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

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Presentation on theme: "1 Complex Numbers Real numbers + Imaginary numbers Dr. Claude S. Moore Danville Community College PRECALCULUS I."— Presentation transcript:

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2 1 Complex Numbers Real numbers + Imaginary numbers Dr. Claude S. Moore Danville Community College PRECALCULUS I

3 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

4 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. Example: Simplifying i n

5 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

6 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

7 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

8 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

9 (3 + 2i ) + (-7 - 5i ) = (3 + -7) + ( )I = i (-6 + 9i ) - (4 - 3i ) = (-6 - 4) + (9 + 3 )i = i Example: Addition & Subtraction

10 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 ) = a 2 + b 2. Complex Conjugates

11 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 ) = a 2 + b 2. (-5 + 6i )(-5 - 6i ) = (-5) 2 + (6) 2 = = 41 Example: Complex Conjugates

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

13 12 Fundamental Theorem of Algebra Dr. Claude S. Moore Danville Community College PRECALCULUS I

14 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

15 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

16 where c 1, c 2, …, c n are complex numbers and a n is leading coefficient of f(x). Linear Factorization continued

17 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

18 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

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

20 RATIONAL FUNCTIONS FRACTION OF TWO POLYNOMIALS

21 DOMAIN DENOMINATOR CAN NOT EQUAL ZERO

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

23 ASYMPTOTES OF RATIONAL FUNCTIONS If q(x) = 0, x = a is VERTICAL. HORIZONTALS: If n < m, y = 0. If n = m, y = a n /b m. NO HORIZONTAL: If n > m.

24 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.

25 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.

26 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 / (x 2 + 1) 3. Graph may cross slant asymptote. f(x) = x 3 / (x 2 + 2)

27 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.

28 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)

29 EXAMPLE 2 2. Graph may cross horizontal asymptote. f(x) = 5x / (x 2 + 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.

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

31 EXAMPLE 3 3. Graph may cross slant asymptote. f(x) = x 3 / (x 2 + 2) Recall how to find a slant asymptote.

32 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.

33 EXAMPLE 3 continued 3. Graph may cross slant asymptote. f(x) = x 3 / (x 2 + 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.

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

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

36 35 Test 2, Wed., No Use of Calculators on Test.

37 36 Test 2, Wed., 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 Descartess Rule of Signs. 7. Simplify complex numbers.

38 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. ?

39 38 PARTIAL FRACTIONS RATIONAL EXPRESSION EQUALS SUM OF SIMPLER RATIONAL EXPRESSIONS

40 39 DECOMPOSTION PROCESS IF FRACTION IS IMPROPER, DIVIDE AND USE REMAINDER OVER DIVISOR TO FORM PROPER FRACTION.

41 40 FACTOR DENOMINATOR COMPLETELY FACTOR DENOMINATOR INTO FACTORS AS LINEAR FORM: (px + q) m and QUADRATIC: (ax 2 + bx + c) n

42 41 C hange improper fraction to proper fraction. Use long division and write remainder over the divisor. EXAMPLE 1

43 42 EXAMPLE 1 continued Find the decomposition of the proper fraction.

44 43 EXAMPLE 1 continued Completely factor the denominator. Write the proper fraction as sum of fractions with factors as denominator s.

45 44 EXAMPLE 1 continued Multiply by LCD to form basic equation: x - 1 = A(x + 3) + B(x + 1)

46 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.

47 46 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.

48 47 EXAMPLE 1 continued Since A = - 1 and B = 2, the proper fraction solution is

49 48 EXAMPLE 1 continued Thus, the partial decomposition of the improper fraction is as shown below.

50 49 EXAMPLE 1: GRAPHS The two graphs are equivalent.

51 50 EXAMPLE 2 Find the partial fraction decomposition of the rational expression:

52 51 Denominator, (x-1) 2, has a repeated factor (exponent of 2). Form two fractions as below. EXAMPLE 2 continued

53 52 EXAMPLE 2 continued Multiply by LCD to form basic equation: 2x - 3 = A(x - 1) + B

54 53 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. 2x - 3 = A(x - 1) + B

55 54 EXAMPLE 2 continued Solution to the basic equation was A = 2 and B = -1. Thus, the decomposition is

56 55 EXAMPLE 2: GRAPHS The two graphs are equivalent.


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