Download 2.5 Complex Numbers and Roots

2.5 Complex Numbers and Roots
Objectives:
N.CN.1: Know there is a complex number i such that i2 = -1, and every complex number has the
form a + bi with a and b real.
N.CN.2: Use the relation i2 = -1, and the commutative, associative, and distributive properties to
add, subtract, and multiply complex numbers.
N.CN.7: Solve quadratic equations with real coefficients that have complex solutions.
For the Board: You will be able to define and use imaginary and complex numbers.
You will be able to solve quadratic equations with complex roots.
Bell Work 2.5:
Simplify each expression.
1.
2.
108
6  24
Find the zeros of each function.
4. f(x) = x2 – 18x + 16
3.
42
 3
5. f(x) = x2 + 6x – 24
Anticipatory Set:
f(x) = x2 + 1 has no real zeros.
Solve for the zeros using square rooting.
x2 + 1 = 0
x2 = -1
x =  1
We can find solutions if we define the square root of a negative number.
This is why imaginary numbers were invented.
The imaginary unit i is defined as  1 .
An imaginary number is the square root of a negative number.
1 = I
 2 = 1 ∙ 2 = i 2
 4 =  1 ∙ 4 = 2i
Imaginary numbers can be written in the form bi, where b is a real number and i is the imaginary unit.
If b is a positive real number, then  b = i b .
The square of an imaginary number is the original negative number.
(  1 )2 = i2 = -1
 b2 =
Open the book to page 95 and read example 1.
Example: Express each number in terms of i.
a. 5  121
5  1 ∙ 121
5i ∙ 11
55i
 1  b 2 = bi
b. -  96
-  1 ∙ 96
-i 16 ∙ 6
-4i 6
(  b )2 = (i  b )2 = -b.
White Board Activity:
Practice: Express each number in terms of i.
a.
 12
 1 ∙ 12
b. 2  36
2  1 36
i 4∙ 3
2i ∙ 6
2i 3
12i
Open the book to page 95 and read example 2.
Example: Solve each equation. Use square rooting.
a. x2 = -144
b. 5x2 + 90 = 0
x =   144
5x2 = -90
x =   1  144
x =   18
x = ±12i
x = ±i 9  2
White Board Activity:
Practice: Solve each equation. Use square rooting.
a. x2 = -36
b. x2 + 48 = 0
x = ±  36
x2 = -48
x = ±  1  36
x = ±  48
x = ±6i
x = ±  1  48
1
 63
3
1

 1  63
3
1
 i 9 7
3
1
 i 3 7
3
i 7
c. 
x2 = -18
x =   1  18
x = ±3i 2
c. 9x2 + 25 = 0
9x2 = -25
25
x2 = 
9
25
x=± 
9
x = ±i 16  3
x = ± 1 
x = ±4i 3
5
x=  i
3
25
9
A complex number is a number that can be written in the form a + bi, where a and b are real numbers
and i =  1 .
Examples: 3 - 5i
-5 + i 2
2 ± 4i 3
The set of real numbers is a subset of the set of complex numbers C.
Open the book to page 95 and study the Venn Diagram.
Every complex number has a real part “a” and an imaginary part “b”.
Real numbers are complex numbers where b = 0.
Imaginary numbers are complex numbers where a = 0 and b ≠ 0.
These are sometimes called pure imaginary numbers.
Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are
equal .
Open the book to page 95 and read example 3.
Example: Find the values of x and y that make the equation 4x + 10i = 2 – (4y)i true.
4x + 10i = 2 + (-4y)i
4x = 2
10 = -4y
X=½
y = 10/-4
y = -2.5
White Board Activity:
Practice: Find the values of x and y that make each equation true.
a. 2x – 6i = -8 + (20y)i
b. -8 + (6y)i = 5x - i 6
2x + (-6)I = -8 + (20y)I
-8 + (6y)I = 5x + (- 6 )i
2x = -8
-6 = 20y
-8 = 5x
6y = - 6
x = -4
y = -6/20
x = -8/5
y= 
6
6
= -3/10
Open the book to page 96 and read example 4.
Example: Find the zeros of each function using completing the square.
a. f(x) = x2 + 10x + 26
b. g(x) = x2 + 4x + 12
x2 + 10x + 26 = 0
x2 + 4x + 12 = 0
2
(x + 10x + _____) = -26 + _____
(x2 + 4x + _____) = -12 + _____
½ (10) = 5; 52 = 25
½ (4) = 2; 22 = 4
(x2 + 10x + 25) = -26 + 25 = -1
(x2 + 4x + 4) = -12 + 4 = -8
2
(x + 5) = -1
(x + 2)2 = -8
x + 5 =  1
x + 2 =   8    1  8  i 4  2
x = -5 ± i
x = -2 ± 2i 2
White Board Activity:
Practice: Find the zeros of each function using completing the square.
a. f(x) = x2 + 4x + 13
b. g(x) = x2 – 8x + 18
2
x + 4x + 13 = 0
x2 - 8x + 18 = 0
(x2 + 4x + _____) = -13 + _____
(x2 - 8x + _____) = -18 + _____
½ (4) = 2; 22 = 4
½ (8) = 4; 42 = 16
2
(x + 4x + 4) = -13 + 4 = -9
(x2 - 8x + 16) = -18 + 16 = -2
(x + 2)2 = -9
(x - 4)2 = -2
x + 2 =  1  9
x - 4 =   2   1  2
x = -2 ± 3i
x=4±i 2
Solutions like all of the above are called complex conjugate pairs.
-5 + I and -5 – I
-2 + 2i 2 and -2 - 2i 2
-2 + 3i and -2 - 3i
4 + i 2 and 4 - i 2
Their real parts are equal and their imaginary parts are opposites.
The complex conjugate of a + bi is a – bi.
Note: If a quadratic equation has non-real roots, then these roots are complex conjugates.
Open the book to page 96 and read example 5.
Example: Find each complex conjugate.
a. 8 + 5i
b. 6i
8 – 5i
White Board Activity:
Practice: Find each complex conjugate.
a. 9 – i
9+i
-6i
b. i + 2
-I +
Assessment:
Question student pairs.
Independent Practice:
Text: pg. 97 – 98 prob. 2 – 35, 43 - 57.
For a Grade:
Text: pg. 97 – 98 prob. 4, 8, 16, 24, 34.
c. -8i
2
8i