Transcript Imaginary Numbers

Complex Numbers - Day 1
 My
introduction
 Syllabus
 Start with add/subtract like variables
(without any brackets)
 Introduce i and complex numbers
(include square root of negative
numbers)
 Add/subtract with i
 Now do distributing variables
 i equals ….
Complex Numbers - Day 2
 Review
previous day in warm-up
(include conjugates)
 Now expand products of i to higher
powers
 Division with i – can not have a square
root in denominator
Complex Numbers - Day 3
 Review
previous day in warm-up
(include conjugates)
 Knowledge check
 Pre-test
Complex Numbers - Day 4
 Intro
to quadratics
 Axis of symmetry
Complex Numbers
Standard MM2N1c: Students will add,
subtract, multiply, and divide complex
numbers
Standard MM2N1b: Write
complex numbers in the
form a + bi.
Complex Numbers Vocabulary
(Name, Desc., Example)
You should be able to define the following
words after today's lesson:
 Complex Number
 Real Number
 Imaginary Number
 Pure Imaginary Number
 Standard Form
Review of old material
49 
7*7  7
25 
5*5  5
72 
6*6*2  6 2
40 
2 * 2 *10  2 10
Complex Numbers
 How
do you think we can reduce:
 36 
6 * 6 * (1)  6 1  6i
81 
9 * 9 * (1)  9 1  9i
12 
2 * 2 * 3* (1)  2i 3
125 
5 * 5 * 5 * (1)  5i 5
Complex Numbers
 The
“i” has special meaning.
 It equals the square root of negative 1.
 We can not really take the square root
of negative 1, so we call it “imaginary”
and give it a symbol of “i”
Guided Practice:
Do problems 21 – 27 odd on page 4
Complex Numbers
 Complex
numbers consist of a “real”
part and an “imaginary” part.
 The standard form of a complex number
is: a + bi, where “a” is the “real” part,
and “bi” is the imaginary part.
Complex Numbers
 Give
some examples of complex
numbers.
 Can “a” and/or “b” equal zero? Yes!!!!
 Give some examples of complex
numbers when “a” and/or “b” equals
zero.
 Can you summarize this into a nice
chart?
Complex Numbers Vocabulary
Real
Numbers
(a + 0i)
-1

Imaginary Numbers
(a + bi, b ≠ 0)
2 + 3i
⅜
23
5 – 5i
Pure Imaginary
Numbers
(0 + bi, b ≠ 0)
-4i
8i
Complex Numbers
 We
are now going to add some complex
numbers.
Review of past material:
Add:
5x + 4 – 3 – 2x =
2x - 7 – 10x – 2 =
3x + 1
-8x - 9
Review of past material:
Add:
(2x + 1) + (4x -3) = 6x - 2
(7x – 5) – (2x + 6) = 5x - 11
And we can change variables:
(3a -2) + (a + 5) = 4a + 3
And we could put them in different order:
(6 + 5i) + (2 - 3i) = 8 + 2i
Complex Numbers
 The
“i” term is the imaginary part of the
complex number, and it can be treated
just like a variable as far as
adding/subtracting like variables.
Complex Numbers
 Simplify
and put in standard form:
 (2 – 3i) + (5 + 2i) = 7 - i
 (7
- 5i) – (3 - 5i) = 4
 Any
questions as far as adding or
subtracting complex numbers?
Complex Numbers
 Just
like you can not add variables (x)
and constants, you can not add the real
and imaginary part of the complex
numbers.
 Solve for x and y:
 x – 3i = 5 + yi
x = 5, y = -3
 -6x + 7yi = 18 + 28i x = -3, y = 4
 Any questions as far as adding or
subtracting complex numbers?
Complex Numbers – Guided
Practice
problems 7 – 13 odd on page 9
 Do problems 35 – 39 odd on page 5
 Do
Warm Up:
Write in standard form:
 49
7i
 27 3i 3
Solve and write solution in standard form:
x  50  0 5i 2
2
Warm-Up
 Simplify
 (3x
– 5) – (7x – 12)
 Solve
 2x
 30
and write in standard form:
for x and y:
+ 8i = 14 – 2yi
minutes to do the “Basic Skills for
Math” NO CALCULATORS!!
(add, subtract, multiply, divide)
Complex Numbers –
Application
 Applications
of Complex Numbers Spring/Mass System
 http://www.picomonster.com/complexnumbers rowing
Review of old material
49 
7*7  7
25 
5*5  5
72 
6*6*2  6 2
40 
2 * 2 *10  2 10
Complex Numbers
 How
do you think we can reduce:
 36 
6 * 6 * (1)  6 1  6i
81 
9 * 9 * (1)  9 1  9i
12 
2 * 2 * 3* (1)  2i 3
125 
5 * 5 * 5 * (1)  5i 5
Guided Practice:
Do problems 21 – 27 odd on page 4
Review of past material:
Multiply:
(2x + 1)(4x -5) = 8x2 - 6x - 5
(7x – 5)(2x + 6) = 14x2 +32x - 30
Complex Numbers
 How
do you think we would do the
following?
 (2 – 3i)(5 + 2i)
 Imaginary numbers may be multiplied
by the distributive rule.
Complex Numbers
 How
do you think we would do the
following?
 (2 – 3i)(5 + 2i)
 = 10 + 4i – 15i – 6i2
 = 10 – 11i – 6i2
 Can we simplify the i2?
 i2 = i * i = -1, so we get:
 = 10 – 11i – 6(-1)
 = 10 – 11i + 6 = 16 – 11i
Complex Numbers
 Simplify:
 (7
+ 5i)(3 - 2i)
 = 21 -14i + 15i – 10i2
 = 21 + i – 10(-1)
 = 21 + i + 10
 = 31 + i
Complex Numbers – Guided
Practice – 5 minutes
 Do
problems 7 – 13 odd on page 13
Complex Numbers
 If i2
= -1, what does i3 equal?
 What does i4 equal?
 How about i5:
 Continue increasing the exponent, and
determine a rule for simplifying i to
some power.
 What would i40 equal?
1
 What would i83 equal?
-i
Imaginary Numbers
1  i
 Definition:
i  1
i 5  i 4i  (1)i  i
i  ( 1)  1
i 4  i 2i 2  (1)(1)  1
i 6  i 4i 2  (1)(i 2 )  1
i8  i 4i 4  (1)(1)  1
2
2
i 3  i 2i  i
i 7  i 4i 3  (1)(i 3 )  i
Reducing Complex Numbers
– Guided Practice – 5 minutes
 Do
problems 7 – 13 odd on page 13
Complex Numbers –
Summary
 Summarize
1.
2.
3.
4.
5.
:
What are complex numbers?
What does their standard form look
like??
What does i equal?
How do we add/subtract them?
How do we take the square root of a
negative number?
Complex Numbers –
Summary
 Summarize
1.
2.
:
How do we multiply them?
How do we simplify higher order
imaginary numbers?
Complex Numbers – Ticket
out the door
 Simplify:
2.
(2 – 3i) – (-5 + 7i)
(4 + i)(3 – 2i)
3.
.  32
4.
i23
1.
Complex Numbers – Warm-up
 Solve
for x and y:
 27 – 8i = -13x + 3yi
x = -2 1/13, y = -2 2/3
 Simplify
+ 3i)(2 – 3i) = 13
 What happened to the “i” term?
 (2
Complex Numbers
Standard MM2N1c: Students will add,
subtract, multiply, and divide complex
numbers
Standard MM2N1a: Write
square roots of negative
numbers in imaginary form.
Dividing by Imaginary
Numbers Vocabulary
(Name, Description, Example)
You should be able to define the following
words after today's lesson:
 Rationalizing the Denominator
 Conjugates
Review of past material:
Simplify:
4x  12

2
4 x 12
  2x  6
2
2
6 x  3x
6x
3x


 3x  1.5
2x
2x 2x
2
2
Dividing by Imaginary
Numbers
 Is
there a problem when we try to divide
complex numbers into real or complex
numbers? HINT: yes, there is a problem
– we can not leave a radical in the
denominator
 Problem
 We
must “rationalize the denominator”,
which means we must eliminate all the
square roots, including i.
Dividing by Imaginary
Numbers
 How
5  2i
3i
can we solve
?
Rationalize the denominator by
multiplying by i/i
(5  2i) i 5i  2i
 2  5i 2 2
 

 1 i
2
3i
i
3i
3
3 3
2
Dividing by Imaginary
Numbers
can we solve 5  2i ? HINT: Look at
2  3i
the warm-up.
 We can rationalize the denominator by
multiplying by it’s conjugate.
 In Algebra, the conjugate is where you
change the sign in the middle of two
terms, like (3x + 5) and (3x – 5)
 Conjugates are (a + bi) and (a – bi), we
just change the sign of the imaginary part
 How
Dividing by Imaginary
Numbers
 How
can we solve
the warm-up.
5  2i
2  3i
? HINT: Look at
(5  2i) (2  3i) 10  15i  4i  6i


2
2  3i (2  3i)
4  6i  6i  9i
Signs are opposite
2
If these do not add
to zero, then you
made a mistake!!!
Dividing by Imaginary
Numbers
 How
can we solve
the warm-up.
5  2i
2  3i
? HINT: Look at
(5  2i) (2  3i) 10  15i  4i  6i


2
2  3i (2  3i)
4  6i  6i  9i
2
10  19i  6(1) 4  19i 4
6

 1 i
4  9(1)
13
13 13
Practice
 Page
13, # 29 – 37 odd