Download complex polynomial function f

Sullivan Algebra and
Trigonometry: Section 5.6
Complex Zeros; Fundamental
Theorem of Algebra
Objectives
• Utilize the Conjugate Pairs Theorem to Find the
Complex Zeros of a Polynomial
• Find a Polynomial Function with Specified Zeros
• Find the Complex Zeros of a Polynomial
A variable in the complex number system is referred
to as a complex variable.
A complex polynomial function f degree n is a
complex function of the form
f ( x )  an x  an 1 x
n
n 1
 a1 x  a0
A complex number r is called a (complex) zero
of a complex function f if f (r) = 0.
Note that real numbers are also complex numbers
in the form a + bi where b = 0. So, this definition
of a complex polynomial function is a
generalization of what was previously introduced.
Fundamental Theorem of Algebra
Every complex polynomial function f (x) of
degree n > 1 has at least one complex zero.
Theorem
Every complex polynomial function f (x) 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.
Conjugate Pairs Theorem
Let f (x) be a complex polynomial 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.
Corollary
A complex polynomial f of odd degree with
real coefficients has at least one real zero.
Find a polynomial f of degree 4 whose coefficients are
real and has zeros 0, -2 and 1 - 3i. Graph f to verify
the solution.
f ( x )  a ( x  0)( x  ( 2 ))( x  (1  3i ))( x  (1  3i ))
f ( x )  x ( x  2)( x 2  (1  3i ) x  (1  3i ) x  (1  3i )(1  3i )
f ( x)  ( x2  2 x)( x2  x  3ix  x  3ix  1  3i  3i  9i 2 )
f ( x )  ( x 2  2 x )( x 2  2 x  10)
f ( x )  x  2 x  10 x  2 x  4 x  20 x
4
3
f ( x )  x  6 x  20 x
4
2
2
3
2
Find the complex zeros of the polynomial
f ( x )  2 x  3 x  40 x  137 x  58.
4
3
2
There are 4 complex zeros.
Using Descartes’ Rule of Sign, there are two,
or no, positive real zeros.
f (  x )  2 x  3x  40 x  137 x  58
4
3
2
Using Descartes’ Rule of Sign, there are two,
or no, negative real zeros.
Now, list all possible rational zeros p/q by factoring
the first and last coefficients of the function.
f ( x )  2 x 4  3 x 3  40 x 2  137 x  58.
p:  1,  2,  29,  58
q:  1,  2
p
1
29
:  1,  2,  29,  58,  , 
q
2
2
Now, begin testing each potential zero using
synthetic division. If a potential zero k is in fact a
zero, then x - k divides into f (remainder will be
zero) and is a factor of f.
f ( x )  2 x  3 x  40 x  137 x  58.
4
3
Test k = -2 - 2 2
2
-3
-4
-7
2
40
137 58
14 -108 - 58
54
29
0
Thus, -2 is a zero of f and x + 2 is a factor of f.
f ( x )  ( x  2)( 2 x 3  7 x 2  54 x  29)
-1
Test k = -1/2
2
2
2
-7
-1
-8
54
4
58
29
- 29
0
Thus, -1/2 is a zero of f and x + 1/2 is a factor of f.
2
1
f ( x )   x  2 x 
2 x  8 x  58
2



f ( x )   x  2 x  1

2

f ( x )  2 x  2  x  1
2
2
x
  8 x  58
2

2
x
  4 x  29
Now, find the complex zeros of the quadratic factor.
 ( 4)  ( 4) 2  4(1)( 29)
x
2(1)
4   100

2
4  10i

2
 2  5i
Zeros: - 2, - 1 , 2  5i , 2 + 5i
2