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**PRECALCULUS I Complex Numbers**

Real numbers + Imaginary numbers Dr. Claude S. Moore Danville Community College

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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.

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**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.

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**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.

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**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.

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**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.

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**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

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**Example: Addition & Subtraction**

(3 + 2i ) + (-7 - 5i ) = (3 + -7) + ( )I = i (-6 + 9i ) - (4 - 3i ) = (-6 - 4) + (9 + 3 )i = i

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**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.

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**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

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**Principal Square Root of Negative**

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

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**PRECALCULUS I Fundamental Theorem of Algebra**

Dr. Claude S. Moore Danville Community College

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**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.

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**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.

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**Linear Factorization continued**

where c1, c2, … , cn are complex numbers and an is leading coefficient of f(x).

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**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).

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**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.

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**Dr. Claude S. Moore Danville Community College**

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

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**FRACTION OF TWO POLYNOMIALS**

RATIONAL FUNCTIONS FRACTION OF TWO POLYNOMIALS

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**DENOMINATOR CAN NOT EQUAL ZERO**

DOMAIN DENOMINATOR CAN NOT EQUAL ZERO

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

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**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.

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**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.

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**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.

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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)

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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.

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**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)

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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.

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**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)

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**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.

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**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.

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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.

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**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)

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**Dr. Claude S. Moore Danville Community College**

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

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**No Use of Calculators on Test.**

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

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**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

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**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

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**RATIONAL EXPRESSION EQUALS SUM OF SIMPLER RATIONAL EXPRESSIONS**

PARTIAL FRACTIONS RATIONAL EXPRESSION EQUALS SUM OF SIMPLER RATIONAL EXPRESSIONS 2

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

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**COMPLETELY FACTOR DENOMINATOR INTO FACTORS AS**

LINEAR FORM: (px + q)m and QUADRATIC: (ax2 + bx + c)n 4

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**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.

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**Find the decomposition of the proper fraction.**

EXAMPLE 1 continued Find the decomposition of the proper fraction.

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**EXAMPLE 1 continued Completely factor the denominator.**

Write the proper fraction as sum of fractions with factors as denominators.

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

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**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

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**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.

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**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

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

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EXAMPLE 1: GRAPHS The two graphs are equivalent.

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**Find the partial fraction decomposition of the rational expression:**

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

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**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.

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

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**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.

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

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EXAMPLE 2: GRAPHS The two graphs are equivalent.

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