Transcript No Slide Title

PRECALCULUS I
Complex Numbers
Real numbers + Imaginary numbers
Dr. Claude S. Moore
Danville Community College
1
Definition of
1
The square root of a negative real
number is not a real number.
Thus, we introduce imaginary
numbers by letting
i = 1
So i 2 = -1, i 3 = - i , and i 4 = +1.
Example: Simplifying i n
Since i 2 = -1, i 3 = - i , and i 4 = +1,
a simplified answer should contain
no exponent of i larger than 1.
Example: i 21 = i 20 i 1 = (+1)( i ) = i
35
32
3
Example: i = i i = (+1)( - i ) = - i
NOTE: 21/4 = 5 with r = 1 and
35/4 = 8 with r = 3.
Definition of Complex Number
For real numbers a and b,
the number
a + bi
is a complex number.
If a = 0 and b  0, the complex
number bi is an imaginary number.
Equality of Complex Numbers
Two complex numbers
a + bi and c + di,
written in standard form,
are equal to each other
a + bi = c + di
if and only if (iff) a = c and b = d.
Example: Equality
If (a + 7) + bi = 9  8i, find a and b.
Since a + bi = c + di
if and only if (iff) a = c and b = d,
a + 7 = 9 and b = -8.
Thus, a = 2 and b = -8.
Addition & Subtraction:
Complex Numbers
Two complex numbers
a + bi and c + di
are added (or subtracted) by adding (or
subtracting) real number parts and
imaginary coefficients, respectively.
(a + bi ) + (c + di ) = (a + c) + (b + d )i
(a + bi )  (c + di ) = (a  c) + (b  d )i
Example: Addition & Subtraction
(3 + 2i ) + (-7 - 5i )
= (3 + -7) + (2 + -5 )I
= -4 - 3i
(-6 + 9i ) - (4 - 3i )
= (-6 - 4) + (9 + 3 )i
= -10 + 12i
Complex Conjugates
Each complex number of the form
a + bi
has a conjugate of the form
a  bi .
NOTE: The product of a complex number
and its conjugate is a real number.
(a + bi )(a  bi ) = a2 + b2.
Example: Complex Conjugates
The conjugate of -5 + 6i is -5 - 6i
The conjugate of 4 + 3i is 4 - 3i
Recall: The product of a complex number
and its conjugate is a real number.
(a + bi )(a  bi ) = a2 + b2.
(-5 + 6i )(-5 - 6i ) = (-5)2 + (6)2
= 25 + 36 = 41
Principal Square Root of Negative
If a is a positive number, the
principal square root of the
negative number -a is defined as
 a  a i
Example:
 16  16  i  4i
PRECALCULUS I
Fundamental Theorem
of Algebra
Dr. Claude S. Moore
Danville Community College
12
The Fundamental Theorem
If f (x) is a polynomial of
degree n, where n > 0,
then f has at least one root (zero)
in the complex number system.
Linear Factorization Theorem
If f (x) is a polynomial of degree n
n
f ( x)  an x + an 1 x
n 1
+    + a0
where n > 0, then f has precisely n
linear factors in the complex
number system.
Linear Factorization continued
n
n 1
f ( x)  an x + an 1 x
+    + a0
f ( x)  an ( x  c1 )( x  c2 )    ( x  cn )
where c1, c2, … , cn are complex
numbers and an is leading
coefficient of f(x).
Complex Roots in Conjugate Pairs
Let f(x) be a polynomial function
with real number coefficients.
If a + bi, where b  0,
is a root of the f(x),
the conjugate a - bi
is also a root of f(x).
Factors of a Polynomial
Every polynomial of degree n > 0
with real coefficients can be
written as the product of linear
and quadratic factors with real
coefficients where the quadratic
factors have no real roots.
PRECALCULUS I
RATIONAL
FUNCTIONS
Dr. Claude S. Moore
Danville Community College
RATIONAL FUNCTIONS
FRACTION OF TWO
POLYNOMIALS
p(x)
f(x) 
q(x)
DOMAIN
DENOMINATOR
CAN NOT
EQUAL ZERO
p(x)
f(x) 
where q(x)  0
q(x)
ASYMPTOTES
• VERTICAL LINE
x = a if
f ( x)  
or
f ( x )  
as x  a
• HORIZONTAL
LINE y = b if
f ( x)  b
as
x
or
x  
ASYMPTOTES OF
RATIONAL FUNCTIONS
n
p(x) an x + 
f(x) 

q(x) bm x m + 
If q(x) = 0, x = a is VERTICAL.
HORIZONTALS:
If n < m, y = 0.
If n = m, y = an/bm.
NO HORIZONTAL: If n > m.
SLANT ASYMPTOTES OF
RATIONAL FUNCTIONS
n
p(x) an x + 
f(x) 

q(x) bm x m + 
If n = m + 1, then slant
asymptote is y = quotient when
p(x) is divided by q(x) using
long division.
GUIDELINES FOR GRAPHING
1. Find f(0) for y-intercept.
2. Solve p(x) = 0 to find x-intercepts.
3. Solve q(x) = 0 to find vertical asymptotes.
4. Find horizontal or slant asymptotes.
5. Plot one or more points between and beyond
x-intercepts and vertical asymptote.
6. Draw smooth curves where appropriate.
IMPORTANT NOTES
1. Graph will not cross vertical asymptote.
f(x) = 2x / (x - 2)
When q(x) = 0, f(x) is undefined.
2. Graph may cross horizontal asymptote.
f(x) = 5x / (x2 + 1)
3. Graph may cross slant asymptote.
f(x) = x3 / (x2 + 2)
EXAMPLE 1
1. Graph will not cross vertical asymptote.
f(x) = 2x / (x - 2)
When q(x) = 0, f(x) is undefined.
If q(x) = 0, x = a is VERTICAL asymptote.
q(x) = x - 2 = 0 yields x = 2 V.A.
EXAMPLE 1: Graph
1. Graph will not
cross vertical
asymptote.
VERTICAL asymptote:
q(x) = x - 2 = 0
yields x = 2 V.A.
Graph of
f(x) = 2x / (x - 2)
EXAMPLE 2
2. Graph may cross horizontal asymptote.
f(x) = 5x / (x2 + 1)
If n < m, y = 0 is HORIZONTAL asymptote.
Since n = 1 is less than m = 2, the graph of
f(x) has y = 0 as H.A.
EXAMPLE 2: Graph
2. Graph may
cross horizontal
asymptote.
If n < m, y = 0 is
HORIZONTAL
asymptote.
Graph of
f(x) = 5x / (x2 + 1)
EXAMPLE 3
3. Graph may cross slant asymptote.
f(x) = x3 / (x2 + 2)
Recall how to find a slant asymptote.
SLANT ASYMPTOTES OF
RATIONAL FUNCTIONS
n
p(x) an x + 
f(x) 

q(x) bm x m + 
If n = m + 1, then slant
asymptote is y = quotient when
p(x) is divided by q(x) using
long division.
EXAMPLE 3 continued
3. Graph may cross slant asymptote.
f(x) = x3 / (x2 + 2)
Since n = 3 is one more than m = 2,
the graph of f(x) has a slant asymptote.
Long division yields y = x as S.A.
EXAMPLE 3: Graph
3. Graph may
cross slant
asymptote.
Long division yields
y = x as S.A.
Graph of
f(x) = x3 / (x2 + 2)
PRECALCULUS I
PARTIAL
FRACTIONS
Dr. Claude S. Moore
Danville Community College
Test 2, Wed., 10-7-98
No Use of
Calculators
on Test.
35
Test 2, Wed., 10-7-98
1. Use leading coefficient test.
2. Use synthetic division.
3. Use long division.
4. Write polynomial given roots.
5. List, find all rational roots.
6. Use Descartes’s Rule of Signs.
7. Simplify complex numbers.
36
Test 2 (continued)
8. Use given root to find all roots.
9. Find horizontal & vertical asymptotes.
10. Find x- and y-intercepts.
11. Write partial fraction decomposition.
12. ?
13. ?
14. ?
37
PARTIAL FRACTIONS
RATIONAL EXPRESSION
EQUALS SUM OF
SIMPLER RATIONAL
EXPRESSIONS
38
DECOMPOSTION PROCESS
N ( x)
R( x )
 Quotient +
D( x )
D( x )
IF FRACTION IS IMPROPER,
DIVIDE AND USE REMAINDER
OVER DIVISOR TO FORM
PROPER FRACTION.
39
FACTOR DENOMINATOR
COMPLETELY FACTOR
DENOMINATOR INTO
FACTORS AS
m
LINEAR FORM: (px + q)
and
2
n
QUADRATIC: (ax + bx + c)
40
EXAMPLE 1
2
3 x + 13 x + 8
2
x + 4x + 3
Change improper fraction to proper
fraction.
Use long division and write remainder
over the divisor.
41
EXAMPLE 1 continued
2
3 x + 13 x + 8
2
x + 4x + 3
 3+
x 1
2
x + 4x + 3
Find the decomposition
of the proper fraction.
42
EXAMPLE 1 continued
2
x + 4 x + 3  (x + 1 )( x + 3)
x 1
A
B

+
2
x + 4 x + 3 (x + 1 ) ( x + 3)
• Completely
factor the
denominator.
• Write the
proper
fraction as
sum of
fractions
with factors
43
as
EXAMPLE 1 continued
x 1
A
B

+
2
x + 4 x + 3 (x + 1 ) ( x + 3)
Multiply by LCD to form
basic equation:
x - 1 = A(x + 3) + B(x + 1)
44
GUIDELINES FOR
LINEAR FACTORS
1. Substitute zeros of each linear factor
into basic equation.
2. Solve for coefficients A, B, etc.
3. For repeated factors, use coefficients
from above and substitute other
values for x and solve.
45
EXAMPLE 1 continued
Solving Basic Equation
x  1  A( x + 3) + B( x + 1)
To solve the basic equation:
Let x = -3 and solve for B = 2.
Let x = -1 and solve for A = -1.
46
EXAMPLE 1 continued
Since A = - 1 and B = 2,
the proper fraction
solution is
x 1
1
2

+
2
x + 4x + 3 x +1 x + 3
47
EXAMPLE 1 continued
Thus, the partial decomposition
of the improper fraction is as
shown below.
2
3x + 13 x + 8
1
2
 3
+
2
x
+
1
x
+
3
x + 4x + 3
48
EXAMPLE 1: GRAPHS
y
2
3x + 13 x + 8
2
x + 4x + 3
1
2
y  3
+
x +1 x + 3
The two graphs are equivalent.
49
EXAMPLE 2
Find the partial fraction
decomposition of the
rational expression:
2x  3
( x  1)
2
50
EXAMPLE 2 continued
Denominator, (x-1)2, has a repeated
factor (exponent of 2).
Form two fractions as below.
2x  3
A
B

+
2
2
( x  1)
(x  1 ) (x  1 )
51
EXAMPLE 2 continued
2x  3
A
B

+
2
2
( x  1)
(x  1 ) (x  1 )
Multiply by LCD to form
basic equation:
2x - 3 = A(x - 1) + B
52
EXAMPLE 2 continued
Solving Basic Equation
2x - 3 = A(x - 1) + B
To solve the basic equation:
Let x = 1 and solve for B.
Let x = 0 and use B = -1 from
above to solve for A = 2.
53
EXAMPLE 2 continued
Solution to the basic equation
was A = 2 and B = -1.
Thus, the decomposition is
2x  3
2
1


2
2
( x  1)
(x  1 ) (x  1 )
54
EXAMPLE 2: GRAPHS
2x  3
y 
2
( x  1)
y
2
(x  1 )

1
2
(x  1 )
The two graphs are equivalent.
55