#### Transcript Complex Numbers: a + bi

```Objectives for Class 3
Complex Numbers.
 Solve Quadratic Equations in the
Complex Number System

Basics of the Complex Number
System
 i2
= -1
 (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
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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)
Complex Numbers: a + bi
‘a’ is called the real part
 ‘b’ is called the imaginary part
 Standard Form: a + bi
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Real number: a
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Ex: 3 – 5i
Real Part is 3
Imaginary part is -5i
Ex: 8
No imaginary part
Pure imaginary: bi
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Ex: 3i
No real part
Equal Complex Numbers:
a + bi = c + di
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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 = -4
so y = -4/5
8x – 9yi = 4 + 2i
 8x
=4
 x = 1/2
 -9y = 2
 y = -2/9
Numbers

Sum of Complex Numbers: (a + bi) + (c
+ di) = (a + c) + (b + d)I
Ex: (4 – 7i) + (12 + 19i)
 (4 + 12) + (-7i + 19i)
 16 + 12i

Try: (8 + 52i) + (-12 – 40i)
 Solution: -4 + 12i
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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)
 (-3 + 8) + (7i – 9i)
 5 – 2i

Try: (25 – 3i) – (13 – 10i)
(25 – 13) + (-3i + 10i)
 12 + 7i
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Examples
– 5i) – (8 + 6i)
 Solution: -6 – 11i
 (2
+ 5i) + (6 – 8i)
 Solution: 9 – 3i
 (3
Product of Complex Numbers

Distribute each term through the next
parenthesis.
 Remember that i2 = -1
 Examples
(5 + 3i)(2 + 7i)
10 + 35i + 6i + 21i2
10 + 35i + 6i – 21
-11 + 41i
(3 – 4i)(2 + i)
6 + 3i – 8i -4i2
6 + 3i – 8i + 4
10 – 5i
3i(-3 + 4i)
-9i - 12
Conjugate
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Same values with opposite middle sign
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Ex: 2 + 3i conjugate: 2 – 3i
_______
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Notation:
a + bi
Find the conjugate of
 3 – 4i
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____
3 + 4i
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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
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Multiply 3 + 2i times its conjugate
What
is the conjugate?
3 – 2i
Do the multiplication.
(3 + 2i)(3 – 2i)
Solution: 13
Dividing Complex Numbers
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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)/(25 + 60i – 60i – 144i2)
Multiply numerators and multiply denominators
(5 + 32i – 48)/(25 + 144)
Simplify
(-43 + 32i) / (169)
(2 – 3i) / (4 – 3i)
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What would you multiply by to eliminate
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4 + 3i
Multiply numerator and multiply
denominator
 ((2 – 3i)(4 + 3i))/((4 – 3i)(4+ 3i))
 Solution: (17 – 6i) / (25)
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Other Examples
If z = 2 – 3i and w = 5 + 2i find each of
the following
_____
 Find
(z + w)
 What does this notation mean?
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Conjugate of the sum
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7 – 1i
so answer is 7 + 1i
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
2. Substitute -1 in for i2 (-1)17i
3. Simplify as far as possible. (-1)i
Simplify: i7 + i25 – i8
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Simplify each power of i
 i7
= (i2)3i = (-1)3i = -1i
 i25 = (i2)12i = (-1)12i = 1i
 i8 = (i2)4 = (-1)4 = 1
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Combine like terms

-1i + 1i – (1) = 1
Complex Number System
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Solve: x2 – 4x = -8
1. Put into standard form.
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x2 – 4x + 8 = 0
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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
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Discriminant: b2 – 4ac
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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)
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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
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b2 – 4ac => 64 – 4(3)(-9)
D>0
2 real number solutions
4x2 + 3 = 5x
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b2 – 4ac => 25 – 4(4)(3)
D>0
2 complex number solutions
Assignment
Page 117
 #9, 13, 19, 23, 25, 27, 31, 35, 39, 49, 51,
55, 59, 63, 75, 77, 83,
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