Complex Numbers: a + bi
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Transcript Complex Numbers: a + bi
Objectives for Class 3
Add, Subtract, Multiply, and Divide
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
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
Real number: a
Ex: 3 – 5i
Real Part is 3
Imaginary part is -5i
Ex: 8
No imaginary part
Pure imaginary: bi
Ex: 3i
No real part
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 = -4
so y = -4/5
8x – 9yi = 4 + 2i
8x
=4
x = 1/2
-9y = 2
y = -2/9
Addition and Subtraction of Complex
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
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
Try: (25 – 3i) – (13 – 10i)
(25 – 13) + (-3i + 10i)
12 + 7i
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
Same values with opposite middle sign
Ex: 2 + 3i conjugate: 2 – 3i
_______
Notation:
a + bi
Find the conjugate of
3 – 4i
____
3 + 4i
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
Multiply 3 + 2i times its conjugate
What
is the conjugate?
3 – 2i
Do the multiplication.
(3 + 2i)(3 – 2i)
Solution: 13
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)/(25 + 60i – 60i – 144i2)
Multiply numerators and multiply denominators
(5 + 32i – 48)/(25 + 144)
Simplify
(-43 + 32i) / (169)
(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)
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
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
Simplify each power of i
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
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
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)
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 => 64 – 4(3)(-9)
D>0
2 real number solutions
4x2 + 3 = 5x
b2 – 4ac => 25 – 4(4)(3)
D>0
2 complex number solutions
Assignment
Page 117
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