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Complex Numbers
Real numbers + Imaginary numbers
Dr. Claude S. MooreDanville Community
College
PRECALCULUS I
The square root of a negative real number is not a real number.
Thus, we introduce imaginary numbers by letting
i =
So i 2 = -1, i 3 = - i , and i 4 = +1.
Definition of 1
1
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
Example: i 35 = i 32 i 3 = (+1)( - i ) = - i
NOTE: 21/4 = 5 with r = 1 and35/4 = 8 with r = 3.
Example: Simplifying i n
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.
Definition of Complex Number
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.
Equality of Complex Numbers
If (a + 7) + bi = 9 i, 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.
Example: Equality
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
Addition & Subtraction: Complex Numbers
(3 + 2i ) + (-7 - 5i )
= (3 + -7) + (2 + -5 )I
= -4 - 3i
(-6 + 9i ) - (4 - 3i )
= (-6 - 4) + (9 + 3 )i
= -10 + 12i
Example: Addition & Subtraction
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.
Complex Conjugates
The conjugate of -5 + 6i is -5 - 6i
The conjugate of 4 + 3i is 4 - 3iRecall: 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
Example: Complex Conjugates
If a is a positive number, the principal square root of the
negative number -a is defined as
Example:
Principal Square Root of Negative
iaa
ii 41616
12
Fundamental Theorem of Algebra
Dr. Claude S. MooreDanville Community
College
PRECALCULUS I
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.
The Fundamental Theorem
If f (x) is a polynomial of degree n
where n > 0, then f has precisely n linear factors in the complex
number system.
Linear Factorization Theorem
01
1)( axaxaxf nn
nn
where c1, c2, … , cn are complex numbers and an is leading
coefficient of f(x).
Linear Factorization continued
01
1)( axaxaxf nn
nn
)())(()( 21 ncxcxcxaxf n
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).
Complex Roots in Conjugate Pairs
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.
Factors of a Polynomial
PRECALCULUS I
RATIONAL FUNCTIONS
RATIONAL FUNCTIONS
Dr. Claude S. MooreDanville Community
College
RATIONAL FUNCTIONS
FRACTION OF TWO POLYNOMIALS
q(x)
p(x)f(x)
DOMAIN
DENOMINATOR CAN NOT
EQUAL ZERO
0q(x)whereq(x)
p(x)f(x)
ASYMPTOTES• HORIZONTAL
LINE y = b if
x
or
x
as
bxf )(
• VERTICAL LINE x = a if
axas
xf
or
xf
)(
)(
ASYMPTOTES OF RATIONAL FUNCTIONS
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.
mm
nn
xb
xa
q(x)
p(x)f(x)
SLANT ASYMPTOTES OF RATIONAL FUNCTIONS
If n = m + 1, then slant asymptote is y = quotient when
p(x) is divided by q(x) using long division.
mm
nn
xb
xa
q(x)
p(x)f(x)
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 off(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
If n = m + 1, then slant asymptote is y = quotient when
p(x) is divided by q(x) using long division.
mm
nn
xb
xa
q(x)
p(x)f(x)
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 FRACTIONSPARTIAL
FRACTIONS
Dr. Claude S. MooreDanville Community
College
35
Test 2, Wed., 10-7-98
No Use of Calculators
on Test.
36
Test 2, Wed., 10-7-981. 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.
37
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. ?
38
PARTIAL FRACTIONS
RATIONAL EXPRESSION EQUALS SUM OF
SIMPLER RATIONAL EXPRESSIONS
39
DECOMPOSTION PROCESS
IF FRACTION IS IMPROPER, DIVIDE AND USE REMAINDER
OVER DIVISOR TO FORM PROPER FRACTION.
)(
)(
)(
)(
xD
xRQuotient
xD
xN
40
FACTOR DENOMINATOR
COMPLETELY FACTOR DENOMINATOR INTO
FACTORS AS
LINEAR FORM: (px + q)m
and
QUADRATIC: (ax2 + bx + c)n
41
Change improper fraction to proper fraction.
Use long division and write remainder over the divisor.
EXAMPLE 1
34
81332
2
xx
xx
42
EXAMPLE 1 continued
34
13
34
813322
2
xx
x
xx
xx
Find the decomposition of the proper fraction.
43
EXAMPLE 1 continued• Completely
factor the denominator.
• Write the proper fraction as sum of fractions with factors as denominators.
)( 31342 x)(xxx
)( 3134
12
x
B
)(x
A
xx
x
44
EXAMPLE 1 continued
Multiply by LCD to form basic equation:
x - 1 = A(x + 3) + B(x + 1)
)( 3134
12
x
B
)(x
A
xx
x
45
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.
46
EXAMPLE 1 continuedSolving Basic Equation
To solve the basic equation:
Let x = -3 and solve for B = 2.
Let x = -1 and solve for A = -1.
)()( 131 xBxAx
47
EXAMPLE 1 continued
Since A = - 1 and B = 2, the proper fraction solution is
3
2
1
1
34
12
xxxx
x
48
EXAMPLE 1 continued
Thus, the partial decomposition of the improper fraction is as shown below.
3
2
1
13
34281323
xxxx
xx
49
EXAMPLE 1: GRAPHS
34281323
xx
xxy
3
2
1
13
xxy
The two graphs are equivalent.
50
EXAMPLE 2
Find the partial fraction decomposition of the rational expression:
21
32
)(
x
x
51
Denominator, (x-1)2, has a repeated factor (exponent of 2).
Form two fractions as below.
EXAMPLE 2 continued
21121
32
)(x
B
)(x
A
x
x
)(
52
EXAMPLE 2 continued
Multiply by LCD to form basic equation:
2x - 3 = A(x - 1) + B
21121
32
)(x
B
)(x
A
x
x
)(
53
EXAMPLE 2 continuedSolving Basic Equation
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.
2x - 3 = A(x - 1) + B
54
EXAMPLE 2 continued
Solution to the basic equation was A = 2 and B = -1.
Thus, the decomposition is
21
1
1
221
32
)(x)(xx
x
)(
55
EXAMPLE 2: GRAPHS
The two graphs are equivalent.
21
32
)(
x
xy
21
1
1
2
)(x)(xy
