A power point lecture design for a digit unit for a high school math course, Algebra 2 or pre calculus

1. What you should learn
1) Use the imaginary unit 𝑖 to
write complex number
2) Add, subtract, and multiply
complex numbers.
3) Use complex conjugates to
write the quotient of two
complex numbers in
standard form
4) Find complex solutions of
quadratic equation
Inspired and base on Larson’s Pre-Calculus textbook,
Eight Edition. Published by Cengage Learning

2. Warm Up
Do the following problems
Question: Find the solutions to the following quadratic
equations
1. 𝑋2 + 3𝑥 + 5= 0
2. 𝑋2 − 36 = 0
3. 𝑋2 + 1= 0

3. Warm Up Answers
Question: Find the solutions to the following quadratic
equations
1. 𝑋2 + 3𝑥 + 5= 0
Method: Use factoring.
Process: Ask what two numbers add to 5 that can multiple to made 3?
Answer: (𝑥 + 2)(𝑥 + 1)
2. 𝑋2 − 36 = 0
Method: same as above.
Answer: (𝑥 − 6)(𝑥 + 6)
3. 𝑋2
+ 1= 0
Method: Use quadratic equation or basic factoring

4. Warm Up Answers Cont.
Question: Find the solutions to the following quadratic
equations
3. X2 + 1= 0
Lets see this in detail
X2 + 1 + −1 = 0 + (−1) Add -1 to both side to get X alone
X2 = −1 Simplify
X2 = −1 Square both side to get rid of the power of two
X = −1 Simplify
Are the solutions X = + −1 and X = − −1 ?

5. Brief Introduction

6. Imaginary Unit 𝑖
X = −1
What is the problem?
The problem is the negative under the square root.
We need to get rid of the negative number some how.
Let use this idea.
𝑖 = −1 and 𝑖2 = −1
The 𝑖 is special, we call it the imaginary unit.

7. Imaginary Unit 𝑖 Video
This is a funny clip, Remember Student, imaginary unit i is
not hard to understand. Your head won’t blow up.

8. Imaginary Unit 𝑖 cont.
Let revisit the problem, here are the last few steps
X2 = −1 Square both side to get rid of the power of two
X = −1 Simplify
Let use our new idea
X = 1𝑖 Imaginary unit substitution
Solutions X = + 1 𝑖 and X = − 1𝑖
The solutions are in the form of COMPLEX NUMBER.
A complex number is any number that has the 𝑖 attached to it.

9. Complex Number
Our last problem we had solutions of X = + 1 𝑖 and X = − 1𝑖
The solutions are in the form of COMPLEX NUMBER.
Let now learn the standard form of a complex number
It looks like this 𝑎 + 𝑏𝑖
Here are some examples, 5 + 2𝑖 and 7 − 3𝑖
Key ideas: the a is the real part because there is no imaginary unit of 𝑖
next to it.
Question: what are the real parts of 5 + 2𝑖 and 7 − 3𝑖
Answer: For 5 + 2𝑖 the 5 is the real part
and 7 − 3𝑖 the 7 is the real part

10. Complex Number cont.
𝑎 + 𝑏𝑖
The a is the real part.
The 𝑏𝑖 is call the imaginary part because the 𝑖 is attached to the letter b
A number like 7 − 3𝑖 is called a complex number
Some definition:
6𝑖 is called a pure imaginary number because there no real part to it.
7 is called a real number because there is no 𝑖.
You just learned how to “Use the imaginary unit 𝑖 to write complex
number”

11. Complex Number Conjugate
This complex number 7 − 3𝑖 has an evil twin, called the conjugate.
The conjugate is 7 + 3𝑖
Idea: So every 𝑎 + 𝑏𝑖 will have a conjugate of 𝑎 − 𝑏𝑖
Question: Write the conjugate for the following complex
numbers
1) 2 + 3𝑖
2) 8 − 7𝑖
3) 𝛼 + 2𝑖
4) 𝜋 − 7𝑖

12. Complex Number Conjugate Cont.
Question: Write the conjugate for the following complex numbers. Also
label the real and imagery part to the conjugate.
1) 2 + 3𝑖
2) 8 − 7𝑖
3) 𝛼 + 2𝑖
4) 𝜋 − 7𝑖
Answer:
1) For 2 + 3𝑖, the conjugate is 2 − 3𝑖
The 2 is the real part, the −3𝑖 is the imagery
2) For 8 − 7𝑖, the conjugate is 8 + 7𝑖
The 8 is the real part, the +7𝑖 is the imagery
3) For 𝛼 + 2𝑖, the conjugate is 𝛼 − 2𝑖
The 𝛼 is the real part, the −2𝑖 is the imagery
4) For 𝜋 − 7𝑖, the conjugate is 𝜋 + 7𝑖
The 𝜋 is the real part, the 7𝑖 is the imagery

13. Operation with Complex Numbers
You remember how to add right?
1+1 =2
Do you remember how to subtract?
5-2= 3
Lets now add, subtract, multiply, divide complex numbers
***Key Ideas***
The sum: 𝑎 + 𝑏𝑖 + 𝑐 + 𝑑𝑖 = 𝑎 + 𝑐 + 𝑏 + 𝑑 𝑖
The difference: 𝑎 + 𝑏𝑖 − 𝑐 + 𝑑𝑖 = 𝑎 − 𝑐 + 𝑏 − 𝑑 𝑖
The multiplication: 𝑎 + 𝑏𝑖 + 𝑐 + 𝑑𝑖 = 𝑎𝑐 + 𝑎𝑑𝑖 + 𝑐𝑏𝑖 − 𝑏𝑑
You remember how to multiply right?
4 *2 =8

14. Operation with Complex Numbers: Sum
The sum: 𝑎 + 𝑏𝑖 + 𝑐 + 𝑑𝑖 = 𝑎 + 𝑐 + 𝑏 + 𝑑 𝑖
Examples
Question: Find the sum of the two complex numbers
4 + 7𝑖 + 1 − 6𝑖
4 + 7𝑖 + 1 − 6𝑖 = 4 + 7𝑖 + 1 − 6𝑖 Remove the parentheses
= 4 + 1 + 7𝑖 − 6𝑖 Group the like term, the imaginary part goes together
and the real parts go together.
= 5 + 𝑖 Simplify and write it in standard form.
The real part is 5 and the imaginary part is 𝑖

16. Operation with Complex Numbers: Difference
The difference: 𝑎 + 𝑏𝑖 − 𝑐 + 𝑑𝑖 = 𝑎 − 𝑐 + 𝑏 − 𝑑 𝑖
Examples
Question: Find the difference of the two complex numbers
6 + 4𝑖 − 1 + 2𝑖
6 + 4𝑖 − 1 + 2𝑖 = 6 + 4𝑖 − 1 − 2𝑖 Remove the parentheses, by
distrusting the negative into the parenthesis
= 6 − 1 + 4𝑖 − 2𝑖 Group the like term, the imaginary part goes together
and the real parts go together.
= 5 + 2𝑖 Simplify and write it in standard form.
The real part is 5 and the imaginary part is 2𝑖

18. Operation with Complex Numbers: Multiplying
The multiplication: 𝑎 + 𝑏𝑖 + 𝑐 + 𝑑𝑖 = 𝑎𝑐 + 𝑎𝑑𝑖 + 𝑐𝑏𝑖 − 𝑏𝑑
Examples
Question: Find the multiplication of the two complex numbers
2 − 4𝑖 4 + 3𝑖
2 − 4𝑖 4 + 3𝑖 = 2 4 + 3𝑖 − 𝑖(4 + 3𝑖) Distributive
= 8 + 6𝑖 − 4𝑖 − 3𝑖2
Remove the parentheses and distributive
= 8 + 6𝑖 − 4𝑖 + 3 Remember that 𝑖2
= -1
= 11 + 2𝑖 Like before, combine the like terms.
The real part is 11 and the imaginary part is 2𝑖

20. Operation with Complex Numbers: Key Idea
Look at your answers to the addition, subtraction, and multiplication
example above.
5 + 𝑖 and 5 + 2𝑖 𝑎𝑛𝑑 11 + 2𝑖
Notice any interest? Discussion among your class mates.
Ask each other this question, when I add or subtract or multiply a complex
number, do I always get a complex number in return?
Why do I always get a complex number in return?
When I add, subtract, or multiply a regular real number, I always get a real
numbers in return. WOW… real numbers are like complex numbers.
You just learned to add, subtract, and multiply complex numbers.

21. Operation with Complex Numbers: Key Idea

22. Multiplying Conjugate
Question: How do I get just a real number form a complex number?
Answer: multiply by its conjugate
Example. 4 + 7𝑖
To get a real number you got to multiply by its conjugate
4 − 7𝑖 is the conjugate
Therefore it becomes 4 + 7𝑖 4 − 7𝑖
4 + 7𝑖 4 − 7𝑖 = 16 − 28𝑖 + 28𝑖 − 49𝑖2 Distribute, (FOIL)ing
= 16 − 28𝑖 + 28𝑖 + 49 Rememeber that 𝑖2 = −1
= 16 + 49 Simplify
= 16 + 49
= 65
Key idea: when you multiple a complex number by its conjugate, you
get a real number, not an imagery number

23. Divide complex numbers
𝑎+𝑏𝑖
𝑐+𝑑𝑖
=
𝑎+𝑏𝑖
𝑐+𝑑𝑖
(
𝑐−𝑑𝑖
𝑐−𝑑𝑖
) The red is just multiplying by 1, but in the form
of the conjugate
=
(𝑎𝑐+𝑏𝑑)+(𝑏𝑐−𝑎𝑑)𝑖
𝑐2+𝑑2 notice how the denominator is a real number,
there is no 𝑖
=
(𝑎𝑐+𝑏𝑑)
𝑐2+𝑑2 +
𝑏𝑐−𝑎𝑑)
𝑐2+𝑑2 𝑖
The real part is
(𝑎𝑐+𝑏𝑑)
𝑐2+𝑑2 , the imaragy part is
𝑏𝑐−𝑎𝑑)
𝑐2+𝑑2 𝑖

24. Divide complex numbers Cont.
Examples
2+3𝑖
4−2𝑖
2+3𝑖
4−2𝑖
=
2+3𝑖
4−2𝑖
4+2𝑖
4+2𝑖
Multiply the numerator ( top) and denominator
(bottom) by complex conjugate of denominator
=
8+4𝑖+12𝑖+6𝑖2
16+4𝑖2 Expand
=
8−6+16𝑖
16+4
Rememeber that 𝑖2
= −1
=
2+16𝑖
20
= =
2
20
+
16
20
𝑖 Simplify and break it into the different parts
The real part is
2
20
, the imaragy part is
16
20
𝑖
You just learn how to use complex conjugates to write the quotient of two
complex numbers in standard form.

26. Putting everything together, Complex solution of
Quadratic Equation
Solve 𝑥2 + 4 = 0
𝑥2 = −4 isolate the 𝑥2
𝑥2 = −4 squared both side to get rid of the power of two
𝑥 = 4𝑖 use the imaginary unit to get rid of the negative
𝑥 = + 4𝑖 and 𝑥 = − 4𝑖 Simplify

27. Putting everything together, Complex solution of
Quadratic Equation Cont.
Solve 3𝑥2
− 2𝑥 + 5 = 0
𝐿𝑒𝑡 𝑎 = 3 𝑏 = −2 𝑐 = 5
𝑥 =
− −2 ± (−2)2−4(3)(5)
2(3)
Use the quadratic Formula
𝑥 =
2±2 −56
6
Simplify
𝑥 =
2±2 56𝑖
6
Write −56 in standard form
𝑥 =
2
6
±
2 56𝑖
6
Part it into the two parts
𝑥 =
1
3
±
14
3
𝑖 Simplify into the standard form
The real part is
1
3
and the imagery part
14
3
𝑖
You just learned to find complex solutions of quadratic equation