Transcript Complex Zeros

Precalculus
Complex Zeros
V. J. Motto
Introduction
We have already seen that an nth-degree
polynomial can have at most n real zeros.
• In the complex number system, an nth-degree
polynomial has exactly n zeros.
• Thus, it can be factored into exactly n linear factors.
• This fact is a consequence of the Fundamental
Theorem of Algebra, which was proved by
the German mathematician C. F. Gauss in 1799.
Fundamental Theorem of Algebra
The following theorem is the basis
for much of our work in:
• Factoring polynomials.
• Solving polynomial equations.
The Fundamental Theorem of Algebra
Every polynomial
P(x) = anxn + an-1xn-1 + . . . + a1x + a0
(n ≥ 0, an ≠ 0)
with complex coefficients has at least
one complex zero.
• As any real number is also a complex number,
it applies to polynomials with real coefficients too.
Fundamental Theorem of Algebra
The Fundamental Theorem of Algebra
and the Factor Theorem together show that
a polynomial can be factored completely
into linear factors—as we now prove.
Complete Factorization Theorem
If P(x) is a polynomial of degree n ≥ 1, then
there exist complex numbers a, c1, c2, . . . , cn
(with a ≠ 0) such that
P(x) = a(x – c1) (x – c2 ) . . . (x – cn)
Complete Factorization Theorem—Proof
By the Fundamental Theorem of Algebra,
P has at least one zero—which we will call c1.
By the Factor Theorem, P(x) can be factored
as:
P(x) = (x – c1) · Q1(x)
where Q1(x) is of degree n – 1.
Complete Factorization Theorem—Proof
Applying the Fundamental Theorem to
the quotient Q1(x) gives us the factorization
P(x) = (x – c1) · (x – c2) · Q2(x)
where:
• Q2(x) is of degree n – 2.
• c2 is a zero of Q1(x).
Complete Factorization Theorem—Proof
Continuing this process for n steps,
we get a final quotient Qn(x) of degree 0—
a nonzero constant that we will call a.
• This means that P has been factored as:
P(x) = a(x – c1)(x – c2) ··· (x – cn)
Complex Zeros
To actually find the complex zeros of
an nth-degree polynomial, we usually:
• First, factor as much as possible.
• Then, use the quadratic formula on parts
that we can’t factor further.
E.g. 1—Factoring a Polynomial Completely
Let P(x) = x3 – 3x2 + x – 3
(a) Find all the zeros of P.
(b) Find the complete factorization of P.
E.g. 1—Factoring Completely
Example (a)
We first factor P as follows.
P  x   x  3x  x  3
3
2
 x  3   x  3
2
  x  3   x  1
x
2
E.g. 1—Factoring Completely
Example (a)
We find the zeros of P by setting each
factor equal to 0:
P(x) = (x – 3)(x2 + 1)
• Setting x – 3 = 0, we see that x = 3 is a zero.
• Setting x2 + 1 = 0, we get x2 = –1; so, x = ±i.
• Thus, the zeros of P are 3, i, and –i.
E.g. 1—Factoring Completely
Example (b)
Since the zeros are 3, i, and i,
by the Complete Factorization Theorem,
P factors as:
P  x    x  3  x  i   x   i  
  x  3  x  i  x  i 
E.g. 2—Factoring a Polynomial Completely
Let P(x) = x3 – 2x + 4.
(a) Find all the zeros of P.
(b) Find the complete factorization of P.
E.g. 2—Factoring Completely
Example (a)
The possible rational zeros are
the factors of 4: ±1, ±2, and ±4.
• Using synthetic division, we find that –2 is
a zero, and the polynomial factors as:
P(x) = (x + 2) (x2 – 2x + 2)
E.g. 2—Factoring Completely
Example (a)
To find the zeros, we set each factor
equal to 0.
• Of course, x + 2 = 0 means x = –2.
• We use the quadratic formula to find
when the other factor is 0.
E.g. 2—Factoring Completely
Example (a)
x  2x  2  0
2
2 48
x
2
2  2i
x
2
x  1 i
• So, the zeros of P are –2, 1 + i, and 1 – i.
E.g. 2—Factoring Completely
Example (b)
Since the zeros are – 2, 1 + i, and 1 – i,
by the Complete Factorization Theorem,
P factors as:
P  x    x   2    x  1  i    x  1  i 
  x  2  x  1  i  x  1  i 
Zeros and Their Multiplicities
Zeros and Their Multiplicities
In the Complete Factorization Theorem,
the numbers c1, c2, . . . , cn are the zeros
of P.
• These zeros need not all be different.
• If the factor x – c appears k times in the complete
factorization of P(x), we say that c is a zero of
multiplicity k.
Zeros and Their Multiplicities
For example, the polynomial
P(x) = (x – 1)3(x + 2)2(x + 3)5
has the following zeros:
• 1(multiplicity 3)
• –2(multiplicity 2)
• –3(multiplicity 5)
Zeros and Their Multiplicities
The polynomial P has the same
number of zeros as its degree.
• It has degree 10 and has 10 zeros—provided
we count multiplicities.
• This is true for all polynomials—as we prove
in the following theorem.
Zeros Theorem
Every polynomial of degree n ≥ 1
has exactly n zeros—provided a zero
of multiplicity k is counted k times.
Zeros Theorem—Proof
Let P be a polynomial of degree n.
• By the Complete Factorization Theorem,
P(x) = a(x – c1)(x – c2) ··· (x – cn)
Zeros Theorem—Proof
Now, suppose that c is a zero of P
other than c1, c2, . . . , cn.
• Then,
P(c) = a(c – c1)(c – c2) ··· (c – cn)
=0
Zeros Theorem—Proof
Thus, by the Zero-Product Property,
one of the factors c – ci must be 0.
• So, c = ci for some i.
• It follows that P has exactly the n zeros
c 1, c 2, . . . , c n.
E.g. 3—Factoring a Polynomial with Complex Zeros
Find the complete factorization and
all five zeros of the polynomial
P(x) = 3x5 + 24x3 + 48x
E.g. 3—Factoring a Polynomial with Complex Zeros
Since 3x is a common factor,
we have:
4
2
P  x   3 x  x  8 x  16 

 3x x  4
2

2
• To factor x2 + 4, note that 2i and –2i
are zeros of this polynomial.
E.g. 3—Factoring a Polynomial with Complex Zeros
Thus, x2 + 4 = (x – 2i )(x + 2i ).
Therefore,
P  x   3 x  x  2i  x  2i  
 3 x  x  2i   x  2i 
2
2
2
• The zeros of P are 0, 2i, and –2i.
• Since the factors x – 2i and x + 2i each occur twice
in the complete factorization, the zeros 2i and –2i
are of multiplicity 2 (or double zeros).
• Thus, we have found all five zeros.
Factoring a Polynomial with Complex Zeros
The table gives further examples of
polynomials with their complete factorizations
and zeros.
E.g. 4—Finding Polynomials with Specified Zeros
(a) Find a polynomial P(x) of degree 4,
with zeros i, –i, 2, and –2 and with
P(3) = 25.
(b) Find a polynomial of degree 4,
with zeros –2 and 0, where –2 is a zero
of multiplicity 3.
Example (a)
E.g. 4—Specified Zeros
The required polynomial has the form
P  x   a  x  i   x   i    x  2   x   2  

 ax

 a x2  1 x2  4
4
 3x  4
2


• We know that P(3) = a(34 – 3 · 32 – 4) = 50a = 25.
• Thus, a = ½ .
• So, P(x) = ½x4 – 3/2x2 – 2
Example (b)
E.g. 4—Specified Zeros
We require:
Q  x   a  x   2  
3
 x  0
 a  x  2 x
3


 a x 3  6 x 2  12 x  8 x
(Special Product Formula 4, Section 1.4)

 a x 4  6 x 3  12 x 2  8 x

E.g. 4—Specified Zeros
Example (b)
We are given no information about Q other
than its zeros and their multiplicity.
So, we can choose any number for a.
• If we use a = 1, we get:
Q(x) = x4 + 6x3 + 12x2 + 8x
E.g. 5—Finding All the Zeros of a Polynomial
Find all four zeros of
P(x) = 3x4 – 2x3 – x2 – 12x – 4
• Using the Rational Zeros Theorem from
Section 3-3, we obtain this list of possible
rational zeros:
±1, ±2, ±4, ±1/3, ±2/3, ±4/3
E.g. 5—Finding All the Zeros of a Polynomial
Checking them using synthetic division,
we find that 2 and -1/3 are zeros, and we
get the following factorization.
P  x   3 x  2x  x  12x  4
4
3

2
  x  2 3 x  4 x  7 x  2
3
2

 3x  3x  6
 3  x  2   x  31   x 2  x  2 
  x  2  x 
1
3
2
E.g. 5—Finding All the Zeros of a Polynomial
The zeros of the quadratic factor
are:
1  1  8
1
7
x
 i
2
2
2
• So, the zeros of P(x) are:
1
1
7
1
7
2,  ,   i
,  i
3
2
2
2
2
Finding All the Zeros of a Polynomial
The figure shows the graph of the
polynomial P in Example 5.
• The x-intercepts
correspond to the real
zeros of P.
• The imaginary zeros
cannot be determined
from the graph.
Complex Zeros Come
in Conjugate Pairs
Complex Zeros Come in Conjugate Pairs
As you may have noticed from the examples
so far, the complex zeros of polynomials with
real coefficients come in pairs.
• Whenever a + bi is a zero, its complex
conjugate a – bi is also a zero.
Conjugate Zeros Theorem
If the polynomial P has real coefficients,
and if the complex number z is a zero of P,
then its complex conjugate z is also a zero
of P.
Conjugate Zeros Theorem—Proof
Let
P(x) = anxn + an-1xn-1 + . . . + a1x + a0
where each coefficient is real.
• Suppose that P(z) = 0.
• We must prove that P  z   0 .
Conjugate Zeros Theorem—Proof
We use the facts that:
• The complex conjugate of a sum of two
complex numbers is the sum of the conjugates.
• The conjugate of a product is the product
of the conjugates.
Conjugate Zeros Theorem—Proof


P z  an z
n

 an 1 z
n 1
     a1 z  a0
 an z n  an 1 z n 1      a1 z  a0
 an z  an 1z
n 1
     a1z  a0
 an z  an 1z
n 1
     a1z  a0
n
n
 P z  0  0
• This shows that z is also a zero of P(x),
which proves the theorem.
E.g. 6—A Polynomial with a Specified Complex Zero
Find a polynomial P(x) of degree 3
that has integer coefficients and zeros
½ and 3 – i.
• Since 3 – i is a zero, then so is 3 + i
by the Conjugate Zeros Theorem.
E.g. 6—A Polynomial with a Specified Complex Zero
That means P(x) has the form
P  x   a  x  21   x   3  i    x   3  i  
 a  x  21   x  3   i   x  3   i 
2

 a  x    x  3  i 2 


1
2
 ax 

1
2
x
2
 a x 3  132 x 2

 13 x  5 
 6 x  10
(Diff. of Squares Formula)
E.g. 6—A Polynomial with a Specified Complex Zero
To make all coefficients integers,
we set a = 2 and get:
P(x) = 2x3 – 13x2 + 26x – 10
• Any other polynomial that satisfies
the given requirements must be
an integer multiple of this one.
E.g. 7—Counting Real and Imaginary Zeros
Without actually factoring, determine how
many positive real zeros, negative real zeros,
and imaginary zeros this polynomial could
have:
P(x) = x4 + 6x3 – 12x2 – 14x – 24
E.g. 7—Counting Real and Imaginary Zeros
There is one change of sign.
• So, by Descartes’ Rule of Signs, P has
one positive real zero.
Also, P(–x) = x4 – 6x3 – 12x2 + 14x – 24
has three changes of sign.
• So, there are either three or one negative real
zero(s).
E.g. 7—Counting Real and Imaginary Zeros
So, P has a total of either four or two
real zeros.
• Since P is of degree 4, it has four zeros in all,
which gives the following possibilities.
Positive Real
Negative Real
Imaginary
1
3
0
1
1
2
Linear and Quadratic Factors
Linear and Quadratic Factors
We have seen that a polynomial factors
completely into linear factors if we use
complex numbers.
If we don’t use complex numbers,
a polynomial with real coefficients can always
be factored into linear and quadratic factors.
Linear and Quadratic Factors
A quadratic polynomial with no real
zeros is called irreducible over the real
numbers.
• Such a polynomial cannot be factored
without using complex numbers.
Linear and Quadratic Factors Theorem
Every polynomial with real coefficients
can be factored into a product of linear
and irreducible quadratic factors with
real coefficients.
Linear and Quadratic Factors Theorem—Proof
We first observe that, if c = a + bi
is a complex number, then
 x  c   x  c    x   a  bi   x   a  bi 
  x  a   bi   x  a   bi 
2
2
  x  a    bi 

 x 2  2ax  a 2  b 2

• The last expression is a quadratic with real
coefficients.
Linear and Quadratic Factors Theorem—Proof
If P is a polynomial with real coefficients, by
the Complete Factorization Theorem,
P(x) = a(x – c1)(x – c2) ··· (x – cn)
• The complex roots occur in conjugate pairs.
• So, we can multiply the factors corresponding to each
such pair to get a quadratic factor with real coefficients.
• This results in P being factored into linear and
irreducible quadratic factors.
E.g. 8—Factoring into Linear and Quadratic Factors
Let P(x) = x4 + 2x2 – 8.
(a) Factor P into linear and irreducible
quadratic factors with real coefficients.
(b) Factor P completely into linear factors
with complex coefficients
E.g. 8—Linear & Quadratic Factors Example (a)
P  x   x  2x  8
4

2

 x 2 x 4

2

2


 x 2 x 2 x 4
2

• The factor x2 + 4 is irreducible since
it has only the imaginary zeros ±2i.
E.g. 8—Linear & Quadratic Factors Example (b)
To get the complete factorization, we
factor the remaining quadratic factor.


 
  x  2  x  2   x  2i  x  2i 
P x  x  2 x  2 x  4
2