1. Objectives for Class 3
Add, Subtract, Multiply, and Divide Complex Numbers.
Solve Quadratic Equations in the Complex Number System

2. Basics of the Complex Number System
(Square root of -1) = i
Simplify the following square roots
(Square root of -25)
Pull the square root of -1 out first to form i(square root of 25)
Break down the square root of 25
(square root of -25) = 5i

3. (square root of -81) 9i (square root of -20) 2i(square root of 5)
Simplify each of the following. Hit “enter” to check your answers. See Mrs. Dorshorst for help if needed.
(square root of -81) 9i
(square root of -20)
2i(square root of 5)

4. Complex Numbers: a + bi ‘a’ is called the real part
‘b’ is called the imaginary part
Standard Form: a + bi
Ex: 3 – 5i
Real Part is 3
Imaginary part is -5i
Real number: a
Ex: No imaginary part
Pure imaginary: bi
Ex: 3i No real part

5. Equal Complex Numbers: a + bi = c + di
a must equal c : Real parts must be equal
b must equal d: Imaginary parts must be equal
Ex: 3x + 5yi = -7 – 4i
Set real parts equal and solve for variable:3x =-7
X = -7/3
Set imaginary parts equal and solve for variable: 5yi = -4i
5y = so y = -4/5

6. 8x – 9yi = 4 + 2i
8x = 4
x = 1/2
-9y = 2
y = -2/9

7. Addition and Subtraction of Complex Numbers
Sum of Complex Numbers: (a + bi) + (c + di) = (a + c) + (b + d)I
Ex: (4 – 7i) + ( i)
(4 + 12) + (-7i + 19i) i
Try: (8 + 52i) + (-12 – 40i)
Solution: i

8. Difference of Complex Numbers: (a + bi) – (c + di) = (a – c) + (b – d)i
Ex: (-3 + 7i) – (-8 + 9i)
Distribute the subtraction through the second complex number
(-3 + 7i) + ( +8 – 9i)
Add like terms
(-3 + 8) + (7i – 9i)
5 – 2i

9. Try: (25 – 3i) – (13 – 10i)
(25 – 13) + (-3i + 10i)
12 + 7i

10. Examples (2 – 5i) – (8 + 6i) Solution: -6 – 11i (3 + 5i) + (6 – 8i)

11. Product of Complex Numbers
Distribute each term through the next parenthesis.
Remember that i2 = -1
Examples
(5 + 3i)(2 + 7i)
i + 6i + 21i2
i + 6i – 21 i

12. (3 – 4i)(2 + i)
6 + 3i – 8i -4i2
6 + 3i – 8i + 4
10 – 5i
3i(-3 + 4i)
-9i - 12

13. Conjugate Same values with opposite middle sign
Ex: i conjugate: 2 – 3i
_______
Notation: a + bi
____
Find the conjugate of i
3 – 4i

14. Multiply 4 – 2i times its conjugate
(4 – 2i)(4 + 2i)
Conjugate is 4 + 2i)
16 + 8i – 8i - 4i2
Distribute 1st parenthesis through second parenthesis
16 + 4
Combine like terms and substitute -1 in for i2 20

15. Multiply 3 + 2i times its conjugate
What is the conjugate?
3 – 2i
Do the multiplication.
(3 + 2i)(3 – 2i)
Solution: 13

16. Dividing Complex Numbers
Multiply both the numerator and the denominator times the conjugate to eliminate the radical in the denominator (recall that ‘i’ is a radical)
Example: (1 + 4i) / (5 – 12i)
((1+4i)(5+12i))/((5-12i)(5+12i))
Multiply by conjugate of denominator
(5 + 12i + 20i +48i2)/( i – 60i – 144i2)
Multiply numerators and multiply denominators
(5 + 32i – 48)/( )
Simplify
( i) / (169)

17. (2 – 3i) / (4 – 3i)
What would you multiply by to eliminate the radical in the denominator?
4 + 3i
Multiply numerator and multiply denominator
((2 – 3i)(4 + 3i))/((4 – 3i)(4+ 3i))
Simplify and find answer
Solution: (17 – 6i) / (25)

18. Other Examples If z = 2 – 3i and w = 5 + 2i find each of the following
_____
Find (z + w)
What does this notation mean?
Conjugate of the sum
Add then state conjugate.
7 – 1i so answer is 7 + 1i

19. Powers of i Recall that i2 = -1
To find high level powers of i: Ex: i35
1.Break down the expression into i2 notation. (i2)17i
Substitute -1 in for i2 (-1)17i
Simplify as far as possible. (-1)i

20. Simplify: i7 + i25 – i8 Simplify each power of i Combine like terms
i7 = (i2)3i = (-1)3i = -1i
i25 = (i2)12i = (-1)12i = 1i
i8 = (i2)4 = (-1)4 = 1
Combine like terms
-1i + 1i – (1) = 1

21. Solving Quadratic Equations in the Complex Number System
Solve: x2 – 4x = -8
1. Put into standard form.
x2 – 4x + 8 = 0
2. Solve using the Quadratic Formula. When simplifying the radical remember that the square root of -1 is i.
(4 + sqrt(16 – 4(1)(8))) / 2
(4 + sqrt -16)/2
(4 + 4i)/2
2 + 2i

22. Discriminant: b2 – 4ac
Value under the radical in the quadratic formula is called the discriminant.
The value of the discriminant determines the type of solution for the equation.
D > 0: discriminant is positive there are 2 real solutions
D < 0: discriminant is negative there are 2 complex solutions
D = 0: discriminant is o there is one real solution (double root)

23. What type of solutions do the following equations have
What type of solutions do the following equations have? Do not solve just determine the type of solutions. (Recall the rules for discriminants on the previous slide.
3x2 – 8x = 9
b2 – 4ac => – 4(3)(-9)
D > real number solutions
4x2 + 3 = 5x
b2 – 4ac => – 4(4)(3)
D > complex number solutions

24. Assignment
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#9, 13, 19, 23, 25, 27, 31, 35, 39, 49, 51, 55, 59, 63, 75, 77, 83,