# PRECALCULUS I Complex Numbers

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PRECALCULUS I Complex Numbers
Real numbers + Imaginary numbers Dr. Claude S. Moore Danville Community College

Definition of 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.

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 Example: i 35 = i 32 i 3 = (+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.

If (a + 7) + bi = 9 - 8i, find a and b. Since a + bi = c + di
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) + ( )I = i (-6 + 9i ) - (4 - 3i ) = (-6 - 4) + (9 + 3 )i = i

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 i is i 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 = = 41

Principal Square Root of Negative
If a is a positive number, the principal square root of the negative number -a is defined as Example:

PRECALCULUS I Fundamental Theorem of Algebra
Dr. Claude S. Moore Danville Community College

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 where n > 0, then f has precisely n linear factors in the complex number system.

Linear Factorization continued
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.

Dr. Claude S. Moore Danville Community College
PRECALCULUS I RATIONAL FUNCTIONS Dr. Claude S. Moore Danville Community College

FRACTION OF TWO POLYNOMIALS
RATIONAL FUNCTIONS FRACTION OF TWO POLYNOMIALS

DENOMINATOR CAN NOT EQUAL ZERO
DOMAIN DENOMINATOR CAN NOT EQUAL ZERO

ASYMPTOTES VERTICAL LINE x = a if HORIZONTAL LINE y = b if

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.

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.

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.

1. Graph will not cross vertical asymptote.
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.

2. Graph may cross horizontal asymptote.
EXAMPLE 2: Graph 2. Graph may cross horizontal asymptote. If n < m, y = 0 is HORIZONTAL asymptote. Graph of f(x) = 5x / (x2 + 1)

Recall how to find a slant asymptote.
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.

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.

3. Graph may cross slant asymptote.
EXAMPLE 3: Graph 3. Graph may cross slant asymptote. Long division yields y = x as S.A. Graph of f(x) = x3 / (x2 + 2)

Dr. Claude S. Moore Danville Community College
PRECALCULUS I PARTIAL FRACTIONS Dr. Claude S. Moore Danville Community College

No Use of Calculators on Test.
Test 2, Wed., No Use of Calculators on Test. 2

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. 2

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. ? 2

RATIONAL EXPRESSION EQUALS SUM OF SIMPLER RATIONAL EXPRESSIONS
PARTIAL FRACTIONS RATIONAL EXPRESSION EQUALS SUM OF SIMPLER RATIONAL EXPRESSIONS 2

DECOMPOSTION PROCESS IF FRACTION IS IMPROPER, DIVIDE AND USE REMAINDER OVER DIVISOR TO FORM PROPER FRACTION. 3

COMPLETELY FACTOR DENOMINATOR INTO FACTORS AS
LINEAR FORM: (px + q)m and QUADRATIC: (ax2 + bx + c)n 4

EXAMPLE 1 Use long division and write remainder over the divisor.
Change improper fraction to proper fraction. Use long division and write remainder over the divisor.

Find the decomposition of the proper fraction.
EXAMPLE 1 continued Find the decomposition of the proper fraction.

EXAMPLE 1 continued Completely factor the denominator.
Write the proper fraction as sum of fractions with factors as denominators.

EXAMPLE 1 continued Multiply by LCD to form basic equation: x - 1 = A(x + 3) + B(x + 1)

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. 7

EXAMPLE 1 continued Solving Basic Equation
To solve the basic equation: Let x = -3 and solve for B = 2. Let x = -1 and solve for A = -1.

Since A = - 1 and B = 2, the proper fraction solution is
EXAMPLE 1 continued Since A = - 1 and B = 2, the proper fraction solution is

EXAMPLE 1 continued Thus, the partial decomposition of the improper fraction is as shown below.

EXAMPLE 1: GRAPHS The two graphs are equivalent.

Find the partial fraction decomposition of the rational expression:
EXAMPLE 2 Find the partial fraction decomposition of the rational expression: 11

Denominator, (x-1)2, has a repeated factor (exponent of 2).
EXAMPLE 2 continued Denominator, (x-1)2, has a repeated factor (exponent of 2). Form two fractions as below.

EXAMPLE 2 continued Multiply by LCD to form basic equation: 2x - 3 = A(x - 1) + B

2x - 3 = A(x - 1) + B EXAMPLE 2 continued Solving 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.

EXAMPLE 2 continued Solution to the basic equation was A = 2 and B = -1. Thus, the decomposition is

EXAMPLE 2: GRAPHS The two graphs are equivalent.

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