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College Algebra Fifth Edition James Stewart Lothar Redlin Saleem Watson.

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Presentation on theme: "College Algebra Fifth Edition James Stewart Lothar Redlin Saleem Watson."— Presentation transcript:

1 College Algebra Fifth Edition James Stewart Lothar Redlin Saleem Watson

2 Equations and Inequalities 1

3 Introduction In Section 1.3, we saw that, if the discriminant of a quadratic equation is negative, the equation has no real solution. For example, the equation x 2 + 4 = 0 has no real solution.

4 Introduction If we try to solve this equation, we get: x 2 = –4 So, However, this is impossiblesince the square of any real number is positive. For example, (–2) 2 = 4, a positive number. Thus, negative numbers dont have real square roots.

5 Complex Numbers 1.4

6 Complex Number System To make it possible to solve all quadratic equations, mathematicians invented an expanded number systemcalled the complex number system.

7 Complex Number First, they defined the new number This means i 2 = –1. A complex number is then a number of the form a + bi, where a and b are real numbers.

8 Complex NumberDefinition A complex number is an expression of the form a + bi where: a and b are real numbers. i 2 = –1.

9 Real and Imaginary Parts The real part of this complex number is a. The imaginary part is b. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.

10 Real and Imaginary Parts Note that both the real and imaginary parts of a complex number are real numbers.

11 E.g. 1Complex Numbers Here are examples of complex numbers. 3 + 4iReal part 3, imaginary part 4 1/2 – 2/3iReal part 1/2, imaginary part –2/3 6iReal part 0, imaginary part 6 –7Real part –7, imaginary part 0

12 Pure Imaginary Number A number such as 6i, which has real part 0, is called: A pure imaginary number.

13 Complex Numbers A real number like –7 can be thought of as: A complex number with imaginary part 0.

14 Complex Numbers In the complex number system, every quadratic equation has solutions. The numbers 2i and –2i are solutions of x 2 = –4 because: (2i) 2 = 2 2 i 2 = 4(–1) = –4 and (–2i) 2 = (–2) 2 i 2 = 4(–1) = –4

15 Complex Numbers Though we use the term imaginary here, imaginary numbers should not be thought of as any less realin the ordinary rather than the mathematical sense of that wordthan negative numbers or irrational numbers. All numbers (except possibly the positive integers) are creations of the human mindthe numbers –1 and as well as the number i.

16 Complex Numbers We study complex numbers as they completein a useful and elegant fashion our study of the solutions of equations. In fact, imaginary numbers are useful not only in algebra and mathematics, but in the other sciences too. To give just one example, in electrical theory, the reactance of a circuit is a quantity whose measure is an imaginary number.

17 Arithmetic Operations on Complex Numbers

18 Complex numbers are added, subtracted, multiplied, and divided just as we would any number of the form a + b The only difference we need to keep in mind is that i 2 = –1.

19 Arithmetic Operations on Complex Numbers Thus, the following calculations are valid. (a + bi)(c + di) = ac + (ad + bc)i + bdi 2 (Multiply and collect all terms) = ac + (ad + bc)i + bd( – 1) (i 2 = –1) = (ac – bd) + (ad + bc)i (Combine real and imaginary parts)

20 Arithmetic Operations on Complex Numbers We therefore define the sum, difference, and product of complex numbers as follows.

21 Adding Complex Numbers (a + bi) + (c + di) = (a + c) + (b + d)i To add complex numbers, add the real parts and the imaginary parts.

22 Subtracting Complex Numbers (a + bi) – (c + di) = (a – c) + (b – d)i To subtract complex numbers, subtract the real parts and the imaginary parts.

23 Multiplying Complex Numbers (a + bi). (c + di) = (ac – bd) + (ad + bc)i Multiply complex numbers like binomials, using i 2 = –1.

24 E.g. 2Adding, Subtracting, and Multiplying Express the following in the form a + bi. (a)(3 + 5i) + (4 – 2i) (b)(3 + 5i) – (4 – 2i) (c)(3 + 5i)(4 – 2i) (d) i 23

25 E.g. 2Adding According to the definition, we add the real parts and we add the imaginary parts. (3 + 5i) + (4 – 2i) = (3 + 4) + (5 – 2)i = 7 + 3i Example (a)

26 E.g. 2Subtracting (3 + 5i) – (4 – 2i) = (3 – 4) + [5 – (– 2)]i = –1 + 7i Example (b)

27 E.g. 2Multiplying (3 + 5i)(4 – 2i) = [3. 4 – 5(– 2)] + [3(–2) + 5. 4]i = 22 + 14i Example (c)

28 E.g. 2Power i 23 = i 22 + 1 = (i 2 ) 11 i = (–1) 11 i = (–1)i = –i Example (d)

29 Dividing Complex Numbers Division of complex numbers is much like rationalizing the denominator of a radical expressionwhich we considered in Section P.8.

30 Complex Conjugates For the complex number z = a + bi, we define its complex conjugate to be:

31 Dividing Complex Numbers Note that: So, the product of a complex number and its conjugate is always a nonnegative real number. We use this property to divide complex numbers.

32 Dividing Complex NumbersFormula To simplify the quotient multiply the numerator and the denominator by the complex conjugate of the denominator:

33 Dividing Complex Numbers Rather than memorize this entire formula, its easier to just remember the first step and then multiply out the numerator and the denominator as usual.

34 E.g. 3Dividing Complex Numbers Express the following in the form a + bi. We multiply both the numerator and denominator by the complex conjugate of the denominator to make the new denominator a real number.

35 E.g. 3Dividing Complex Numbers The complex conjugate of 1 – 2i is: Example (a)

36 E.g. 3Dividing Complex Numbers The complex conjugate of 4i is –4i. Example (b)

37 Square Roots of Negative Numbers

38 Just as every positive real number r has two square roots, every negative number has two square roots as well. If -r is a negative number, then its square roots are, because: and

39 Square Roots of Negative Numbers If –r is negative, then the principal square root of –r is: The two square roots of –r are: We usually write instead of to avoid confusion with.

40 E.g. 4Square Roots of Negative Numbers

41 Square Roots of Negative Numbers Special care must be taken when performing calculations involving square roots of negative numbers. Although when a and b are positive, this is not true when both are negative.

42 Square Roots of Negative Numbers For example, However, Thus,

43 Square Roots of Negative Numbers When multiplying radicals of negative numbers, express them first in the form (where r > 0) to avoid possible errors of this type.

44 E.g. 5Using Square Roots of Negative Numbers Evaluate and express in the form a + bi.

45 Complex Solutions of Quadratic Equations

46 Complex Roots of Quadratic Equations We have already seen that, if a 0, then the solutions of the quadratic equation ax 2 + bx + c = 0 are:

47 Complex Roots of Quadratic Equations If b 2 – 4ac < 0, the equation has no real solution. However, in the complex number system, the equation will always have solutions. This is because negative numbers have square roots in this expanded setting.

48 E.g. 6Quadratic Equations with Complex Solutions Solve each equation. (a) x 2 + 9 = 0 (b) x 2 + 4x + 5 = 0

49 E.g. 6Complex Solutions The equation x 2 + 9 = 0 means x 2 = –9. So, The solutions are therefore 3i and –3i. Example (a)

50 E.g. 6Complex Solutions By the quadratic formula, we have: So, the solutions are –2 + i and –2 – i. Example (b)

51 E.g. 7Complex Conjugates as Solutions of a Quadratic Show that the solutions of the equation 4x 2 – 24x + 37 = 0 are complex conjugates of each other.

52 E.g. 7Complex Conjugates as Solutions of a Quadratic We use the quadratic formula to get: So, the solutions are 3 + ½i and 3 – ½i. These are complex conjugates.


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