The document discusses complex numbers. It explains that the equation x^2 = -1 has no real solutions, so an imaginary number i is defined where i^2 = -1. A complex number is defined as a + bi, where a is the real part and bi is the imaginary part. Complex numbers can be added or multiplied by treating i as a variable and using rules like i^2 = -1. Examples show how to solve equations and perform operations with complex numbers.

1. Complex Numbers

2. Complex Numbers
Because the square of any real number can't be negative,
the equation x2 = –1 does not have any real solution.

3. Complex Numbers
Because the square of any real number can't be negative,
the equation x2 = –1 does not have any real solution.
We make up a new number called an imaginary number
–1
to be a solution of this equation

4. Complex Numbers
Because the square of any real number can't be negative,
the equation x2 = –1 does not have any real solution.
We make up a new number called an imaginary number
–1 ↔ i
to be a solution of this equation and we name it “ i ”,

5. Complex Numbers
Because the square of any real number can't be negative,
the equation x2 = –1 does not have any real solution.
We make up a new number called an imaginary number
–1 ↔ i
to be a solution of this equation and we name it “ i ”, i.e.
(±i)2 = –1

6. Complex Numbers
Using i, the “solutions” of the equations of the form
Because the square of any real number can't be negative,
the equation x2 = –1 does not have any real solution.
We make up a new number called an imaginary number
–1 ↔ i
to be a solution of this equation and we name it “ i ”, i.e.
(±i)2 = –1
x2 = –r

7. Complex Numbers
Using i, the “solutions” of the equations of the form
Because the square of any real number can't be negative,
the equation x2 = –1 does not have any real solution.
We make up a new number called an imaginary number
–1 ↔ i
to be a solution of this equation and we name it “ i ”, i.e.
(±i)2 = –1
x2 = –r
are x = ± ir

8. Complex Numbers
Using i, the “solutions” of the equations of the form
Example A. Solve x2 + 49 = 0 using imaginary numbers.
Because the square of any real number can't be negative,
the equation x2 = –1 does not have any real solution.
We make up a new number called an imaginary number
–1 ↔ i
to be a solution of this equation and we name it “ i ”, i.e.
(±i)2 = –1
x2 = –r
are x = ± ir

9. Complex Numbers
Using i, the “solutions” of the equations of the form
Example A. Solve x2 + 49 = 0 using imaginary numbers.
Using the square-root method:
x2 + 49 = 0 → x2 = –49
Because the square of any real number can't be negative,
the equation x2 = –1 does not have any real solution.
We make up a new number called an imaginary number
–1 ↔ i
to be a solution of this equation and we name it “ i ”, i.e.
(±i)2 = –1
x2 = –r
are x = ± ir

10. Complex Numbers
Using i, the “solutions” of the equations of the form
Example A. Solve x2 + 49 = 0 using imaginary numbers.
Using the square-root method:
x2 + 49 = 0 → x2 = –49 so
x = ±–49
Because the square of any real number can't be negative,
the equation x2 = –1 does not have any real solution.
We make up a new number called an imaginary number
–1 ↔ i
to be a solution of this equation and we name it “ i ”, i.e.
(±i)2 = –1
x2 = –r
are x = ± ir

11. Complex Numbers
Using i, the “solutions” of the equations of the form
Example A. Solve x2 + 49 = 0 using imaginary numbers.
Using the square-root method:
x2 + 49 = 0 → x2 = –49 so
x = ±–49
x = ±49–1
x = ±7i
Because the square of any real number can't be negative,
the equation x2 = –1 does not have any real solution.
We make up a new number called an imaginary number
–1 ↔ i
to be a solution of this equation and we name it “ i ”, i.e.
(±i)2 = –1
x2 = –r
are x = ± ir

12. A complex number is a number of the form
a + bi
where a and b are real numbers,
Complex Numbers

13. A complex number is a number of the form
a + bi
where a and b are real numbers, a is called the real part
Complex Numbers

14. A complex number is a number of the form
a + bi
where a and b are real numbers, a is called the real part and
bi is called the imaginary part of the complex number.
Complex Numbers

15. A complex number is a number of the form
a + bi
where a and b are real numbers, a is called the real part and
bi is called the imaginary part of the complex number.
Complex Numbers
Example B. 5 – 3i, 6i, –17 are complex numbers.

16. A complex number is a number of the form
a + bi
where a and b are real numbers, a is called the real part and
bi is called the imaginary part of the complex number.
Complex Numbers
Example B. 5 – 3i, 6i, –17 are complex numbers.
The imaginary part of 5 – 3i is –3i.

17. A complex number is a number of the form
a + bi
where a and b are real numbers, a is called the real part and
bi is called the imaginary part of the complex number.
Complex Numbers
Example B. 5 – 3i, 6i, –17 are complex numbers.
The imaginary part of 5 – 3i is –3i. The real part of 6i is 0.

18. A complex number is a number of the form
a + bi
where a and b are real numbers, a is called the real part and
bi is called the imaginary part of the complex number.
Complex Numbers
Example B. 5 – 3i, 6i, –17 are complex numbers.
The imaginary part of 5 – 3i is –3i. The real part of 6i is 0.
Any real number a is also complex because a = a + 0i
hence –17 = –17 + 0i.

19. A complex number is a number of the form
a + bi
where a and b are real numbers, a is called the real part and
bi is called the imaginary part of the complex number
Example B. 5 – 3i, 6i, –17 are complex numbers.
The imaginary part of 5 – 3i is –3i. The real part of 6i is 0.
Any real number a is also complex because a = a + 0i
hence –17 = –17 + 0i.
Complex Numbers
(Addition and subtraction of complex numbers)
Treat the "i" as a variable when adding or subtracting complex
numbers.

20. A complex number is a number of the form
a + bi
where a and b are real numbers, a is called the real part and
bi is called the imaginary part of the complex number
Complex Numbers
(Addition and subtraction of complex numbers)
Treat the "i" as a variable when adding or subtracting complex
numbers.
Example C.
(7 + 4i) + (5 – 3i)
Example B. 5 – 3i, 6i, –17 are complex numbers.
The imaginary part of 5 – 3i is –3i. The real part of 6i is 0.
Any real number a is also complex because a = a + 0i
hence –17 = –17 + 0i.

21. A complex number is a number of the form
a + bi
where a and b are real numbers, a is called the real part and
bi is called the imaginary part of the complex number
Complex Numbers
(Addition and subtraction of complex numbers)
Treat the "i" as a variable when adding or subtracting complex
numbers.
Example C.
(7 + 4i) + (5 – 3i) = 7 + 4i + 5 – 3i
Example B. 5 – 3i, 6i, –17 are complex numbers.
The imaginary part of 5 – 3i is –3i. The real part of 6i is 0.
Any real number a is also complex because a = a + 0i
hence –17 = –17 + 0i.

22. A complex number is a number of the form
a + bi
where a and b are real numbers, a is called the real part and
bi is called the imaginary part of the complex number
Complex Numbers
(Addition and subtraction of complex numbers)
Treat the "i" as a variable when adding or subtracting complex
numbers.
Example C.
(7 + 4i) + (5 – 3i) = 7 + 4i + 5 – 3i = 12 + i
Example B. 5 – 3i, 6i, –17 are complex numbers.
The imaginary part of 5 – 3i is –3i. The real part of 6i is 0.
Any real number a is also complex because a = a + 0i
hence –17 = –17 + 0i.

23. A complex number is a number of the form
a + bi
where a and b are real numbers, a is called the real part and
bi is called the imaginary part of the complex number
Complex Numbers
(Addition and subtraction of complex numbers)
Treat the "i" as a variable when adding or subtracting complex
numbers.
Example C.
(7 + 4i) + (5 – 3i) = 7 + 4i + 5 – 3i = 12 + i
(7 + 4i) – (5 – 3i)
Example B. 5 – 3i, 6i, –17 are complex numbers.
The imaginary part of 5 – 3i is –3i. The real part of 6i is 0.
Any real number a is also complex because a = a + 0i
hence –17 = –17 + 0i.

24. A complex number is a number of the form
a + bi
where a and b are real numbers, a is called the real part and
bi is called the imaginary part of the complex number
Complex Numbers
(Addition and subtraction of complex numbers)
Treat the "i" as a variable when adding or subtracting complex
numbers.
Example C.
(7 + 4i) + (5 – 3i) = 7 + 4i + 5 – 3i = 12 + i
(7 + 4i) – (5 – 3i) = 7 + 4i – 5 + 3i
Example B. 5 – 3i, 6i, –17 are complex numbers.
The imaginary part of 5 – 3i is –3i. The real part of 6i is 0.
Any real number a is also complex because a = a + 0i
hence –17 = –17 + 0i.

25. A complex number is a number of the form
a + bi
where a and b are real numbers, a is called the real part and
bi is called the imaginary part of the complex number
Complex Numbers
(Addition and subtraction of complex numbers)
Treat the "i" as a variable when adding or subtracting complex
numbers.
Example C.
(7 + 4i) + (5 – 3i) = 7 + 4i + 5 – 3i = 12 + i
(7 + 4i) – (5 – 3i) = 7 + 4i – 5 + 3i = 2 + 7i
Example B. 5 – 3i, 6i, –17 are complex numbers.
The imaginary part of 5 – 3i is –3i. The real part of 6i is 0.
Any real number a is also complex because a = a + 0i
hence –17 = –17 + 0i.

26. (Multiplication of complex numbers)
Complex Numbers

27. (Multiplication of complex numbers)
To multiply complex numbers, use FOIL, then set i2 to be (-1)
and simplify the result.
Complex Numbers

28. (Multiplication of complex numbers)
To multiply complex numbers, use FOIL, then set i2 to be (-1)
and simplify the result.
Complex Numbers
Example D. (4 – 3i)(2 + 7i)

29. (Multiplication of complex numbers)
To multiply complex numbers, use FOIL, then set i2 to be (-1)
and simplify the result.
Complex Numbers
Example D. (4 – 3i)(2 + 7i) FOIL
= 8 – 6i + 28i – 21i2

30. (Multiplication of complex numbers)
To multiply complex numbers, use FOIL, then set i2 to be (-1)
and simplify the result.
Complex Numbers
Example D. (4 – 3i)(2 + 7i) FOIL
= 8 – 6i + 28i – 21i2 set i2 = (-1)
= 8 – 6i + 28i + 21

31. (Multiplication of complex numbers)
To multiply complex numbers, use FOIL, then set i2 to be (-1)
and simplify the result.
Complex Numbers
Example D. (4 – 3i)(2 + 7i) FOIL
= 8 – 6i + 28i – 21i2 set i2 = (-1)
= 8 – 6i + 28i + 21
= 29 + 22i

32. (Multiplication of complex numbers)
To multiply complex numbers, use FOIL, then set i2 to be (-1)
and simplify the result.
Complex Numbers
Example D. (4 – 3i)(2 + 7i) FOIL
= 8 – 6i + 28i – 21i2 set i2 = (-1)
= 8 – 6i + 28i + 21
= 29 + 22i
The conjugate of (a + bi) is (a – bi) and vice–versa.

33. (Multiplication of complex numbers)
To multiply complex numbers, use FOIL, then set i2 to be (-1)
and simplify the result.
Complex Numbers
Example D. (4 – 3i)(2 + 7i) FOIL
= 8 – 6i + 28i – 21i2 set i2 = (-1)
= 8 – 6i + 28i + 21
= 29 + 22i
The conjugate of (a + bi) is (a – bi) and vice–versa.
The most important complex number multiplication formula is
the product of a pair of conjugate numbers.

34. (Multiplication of complex numbers)
To multiply complex numbers, use FOIL, then set i2 to be (-1)
and simplify the result.
Complex Numbers
Example D. (4 – 3i)(2 + 7i) FOIL
= 8 – 6i + 28i – 21i2 set i2 = (-1)
= 8 – 6i + 28i + 21
= 29 + 22i
(Conjugate Multiplication)
The conjugate of (a + bi) is (a – bi) and vice–versa.
The most important complex number multiplication formula is
the product of a pair of conjugate numbers.

35. (Multiplication of complex numbers)
To multiply complex numbers, use FOIL, then set i2 to be (-1)
and simplify the result.
Complex Numbers
Example D. (4 – 3i)(2 + 7i) FOIL
= 8 – 6i + 28i – 21i2 set i2 = (-1)
= 8 – 6i + 28i + 21
= 29 + 22i
(Conjugate Multiplication) The nonzero conjugate product is
(a + bi)(a – bi) = a2 + b2 which is always positive.
The conjugate of (a + bi) is (a – bi) and vice–versa.
The most important complex number multiplication formula is
the product of a pair of conjugate numbers.

36. (Multiplication of complex numbers)
To multiply complex numbers, use FOIL, then set i2 to be (-1)
and simplify the result.
Complex Numbers
Example D. (4 – 3i)(2 + 7i) FOIL
= 8 – 6i + 28i – 21i2 set i2 = (-1)
= 8 – 6i + 28i + 21
= 29 + 22i
Example E.
(4 – 3i)(4 + 3i)
The conjugate of (a + bi) is (a – bi) and vice–versa.
The most important complex number multiplication formula is
the product of a pair of conjugate numbers.
(Conjugate Multiplication) The nonzero conjugate product is
(a + bi)(a – bi) = a2 + b2 which is always positive.

37. (Multiplication of complex numbers)
To multiply complex numbers, use FOIL, then set i2 to be (-1)
and simplify the result.
Complex Numbers
Example D. (4 – 3i)(2 + 7i) FOIL
= 8 – 6i + 28i – 21i2 set i2 = (-1)
= 8 – 6i + 28i + 21
= 29 + 22i
Example E.
(4 – 3i)(4 + 3i) = 42 + 32 = 25
The conjugate of (a + bi) is (a – bi) and vice–versa.
The most important complex number multiplication formula is
the product of a pair of conjugate numbers.
(Conjugate Multiplication) The nonzero conjugate product is
(a + bi)(a – bi) = a2 + b2 which is always positive.

38. (Multiplication of complex numbers)
To multiply complex numbers, use FOIL, then set i2 to be (-1)
and simplify the result.
Complex Numbers
Example D. (4 – 3i)(2 + 7i) FOIL
= 8 – 6i + 28i – 21i2 set i2 = (-1)
= 8 – 6i + 28i + 21
= 29 + 22i
Example E.
(4 – 3i)(4 + 3i) = 42 + 32 = 25
(5 – 7i)(5 + 7i)
The conjugate of (a + bi) is (a – bi) and vice–versa.
The most important complex number multiplication formula is
the product of a pair of conjugate numbers.
(Conjugate Multiplication) The nonzero conjugate product is
(a + bi)(a – bi) = a2 + b2 which is always positive.

39. (Multiplication of complex numbers)
To multiply complex numbers, use FOIL, then set i2 to be (-1)
and simplify the result.
Complex Numbers
Example D. (4 – 3i)(2 + 7i) FOIL
= 8 – 6i + 28i – 21i2 set i2 = (-1)
= 8 – 6i + 28i + 21
= 29 + 22i
Example E.
(4 – 3i)(4 + 3i) = 42 + 32 = 25
(5 – 7i)(5 + 7i) = (5)2 + 72
The conjugate of (a + bi) is (a – bi) and vice–versa.
The most important complex number multiplication formula is
the product of a pair of conjugate numbers.
(Conjugate Multiplication) The nonzero conjugate product is
(a + bi)(a – bi) = a2 + b2 which is always positive.

40. (Multiplication of complex numbers)
To multiply complex numbers, use FOIL, then set i2 to be (-1)
and simplify the result.
Complex Numbers
Example D. (4 – 3i)(2 + 7i) FOIL
= 8 – 6i + 28i – 21i2 set i2 = (-1)
= 8 – 6i + 28i + 21
= 29 + 22i
Example E.
(4 – 3i)(4 + 3i) = 42 + 32 = 25
(5 – 7i)(5 + 7i) = (5)2 + 72 = 54
The conjugate of (a + bi) is (a – bi) and vice–versa.
The most important complex number multiplication formula is
the product of a pair of conjugate numbers.
(Conjugate Multiplication) The nonzero conjugate product is
(a + bi)(a – bi) = a2 + b2 which is always positive.

41. Complex Numbers
(Division of Complex Numbers)

42. Complex Numbers
(Division of Complex Numbers)
To divide complex numbers, we write the division as a fraction,
then multiply the top and the bottom of the fraction by the
conjugate of the denominator.

43. Complex Numbers
(Division of Complex Numbers)
To divide complex numbers, we write the division as a fraction,
then multiply the top and the bottom of the fraction by the
conjugate of the denominator.
3 – 2i
4 + 3i
Example F. Simplify

44. Complex Numbers
(Division of Complex Numbers)
To divide complex numbers, we write the division as a fraction,
then multiply the top and the bottom of the fraction by the
conjugate of the denominator.
3 – 2i
4 + 3i
Example F. Simplify
Multiply the conjugate of the denominator (4 – 3i) to the top
and the bottom.
(3 – 2i)
(4 + 3i)

45. Complex Numbers
(Division of Complex Numbers)
To divide complex numbers, we write the division as a fraction,
then multiply the top and the bottom of the fraction by the
conjugate of the denominator.
3 – 2i
4 + 3i
Example F. Simplify
Multiply the conjugate of the denominator (4 – 3i) to the top
and the bottom.
(3 – 2i)
(4 + 3i)
=
(4 – 3i)
(4 – 3i)*

46. Complex Numbers
(Division of Complex Numbers)
To divide complex numbers, we write the division as a fraction,
then multiply the top and the bottom of the fraction by the
conjugate of the denominator.
3 – 2i
4 + 3i
Example F. Simplify
Multiply the conjugate of the denominator (4 – 3i) to the top
and the bottom.
(3 – 2i)
(4 + 3i)
=
(4 – 3i)
(4 – 3i)* 42 + 32

47. Complex Numbers
(Division of Complex Numbers)
To divide complex numbers, we write the division as a fraction,
then multiply the top and the bottom of the fraction by the
conjugate of the denominator.
3 – 2iExample F. Simplify
Multiply the conjugate of the denominator (4 – 3i) to the top
and the bottom.
(3 – 2i)
(4 + 3i)
=
(4 – 3i)
(4 – 3i)* 42 + 32 =
25
4 + 3i

48. Complex Numbers
(Division of Complex Numbers)
To divide complex numbers, we write the division as a fraction,
then multiply the top and the bottom of the fraction by the
conjugate of the denominator.
3 – 2i
4 + 3i
Example F. Simplify
Multiply the conjugate of the denominator (4 – 3i) to the top
and the bottom.
(3 – 2i)
(4 + 3i)
=
(4 – 3i)
(4 – 3i)* 42 + 32 =
25
12 – 8i – 9i + 6i2

49. Complex Numbers
(Division of Complex Numbers)
To divide complex numbers, we write the division as a fraction,
then multiply the top and the bottom of the fraction by the
conjugate of the denominator.
3 – 2i
4 + 3i
Example F. Simplify
Multiply the conjugate of the denominator (4 – 3i) to the top
and the bottom.
(3 – 2i)
(4 + 3i)
=
(4 – 3i)
(4 – 3i)* 42 + 32 =
25
12 – 8i – 9i + 6i2
–6

50. Complex Numbers
(Division of Complex Numbers)
To divide complex numbers, we write the division as a fraction,
then multiply the top and the bottom of the fraction by the
conjugate of the denominator.
3 – 2i
4 + 3i
Example F. Simplify
Multiply the conjugate of the denominator (4 – 3i) to the top
and the bottom.
(3 – 2i)
(4 + 3i)
=
(4 – 3i)
(4 – 3i)* 42 + 32 =
25
12 – 8i – 9i + 6i2
–6
6 – 17i
=
25
6
25
17i
–

51. Complex Numbers
(Division of Complex Numbers)
To divide complex numbers, we write the division as a fraction,
then multiply the top and the bottom of the fraction by the
conjugate of the denominator.
3 – 2i
4 + 3i
Example F. Simplify
Multiply the conjugate of the denominator (4 – 3i) to the top
and the bottom.
(3 – 2i)
(4 + 3i)
=
(4 – 3i)
(4 – 3i)* 42 + 32 =
25
12 – 8i – 9i + 6i2
–6
6 – 17i
=
25
6
25
17i
–
Using the quadratic formula, we can solve all 2nd degree
equations and obtain their complex number solutions.

52. Complex Numbers
(Division of Complex Numbers)
To divide complex numbers, we write the division as a fraction,
then multiply the top and the bottom of the fraction by the
conjugate of the denominator.
3 – 2i
4 + 3i
Example F. Simplify
Multiply the conjugate of the denominator (4 – 3i) to the top
and the bottom.
(3 – 2i)
(4 + 3i)
=
(4 – 3i)
(4 – 3i)* 42 + 32 =
25
12 – 8i – 9i + 6i2
–6
6 – 17i
=
25
6
25
17i
–
Example G. Solve 2x2 – 2x + 3 = 0 and simplify the answers.
Using the quadratic formula, we can solve all 2nd degree
equations and obtain their complex number solutions.

53. Complex Numbers
(Division of Complex Numbers)
To divide complex numbers, we write the division as a fraction,
then multiply the top and the bottom of the fraction by the
conjugate of the denominator.
3 – 2i
4 + 3i
Example F. Simplify
Multiply the conjugate of the denominator (4 – 3i) to the top
and the bottom.
(3 – 2i)
(4 + 3i)
=
(4 – 3i)
(4 – 3i)* 42 + 32 =
25
12 – 8i – 9i + 6i2
–6
6 – 17i
=
25
6
25
17i
–
Example G. Solve 2x2 – 2x + 3 = 0 and simplify the answers.
To find b2 – 4ac first: a = 2, b = –2, c = 3,
Using the quadratic formula, we can solve all 2nd degree
equations and obtain their complex number solutions.

54. Complex Numbers
(Division of Complex Numbers)
To divide complex numbers, we write the division as a fraction,
then multiply the top and the bottom of the fraction by the
conjugate of the denominator.
3 – 2i
4 + 3i
Example F. Simplify
Multiply the conjugate of the denominator (4 – 3i) to the top
and the bottom.
(3 – 2i)
(4 + 3i)
=
(4 – 3i)
(4 – 3i)* 42 + 32 =
25
12 – 8i – 9i + 6i2
–6
6 – 17i
=
25
6
25
17i
–
Example G. Solve 2x2 – 2x + 3 = 0 and simplify the answers.
To find b2 – 4ac first: a = 2, b = –2, c = 3, so b2 – 4ac = –20.
Using the quadratic formula, we can solve all 2nd degree
equations and obtain their complex number solutions.

55. Complex Numbers
(Division of Complex Numbers)
To divide complex numbers, we write the division as a fraction,
then multiply the top and the bottom of the fraction by the
conjugate of the denominator.
3 – 2i
4 + 3i
Example F. Simplify
Multiply the conjugate of the denominator (4 – 3i) to the top
and the bottom.
(3 – 2i)
(4 + 3i)
=
(4 – 3i)
(4 – 3i)* 42 + 32 =
25
12 – 8i – 9i + 6i2
–6
6 – 17i
=
25
6
25
17i
–
Example G. Solve 2x2 – 2x + 3 = 0 and simplify the answers.
To find b2 – 4ac first: a = 2, b = –2, c = 3, so b2 – 4ac = –20.
x =
2 ± –20
4
Using the quadratic formula, we can solve all 2nd degree
equations and obtain their complex number solutions.
Hence

56. Complex Numbers
(Division of Complex Numbers)
To divide complex numbers, we write the division as a fraction,
then multiply the top and the bottom of the fraction by the
conjugate of the denominator.
3 – 2i
4 + 3i
Example F. Simplify
Multiply the conjugate of the denominator (4 – 3i) to the top
and the bottom.
(3 – 2i)
(4 + 3i)
=
(4 – 3i)
(4 – 3i)* 42 + 32 =
25
12 – 8i – 9i + 6i2
–6
6 – 17i
=
25
6
25
17i
–
Example G. Solve 2x2 – 2x + 3 = 0 and simplify the answers.
To find b2 – 4ac first: a = 2, b = –2, c = 3, so b2 – 4ac = –20.
x =
2 ± –20
4 =
2 ± 2–5
4
Using the quadratic formula, we can solve all 2nd degree
equations and obtain their complex number solutions.
Hence

57. Complex Numbers
(Division of Complex Numbers)
To divide complex numbers, we write the division as a fraction,
then multiply the top and the bottom of the fraction by the
conjugate of the denominator.
3 – 2i
4 + 3i
Example F. Simplify
Multiply the conjugate of the denominator (4 – 3i) to the top
and the bottom.
(3 – 2i)
(4 + 3i)
=
(4 – 3i)
(4 – 3i)* 42 + 32 =
25
12 – 8i – 9i + 6i2
–6
6 – 17i
=
25
6
25
17i
–
Example G. Solve 2x2 – 2x + 3 = 0 and simplify the answers.
To find b2 – 4ac first: a = 2, b = –2, c = 3, so b2 – 4ac = –20.
x =
2 ± –20
4 =
2 ± 2–5
4 =
2(1 ± i5)
4 =
1 ± i5
2
Using the quadratic formula, we can solve all 2nd degree
equations and obtain their complex number solutions.
Hence

58. Powers of i
The powers of i go in a cycle as shown below:

59. The powers of i go in a cycle as shown below:
i
-1 = i2
Powers of i

60. The powers of i go in a cycle as shown below:
i
-1 = i2
-i = i3
Powers of i

61. The powers of i go in a cycle as shown below:
i
-1 = i2
-i = i3
1 = i4
Powers of i

62. The powers of i go in a cycle as shown below:
i = i5
-1 = i2
-i = i3
1 = i4
Powers of i

63. Complex Numbers
Exercise D. Divide by rationalizing the denominators.
2 + 3i
i24.
3 – 4i
i25.
3 – 4i
i26.
1 + i
1 – i27. 2 – i
3 – i28. 3 – 2i
2 + i29.
2 + 3i
2 – 3i
30.
3 – 4i
3 – 2i
31.
3 – 4i
2 + 5i
32.
33. Is there a difference between √4i and 2i?

64. The powers of i go in a cycle as shown below:
i = i5
-1 = i2 = i6 ..
-i = i3
1 = i4
Powers of i

65. The powers of i go in a cycle as shown below:
i = i5
-1 = i2 = i6 ..
-i = i3 = i7 ..
1 = i4
Powers of i

66. The powers of i go in a cycle as shown below:
i = i5
-1 = i2 = i6 ..
-i = i3 = i7 ..
1 = i4 = i8 ..
Powers of i

67. The powers of i go in a cycle as shown below:
i = i5 = i9 ..
-1 = i2 = i6 ..
-i = i3 = i7 ..
1 = i4 = i8 ..
Powers of i

68. The powers of i go in a cycle as shown below:
i = i5 = i9 ..
-1 = i2 = i6 ..
-i = i3 = i7 ..
1 = i4 = i8 ..
Example H. Simplify i59
Powers of i

69. The powers of i go in a cycle as shown below:
i = i5 = i9 ..
-1 = i2 = i6 ..
-i = i3 = i7 ..
1 = i4 = i8 ..
Example H. Simplify i59
59 = 4*14 + 3,
Powers of i

70. The powers of i go in a cycle as shown below:
i = i5 = i9 ..
-1 = i2 = i6 ..
-i = i3 = i7 ..
1 = i4 = i8 ..
Example H. Simplify i59
59 = 4*14 + 3,
hence i59 = i4*14+3
Powers of i

71. The powers of i go in a cycle as shown below:
i = i5 = i9 ..
-1 = i2 = i6 ..
-i = i3 = i7 ..
1 = i4 = i8 ..
Example H. Simplify i59
59 = 4*14 + 3,
hence i59 = i4*14+3 = i4*14+3
Powers of i

72. The powers of i go in a cycle as shown below:
i = i5 = i9 ..
-1 = i2 = i6 ..
-i = i3 = i7 ..
1 = i4 = i8 ..
Example H. Simplify i59
59 = 4*14 + 3,
hence i59 = i4*14+3 = i4*14+3 = (i4)14 i3
Powers of i

73. The powers of i go in a cycle as shown below:
i = i5 = i9 ..
-1 = i2 = i6 ..
-i = i3 = i7 ..
1 = i4 = i8 ..
Example H. Simplify i59
59 = 4*14 + 3,
hence i59 = i4*14+3 = i4*14+3 = (i4)14 i3 = 114 i3
Powers of i

74. The powers of i go in a cycle as shown below:
i = i5 = i9 ..
-1 = i2 = i6 ..
-i = i3 = i7 ..
1 = i4 = i8 ..
Example H. Simplify i59
59 = 4*14 + 3,
hence i59 = i4*14+3 = i4*14+3 = (i4)14 i3 = 114 i3 = i3 = -i
Powers of i

75. Quadratic Formula and Complex Numbers
and
From example G, the solutions of 2x2 – 2x + 3 = 0 are
x =
1 + i5
2
x = 1 – i5
2
because b2 – 4ac = –20 < 0.

76. Quadratic Formula and Complex Numbers
and
From example G, the solutions of 2x2 – 2x + 3 = 0 are
x =
1 + i5
2
x = 1 – i5
2
because b2 – 4ac = –20 < 0.
Therefore we have a complex conjugate pair as solutions.

77. Quadratic Formula and Complex Numbers
and
From example G, the solutions of 2x2 – 2x + 3 = 0 are
x =
1 + i5
2
x = 1 – i5
2
In general, for ax2 + bx + c = 0 with a, b, and c real numbers,
and b2 – 4ac < 0,
because b2 – 4ac = –20 < 0.
Therefore we have a complex conjugate pair as solutions.

78. Quadratic Formula and Complex Numbers
x =
–b +b2 – 4ac
2a
x =
–b –b2 – 4ac
2a
and
From example G, the solutions of 2x2 – 2x + 3 = 0 are
x =
1 + i5
2
x = 1 – i5
2
and
In general, for ax2 + bx + c = 0 with a, b, and c real numbers,
and b2 – 4ac < 0, then its two roots:
because b2 – 4ac = –20 < 0.
are of the form A + Bi and A – Bi, a conjugate pair.
Therefore we have a complex conjugate pair as solutions.

79. Quadratic Formula and Complex Numbers
x =
–b +b2 – 4ac
2a
x =
–b –b2 – 4ac
2a
and
If a, b, and c are real numbers, then the complex roots* for
ax2 + bx + c = 0
are a complex conjugates pair. ( * if b2 – 4ac < 0).
From example G, the solutions of 2x2 – 2x + 3 = 0 are
x =
1 + i5
2
x = 1 – i5
2
and
In general, for ax2 + bx + c = 0 with a, b, and c real numbers,
and b2 – 4ac < 0, then its two roots:
because b2 – 4ac = –20 < 0.
are of the form A + Bi and A – Bi, a conjugate pair.
Therefore we have a complex conjugate pair as solutions.

80. Quadratic Formula and Complex Numbers
x =
–b +b2 – 4ac
2a
x =
–b –b2 – 4ac
2a
and
If a, b, and c are real numbers, then the complex roots* for
ax2 + bx + c = 0
are a complex conjugates pair. ( * if b2 – 4ac < 0).
From example G, the solutions of 2x2 – 2x + 3 = 0 are
x =
1 + i5
2
For example, if x = i is a solution of #x2 + # x + # = 0,
then automatically x = – i is the other root (# real numbers.)
x = 1 – i5
2
and
In general, for ax2 + bx + c = 0 with a, b, and c real numbers,
and b2 – 4ac < 0, then its two roots:
because b2 – 4ac = –20 < 0.
are of the form A + Bi and A – Bi, a conjugate pair.
Therefore we have a complex conjugate pair as solutions.

81. Complex Numbers
In what sense are the complex numbers, numbers?
Real numbers are physically measurable quantities (or the
lack of such quantities in the case of the negative numbers).
Theoretically, we can forever improve upon the construction of
a stick with length exactly 2. But how do we make a stick of
length 3i, or a cookie that weighs 3i oz? Well, we can’t.
Imaginary numbers and complex numbers in general are not
physically measurable in the traditional sense. Only the real
numbers, which are a part of the complex numbers, are
tangible in the traditional sense.
Complex numbers are directional measurements.
They keep track of measurements and directions,
i.e. how much and in what direction (hence the two–
component form of the complex numbers).
Google the terms “complex numbers, 2D vectors” for further
information.

82. Complex Numbers
Exercise A. Write the complex numbers in i’s. Combine the
following expressions.
1. 2 – 2i + 3 + √–4 2. 4 – 5i – (4 – √–9) 3. 3 + 2i + (4 – i√5)
4. 4 – 2i + (–6 + i√3) 5. 4 – √–25 – (9 – √–16)
6. 11 – 9i + (–7 + i√12) 7. ½ – (√–49)/3 – (3/4 – √–16)
Exercise B. Do by inspection.
8. (1 – 2i)(1 + 2i) 9. (1 + 3i)(1 – 3i) 10. (2 + 3i)(2 – 3i)
11. (3 – 4i)(3 + 4i) 12. (9 + i√3)(9 – √3i) 13. (7 – i√5)(7 + i√5)
14. (9 + i√3) (7 – i√5)(9 – i√3) (7 + i√5)
15. (√3 + i√3) (√7 – i√5)(√3 – i√3)(√7 + i√5)
Exercise C. Expand and simplify.
16. (1 – 3i)(1 + 2i) 17. (2 + 3i)(1 – 3i) 18. (2 + 3i)(3 – 2i)
19. (4 – 3i)(3 – 4i) 20. (5 + 3i)(5 + 3i) 21. (1 – i)2
22. (2 + 3i)2 23. (5 + 2i)2

83. Complex Numbers
Exercise D. Divide by rationalizing the denominators.
2 + 3i
i24.
3 – 4i
i25.
3 + 4i
i26.
1 + i
1 – i27. 2 – i
3 – i28. 3 – 2i
2 + i29.
2 + 3i
2 – 3i
30.
3 – 4i
3 – 2i
31.
3 – 4i
2 + 5i
32.
Simplify
33. i92
38. Find a and b if (a + bi) 2 = i.
34. i –25 36. i 205
37. i –102
39. Is there a difference between √4i and 2i?

84. Complex Numbers
(Answers to odd problems) Exercise A.
1. 5 3. 7 + (2 – √5) i
Exercise B.
9. 10 11. 5
Exercise C.
17. 11 – 3i 19. – 25i
5. –5 – i 7. –1/4 + 5/3 i
13. 54 15. 72
21. – 2i 23. 21 + 20i
Exercise D.
– 4 – 3i25. 27.
33. 1
i 29.
1
5
(4 – 7i) 31.
1
13
(17 – 6i)
37. – 1 39. There is no difference