1. P.6 Complex Numbers
Pre-calculus

2. Complex number:
Sometimes an equation such as 𝑓 𝑥 = 𝑥 2 +1 has no real zeros, and therefore, no real number solutions.
To fix this, we need to be able to take −1
Imaginary unit: i= −1
For any negative number 𝑎: 𝑎 = 𝑎 ∙𝑖
Example: −4 = 4 ∙𝑙𝑖=2𝑖

3. The pattern of 𝑖:
𝑖= −1 =𝑖 𝑖=𝑖 𝑖 2 = −1 ∙ −1 =−1 𝑖 2 =−1 𝑖 3 = 𝑖 2 ∙ −1 =−1𝑖=−𝑖 𝑖 3 =−𝑖 𝑖 4 = 𝑖 2 ∙ 𝑖 2 =−1∙−1=1 𝑖 4 =1 𝑖 5 = 𝑖 4 ∙𝑖=1∙𝑖=𝑖 𝑖 5 =𝑖 𝑖 6 = 𝑖 4 ∙ 𝑖 2 =−1∙1=−1 𝑖 6 =−1 ⋮ 𝑖 7 =−𝑖 𝑖 8 =1
This pattern Repeats forever

4. Complex Numbers: Any number that can be written in the form: 𝑎+𝑏𝑖
*** a and b are both numbers. The 𝑖 is the imaginary part
Examples: − 𝑐𝑎𝑛 𝑏𝑒 𝑤𝑟𝑖𝑡𝑡𝑒𝑛 𝑎𝑠 −6+0𝑖
5𝑖 −7𝑖 (can be written as 𝑖)
𝑖 −2+3𝑖 etc

5. Adding and subtracting Complex numbers
𝑎+𝑏𝑖 + 𝑐+𝑑𝑖 = 𝑎+𝑐 + 𝑏+𝑑 𝑖
Example:
7−3𝑖 + 4+5𝑖 =
7+4 + −3+5 𝑖=
11+2𝑖

6. You try:
8−4𝑖 + 4+3𝑖 = (answer on next click) 12−𝑖

7. Adding and subtracting Complex numbers
𝑎+𝑏𝑖 − 𝑐+𝑑𝑖 = 𝑎−𝑐 + 𝑏−𝑑 𝑖
Example:
2−𝑖 − 8+3𝑖 =
2−8 + −1−3 𝑖=
−6−4𝑖

8. You try:
−8+5𝑖 − 4−2𝑖 = (Answer on next click) −12+7𝑖

9. What is: 𝑎+𝑏𝑖 + −𝑎−𝑏𝑖 ?
Answer: 𝑎−𝑎 + 𝑏−𝑏 𝑖=
0+0𝑖=

10. Multiplying Complex Numbers
F O I L
𝑎+𝑏𝑖 𝑐+𝑑𝑖
𝑎𝑐+𝑎𝑑𝑖+𝑏𝑐𝑖+𝑏𝑑 𝑖 2
𝑎𝑐+ 𝑎𝑑+𝑏𝑐 𝑖+𝑏𝑑 −1
(remember 𝑖 2 =−1 )

11. Practice Problem: 2+3𝑖 5−𝑖 = (2)(5)+(2) −𝑖 + 3𝑖 5 +(3𝑖)(−𝑖)
2+3𝑖 5−𝑖 =
(2)(5)+(2) −𝑖 + 3𝑖 5 +(3𝑖)(−𝑖)
10−2𝑖+15𝑖−3 𝑖 2
10+13𝑖−3 −1
10+13𝑖+3
13+13𝑖

12. Your turn:
Simplify:
7−3𝑖 3+4𝑖
(Answer on next click)
33+19𝑖

13. Complex conjugate
Conjugate: Sometimes we want to get rid of the imaginary part of a number, and so we times by the conjugate.
If we have an imaginary number: 𝑎+𝑏𝑖, then the conjugate is 𝑎−𝑏𝑖

14. Dividing Complex Numbers
Example: Write the complex number in standard form:
2 3−𝑖
2 3−𝑖 ∙ 3+𝑖 3+𝑖
(multiply both the top and the bottom by the conjugate of the bottom)
2(3+𝑖) 9+3𝑖−3𝑖− 𝑖 2
6+2𝑖 9−(−1)
6+2𝑖 10 = 3+𝑖 5

15. Your turn:
Write the complex number in standard form: 5 2−3𝑖 Answer on next click 10+15𝑖 13

16. Homework
Textbook:
P6, pg 57: 1-4, 7, 9-13, 17-20, odd