number in the form a + bi where a and b are real
numbers and i = −1 . This form, a + bi is referred to as standard form.
{a + bi | a and b are real numbers and i = −1 }
Notes:
1)
If b€= 0, then the complex number is a real number.
2)
If b ≠ 0 and a = 0, then the complex number is an imaginary
€ imaginary number).
number (some books call this a pure
3)
The number a represents the real part of the complex number. The
number b represents the imaginary part of the complex number.
4)
The complex number a – bi is the complex conjugate of a + bi .
Identify the real number part and the imaginary part of each number:
Ex. 1a
– 7 + 6i
Ex. 1b
– 8.1i
Ex. 1c
9
Solution:
a)
The real part is – 7 and the imaginary part is 6.
b)
Since – 8.1i = 0 – 8.1i , the real part is 0 and the imaginary
part is – 8.1.
c)
Since 9 = 9 + 0i , The real part is 9 and the imaginary part is 0.
133
Definition
A complex polynomial function f with degree n is a function in the form:
f(x) = anxn + an – 1xn – 1 + … + a1x + a0
where a0, a1, … , an are complex numbers, an ≠ 0 and n is a whole number.
The variable x is a complex variable. A complex number r is a complex
zero of f if f(r) = 0.
Any irreducible quadratic equations in the real numbers can be factored in
the complex numbers. This suggests that every complex non-constant
polynomial has at least one complex zero. This is know as the:
Fundamental Theorem of Algebra
Every complex polynomial function with degree n ≥ 1 has at least one
complex zero.
Using this theorem in conjunction with the Factor Theorem leads to the
following theorem:
Theorem
Every complex polynomial function f of degree n ≥ 1 can be factored into n
linear factors (not necessarily distinct) of the form:
f(x) = an(x – r1)(x – r2)•…•(x – rn)
where an, r1, r2, …, rn are complex numbers. In other words, every nonconstant polynomial has exact n complex zeros including repeating zeros.
Proof:
Let f be a complex polynomial function with degree n ≥ 1. So, f is in the
form:
f(x) = anxn + an – 1xn – 1 + … + a1x + a0
By the Fundamental Theorem of Algebra, there exists at least one zero r1
of f. By the factor theorem, x – r1 is a factor and f can be rewritten as
f(x) = (x – r1)q1(x) where q1(x) is a complex function with degree
n – 1. Repeating this process n times we get:
f(x) = (x – r1)(x – r2)•…•(x – rn)qn(x) where q1(x) is a complex
function with degree n – n = 0 meaning that it is a constant. Since the
product of (x – r1)(x – r2)•…•(x – rn) has a leading coefficient of 1, then
qn(x) = an, meaning f(x) = an(x – r1)(x – r2)•…•(x – rn). This implies that every
complex polynomial of degree n ≥ 1 has exactly n zeros (not necessarily
distinct).
134
Objective #1:
Using the Conjugate Pairs Theorem
Complex Conjugates
If r = a + bi is a complex number, then r = a – bi is the complex conjugate.
Complex conjugate pairs have the following property:
r•r = (a + bi )(a – bi ) = a2 + b2
Thus, (5 + 3i )(5 – 3i ) = 25 – 15i + 15i – 9i 2 = 25 – 9(– 1) = 25 + 9 = 34.
r and r are called conjugate pairs.
Conjugate Pairs Theorem
Let f be a polynomial function whose coefficients are real numbers. If
r = a + bi is a zero of f, then the complex conjugate r = a – bi is also a zero
of f so long as b ≠ 0.
Proof:
Let f be a complex polynomial function in the form:
f(x) = anxn + an – 1xn – 1 + … + a1x + a0
where a1, a2, …, an are real numbers and an ≠ 0.
Let r = a + bi be a zero of f. This implies that f(r) = f(a + bi ) = 0, thus
f(r) = anrn + an – 1rn – 1 + … + a1r + a0 = 0
Take the conjugate of both sides:
a n rn + a n – 1 rn – 1 + … + a 1 r + a 0 = 0
a n rn + a n – 1 rn – 1 + … + a 1 r + a 0 = 0
an(r)n + an – 1(r)n – 1 + … + a1(r) + a0 = 0
an(r)n + an – 1(r)n – 1 + … + a1(r) + a0 = 0
(conjugate of the sum = sum
of the conjugates)
(conjugate of the product =
product of the conjugates)
(conjugate of a real number
= the real number)
This implies that f(r) = 0 which means r = a – bi is a zero of f.
We can now prove a couple of theorems we had from the last section:
Theorem
A polynomial function of odd degree and real coefficients has at least one
real zero.
Proof:
Since complex zeros of the form a + bi , where b ≠ 0 also occur as
conjugate pairs. Thus, if a + bi is a zero, then a – bi is also a zero. This
means that there will always be an even number of complex zeros with
b ≠ 0. Since the degree of f is odd, one of its zeros has to be real.
135
Theorem
Every non-constant polynomial function f with real coefficients can be
uniquely factored into a product of linear and/or irreducible quadratic
factors.
Proof:
Every non-constant complex polynomial of degree n can be written as the
product of n linear factors. Since the coefficients are real, the complex
zeros will occur in conjugate pairs where b ≠ 0. Hence r = a + bi is a
complex zero, then r = a – bi is also a zero provided b ≠ 0. This means that
(x – r) and (x – r) are factors of f. If we multiply these two factors together,
we get:
(x – r)(x – r) = x2 – (r + r)x + r•r
= x2 – (a + bi + a – bi )x + (a + bi )(a – bi ) = x2 – 2ax + a2 + b2
Since this second degree polynomial has real coefficients and has no real
zeros, it is irreducible over the real numbers. Hence, f can be uniquely
factored into a product of linear and/or irreducible quadratic factors.
Find the missing zeros of the polynomial with real coefficients with
the given zeros:
Ex. 2 Degree 5; Zeros: – 2, 3 – 5i , 4i
Solution:
Since the polynomial has a degree of 5, then it has five zeros.
The conjugate pair of 4i is – 4i which is also a zero and the
conjugate pair of 3 – 5i is 3 + 5i which is a zero as well. So, the
missing zeros are – 4i and 3 + 5i .
Objective #2
Finding a Complex Polynomial Function.
Find a polynomial function with the given information:
Ex. 3 Degree 5; Zeros: – 2, 3 – 5i , 4i
Solution:
From the last example, the zeros are – 2, 3 – 5i , 3 + 5i , – 4i , & 4i .
This means that (x – (– 2)), (x – (3 – 5i )), (x – (3 + 5i )), (x – (– 4i )),
& (x – (4i )) are factors. Thus, f can be written as:
f(x) = a(x – (– 2))(x – (3 – 5i ))(x – (3 + 5i ))(x – (– 4i ))(x – (4i ))
where a is any nonzero real number.
f(x) = a(x + 2)[x – (3 – 5i )][x – (3 + 5i )](x + 4i )(x – 4i ) (expand)
f(x) = a(x + 2)[x2 – (3 – 5i + 3 + 5i )x + (3 – 5i )(3 + 5i )](x2 – 16i 2)
136
f(x) = a(x + 2)[x2 – 6x + 9 – 25i 2](x2 + 16)
f(x) = a(x + 2)[x2 – 6x + 34](x2 + 16)
(multiply)
3
2
2
f(x) = a(x – 4x + 22x + 68)(x + 16) (multiply)
f(x) = a(x5 – 4x4 + 38x3 + 4x2 + 352x + 1088)
Objective #3:
Finding the Complex Zeros of a Polynomial Function.
Find all the complex zeros of the P(x). Then write P in factored form:
Ex. 4
P(x) = x4 + 2x3 + 22x2 + 50x – 75
Solution:
The possible values of p are ± 1, ± 3, ± 5, ± 15, ± 25, ± 75 and the
possible values of q are ± 1. Thus, the possible combinations of
p over q are ± 1, ± 3, ± 5, ± 15, ± 25, ± 75 which gives us the possible
rational zeros. Now, we will use synthetic division:
Let's try x = 1:
1
1
2
22
50 – 75
1
3
25
– 75
1
3
25
75
0
Yes
Try x = – 1
–1
1
1
3
–1
2
25
–2
23
75
– 23
52
No
Try x = 3: (boundness theorem says 3 would not work anyway)
3
1
3
25
75
3
18 129
1
6
43 204
No
Try x = – 3
–3
1
1
3
–3
0
25
0
25
75
– 75
0
Now, factor x2 + 25 in the complex numbers:
x2 + 25 = (x + 5i )(x – 5i )
Solving (x + 5i )(x – 5i ) = 0 yields x = ± 5i .
Thus, the zeros are {– 5i , 5i , – 3, 1}.
As a product of linear factors, we get:
P(x) = (x + 5i )(x – 5i )(x + 3)(x – 1)
Yes
137
Solve the following:
Ex. 5
f(x) = 2x4 + 5x3 + 9x2 + 2x – 4 = 0
Solution:
The possible values of p are ± 1, ± 2, ± 4 and the possible values of q
are ± 1, ± 2. Thus, the possible combinations of p over q are
1
± 1, ± , ± 2, ± 4.
2
Now, we will use synthetic division
Let's try x = 1:
1
2
5
9
2
€
2
7
16
2
7
16
18
Try x = – 1
–1
2
5
–2
3
9
–3
6
2
–6
–4
Try x = – 1 again
–1
2
3
–2
2
1
6
–1
5
–4
–5
–9
2
Try x =
–4
18
14
No
–4
4
0
Yes
No
1
2
1
2
€
2
3
6
–4
2
1
4
2
8
4
0
Yes
€ Working with the depressed equation 2x2 + 4x + 8 = 0, we can factor
out a 2 to get 2(x2 + 2x + 4) and then we will need to use the
quadratic formula on x2 + 2x + 4:
x=
−b± b2 −4ac
2a
=
−(2) ± (2)2 −4(1)(4)
2(1)
So, the solution is {– 1 –
€
3i,–1+
=
−2±
−12
2
3 i , – 1,
=–1±
1
}.
2
€
€ form, we would get:
€ write f(x) in factored
If we had to
f(x) = 2(x – (– 1 €
– 3 i ))(x –€(– 1 + 3€
i ))(x + 1)(x –
€
€
€
1
)
2
3i
138
Use the given zero to find the remaining zeros of the function:
Ex. 6
f(x) = 4x4 – 17x3 + 51x2 + 7x – 65; Zero: 2 – 3i
Solution:
Using synthetic division, we get:
2 – 3i
4
4
– 17
8 – 12i
– 9 – 12i
51
– 54 + 3i
– 3 + 3i
7
3 + 15i
10 + 15i
– 65
65
0
Since 2 – 3i is a zero of f(x), its complex conjugate, 2 + 3i, is also
a zero.
2 + 3i
4
4
– 9 – 12i
8 + 12i
–1
– 3 + 3i
– 2 – 3i
–5
10 + 15i
– 10 – 15i
0
Now, factor 4x2 – x – 5 to solve the depressed equation:
4x2 – x – 5 = 0
(4x – 5)(x + 1) = 0
5
x=
and x = – 1
4
So, the solution is {2 – 3i, 2 + 3i, – 1,
Ex. 7
5
4
}.
g(x) = x4 – 6x3 + 85x2 – 294x + 1764;
€
Solution:
€
Using synthetic division, we get:
– 7i
1
–6
85
– 7i
– 49 + 42i
1
– 6 – 7i
36 + 42i
Zero: – 7i
– 294
294 – 252i
– 252i
1764
– 1764
0
Since – 7i is a zero of f(x), its complex conjugate, 7i, is also a zero.
7i
1
– 6 – 7i
36 + 42i
– 252i
7i
– 42i
252i
1
–6
36
0
Working with the depressed equation x2 – 6x + 36 = 0, we will need to
use the quadratic formula:
x=
−b± b2 −4ac
2a
=
−(−6) ± (−6)2 −4(1)(36)
2(1)
=
6±
−108
2
So, the solution is {– 7i, 7i, 3 – 3 3 i, 3 + 3 3 i } .
€
€
€
€
€
€
=3±3 3i