1. College Algebra Fifth Edition
James Stewart  Lothar Redlin  Saleem Watson

2. 1
Equations and Inequalities

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 x2 + 4 = 0 has no real solution.

4. If we try to solve this equation, we get: x2 = –4 So,
Introduction
If we try to solve this equation, we get: x2 = –4
So,
However, this is impossible—since the square of any real number is positive.
For example, (–2)2 = 4, a positive number.
Thus, negative numbers don’t have real square roots.

5. 1.4
Complex Numbers

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

7. First, they defined the new number
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 Number—Definition
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. Here are examples of complex numbers.
E.g. 1—Complex Numbers
Here are examples of complex numbers.
3 + 4i Real part 3, imaginary part 4
1/2 – 2/3i Real part 1/2, imaginary part –2/3
6i Real part 0, imaginary part 6
–7 Real part –7, imaginary part 0

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

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

14. In the complex number system, every quadratic equation has solutions.
The numbers 2i and –2i are solutions of x2 = –4 because: (2i)2 = 22i2 = 4(–1) = –4 and (–2i)2 = (–2)2i2 = 4(–1) = –4

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

16. Complex Numbers
We study complex numbers as they complete—in 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. Arithmetic Operations on Complex Numbers
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 i2 = –1.

19. Arithmetic Operations on Complex Numbers
Thus, the following calculations are valid.
(a + bi)(c + di)
= ac + (ad + bc)i + bdi2 (Multiply and collect all terms)
= ac + (ad + bc)i + bd(–1) (i2 = –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. 2—Adding, Subtracting, and Multiplying
Express the following in the form a + bi.
(3 + 5i) + (4 – 2i)
(3 + 5i) – (4 – 2i)
(3 + 5i)(4 – 2i)
i 23

25. E.g. 2—Adding
Example (a)
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

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

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

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

29. Dividing Complex Numbers
Division of complex numbers is much like rationalizing the denominator of a radical expression—which 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 Numbers—Formula
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, it’s easier to just remember the first step and then multiply out the numerator and the denominator as usual.

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

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

37. Square Roots of Negative Numbers

38. Square Roots of Negative Numbers
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. 4—Square 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. 5—Using 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 ax2 + bx + c = 0 are:

47. Complex Roots of Quadratic Equations
If b2 – 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. 6—Quadratic Equations with Complex Solutions
Solve each equation.
x2 + 9 = 0
x2 + 4x + 5 = 0

49. The equation x2 + 9 = 0 means x2 = –9.
E.g. 6—Complex Solutions
Example (a)
The equation x2 + 9 = 0 means x2 = –9.
So,
The solutions are therefore 3i and –3i.

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

51. E.g. 7—Complex Conjugates as Solutions of a Quadratic
Show that the solutions of the equation x2 – 24x + 37 = 0 are complex conjugates of each other.

52. E.g. 7—Complex 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.