Transcript Aim: What are imaginary and complex numbers?

What are imaginary and complex numbers?
Graph it
Do Now:
Solve for x: x2 + 1 = 0
2
x  1
2
x   1
?
What number when
multiplied by itself
gives us a negative one?
No such real number
parabola does
not intersect
x-axis NO REAL ROOTS
Imaginary Numbers
1  is not a real number,
then 1  is a non-real or
If
imaginary number.
Definition: A pure imaginary number is
any number that can be expressed in the
form bi, where b is a real number such
that b ≠ 0, and i is the imaginary unit.
1 
ab  a  b
i
1
1)  5 25
15i 5 1  5i b = 5
25  25
25 (25
25
1 
1
7
7 77 1
1 7i
7i ii 77
b 7
In general, for any real number b, where b > 0:
2
b  b
2
1  bi
1  i
Powers of i
 
1
If
i2
2
 i2 = –1

 1

2
 i2 = –1
i3
= – 1, then = ?
i3 = i2 • i = –1( 1 ) = –i
i4 = i2 • i2 = (–1)(–1) = 1
i5 = i4 • i = 1( 1) = i
i6 = i4 • i2 = (1)(–1) = –1
i7 = i6 • i = -1( 1 ) = –i
i8 = i6 • i2 = (–1)(–1) = 1
What is i82 in simplest form?
82 ÷ 4 = 20 remainder 2
i82 equivalent to i2 = –1
i0 = 1
i1 = i
i2 = –1
i3 = –i
i4 = 1
i5 = i
i6 = –1
i7 = –i
i8 = 1
i9 = i
i10 = –1
i11 = –i
i12 = 1
1  i
Properties of i
16  9 
Addition:
16 1  9 1  4i + 3i = 7i
Subtraction: 25  16 
25 1  16 1  5i – 4i = i
Multiplication:
36 4 
36 1  4 1  (6i)(2i) = 12i2 = –12
note :
Division:
36 4  144
16  4 
16

4
16 1 4i

2
4 1 2i
Complex Numbers
Definition: A complex number is
any number that can be expressed in the
form a + bi, where a and b are real numbers
and i is the imaginary unit.
a + bi
real numbers
pure imaginary
number
Any number can be expressed as a complex
number:
7 + 0i = 7
a + bi
0 + 2i = 2i
The Number System
5 76
-i
Complex Numbers
Real Numbers
i
i3
-i Irrational
Numbers
i9
i
2 + 3i
47
Rational
Numbers
i
Integers
i
i75
Whole Numbers
Counting
Numbers
1/2 – 12i
-6 – 3i
i
-i47
i
Graphing Complex Numbers
(x, y)
Complex Number Plane a + bi
pure imaginaries
5i
(4 + 5i)
4i
3i
(0 + 3i)
2i
i
-5
-4 -3 -2 -1
(–5 – 2i)
-i
(0 + 0i)
0
1
2
-2i
-5i
-6i
3
4
5
(3 – 2i)
-3i
-4i
reals
(0 – 4i)
6
Vectors
Vector - a directed line segment that represents
directed force notation: OP
pure imaginaries
5i
3  4i  32  42
P
(3 + 4i)
4i
3i
 25  5
2i
i
OP
reals
O
-4 -3 -2 -1
0 1 2 3 4 5 6
The-5length
of vectors
-i is found by using the
Pythagorean Theorem
& is always positive.
-2i
-3i
The length of a vector representing
a complex number is
the absolute value of -4i
the complex number a  bi
represented-5iby the equation
-6i
2
a  bi  a  b 2  a 2  b2
Model Problems
Express in terms of i and simplify:
100 = 10i  16 = 4/5i  1 300  5i 3
2
25
Write each given power of i in simplest terms:
i49 = i
i54 = -1 i300 = 1
i2001 = i
Add:
4 18  50  4i 9 2  i 25 2
 12i 2  5i 2  17i 2
Multiply: 4 5  80
 4i 5  4i 5  16i
2
 5
2
 4i 5  i 16 5
 16 5  80
Simplify: 72  32  3 8
 i 36 2  i 16 2  3i 4 2
 4i 2
 6i 2  4i 2  6i 2
Model Problems
Which number is included in the shaded region?
1)
2)
3)
4)
(-1.5 + 3.5i)
(1.5 – 3.5i)
(3.5)
(1)
(4.5i)
5i
4i
yi
(4)
3i
2i
i
-5
-4 -3 -2 -1
-i
(3)
0
1
2
-2i
-3i
-4i
-5i
-6i
(2)
3
4
x
5
6
How do we add and subtract complex numbers?
Do Now:
Simplify:
3 45  125  2 20
3 9 5  25 5  2 4 5
9 5 5 54 5
10 5
Adding Complex Numbers
(2 + 3i) + (5 + i)
= (2 + 5) + (3i + i)
= 7 + 4i
In general, addition of complex numbers:
(a + bi) + (c + di) = (a + c) + (b + d)i
Combine the real parts and the imaginary parts separately.
Find the sum of
convert to
complex numbers
combine reals and
imaginary parts
separately
(5  36) and (3  16)
(5  36)  (3  16)
(5  i 36)  (3  i 16)
(5  6i)  (3  4i)
(5  3)  (6i  4i)
8  2i
Subtracting Complex Numbers
What is the additive inverse of 2 + 3i?
-(2 + 3i) or -2 – 3i
Subtraction is the addition of an additive inverse
(1 + 3i) – (3 + 2i)
= (1 + 3i) + (-3 – 2i)
= -2 + i
In general, subtraction of complex numbers:
(a + bi) – (c + di) = (a – c) + (b – d)i
Subtract
6  2i 3 from 5  3i 3
5

3i
3

6

2i
3




change to addition
problem
5  3i 3  6  2i 3 
combine reals and 5  (6)  (3i 3)  2i 3


imaginary parts
separately
1  i 3
Adding Complex Numbers Graphically
(2 + 3i) + (3 + 0i)
= (2 + 3) + (3i + 0i) =
= 5 + 3i
yi
5i
4i
vector: 2 + 3i
(2 + 3i)
3i
vector: 3 + 0i
i
vector: 5 + 3i
-5
(5 + 3i)
2i
-4 -3 -2 -1
-i
-2i
-3i
-4i
-5i
-6i
(3 + 0i)
0
1
2
3
4
5
x
6
Adding Vectors
Vector - a directed line segment that represents
directed force notation:
OS
S
P
resultant force
O
R
The vectors that represent the applied forces
form two adjacent sides of a parallelogram,
and the vector that represents the resultant
force is the diagonal of this parallelogram.
Subtracting Complex Numbers Graphically
(1 + 3i) – (3 + 2i)
= (1 + 3i) + (-3 – 2i) =
yi
-2 + i
5i
4i
(1 + 3i)
3i
2i
(-2 + i)
(3 + 2i)
i
x
-5
-4 -3 -2 -1
(-3 – 2i)
-i
0
1
2
3
4
5
-2i
-3i
The vector representing-4ithe additive inverse is
the image of the vector reflected through the
-5i
origin. Or the image under a rotation about
-6i
the origin of 1800.
6
Model Problems
Add/Subtract and simplify:
(10 + 3i) + (5 + 8i)
= 15 + 11i
(4 – 2i) + (-3 + 2i)
=1
2

3
i  1
 
4  6
1 
i  4 i  1 2i 
   
2  6 4  6 4 
5 3i 
 
6 4 
 

80  3  20  4  6i 5
Express the difference of
1 
 
 in form a + bi
3  4 5  9 2 i
80  2  162