Download 2_5 Complex Numbers - Kenwood Academy High School


Warm Up
◦ Put the factors into standard form
( x  2)( x  3)
◦ Solve for x
x 1  0
2

Announcements
◦ Assignment
◦ p. 258
◦ # 11 – 14, 23 – 26, 32 - 35
2.5 Complex Numbers
•How
to use the imaginary unit i to write complex
numbers
•How
to add, subtract, and multiply complex numbers
•How
to use complex conjugates to write the quotient of
two complex numbers in standard form
•How
to express solutions as complex numbers
The Imaginary Unit i
 x2 + 1 = 0
 x2 = -1
 x =±(√-1)
solution.
does not have a real
◦ We need a #, that when you
square it, is –1.
i is called the imaginary unit.
 i2 = -1.
What is this nonsense?
 Any
number can be written as
a + bi or a + b(√-1)
◦ a is called the real part
◦ bi is the imaginary part.
◦ Examples:
 3 + 4i
 2 = 2 +0i
To add two complex numbers, we add the
two real parts then add the two imaginary
parts. That is, combine like terms
 (a + bi) + (c + di) = (a + c) + (b + d )i
 (2 + 3i) + (4 + 5i) = (2 + 4) + (3 + 5 )i

= 6 + 8i
Operations with complex numbers
a)
6  3i   1  2i 
6  3i   1  2i   6 1   3  2i  7  i
b)
5  i   2  4i 
5  i  2  4i  5 i  2  4i  3  3i
Example 1. Add or subtract the
following complex numbers.
Distribution
 (i)(-3i + 2) = -3i2 + 2i
= -3(-1) + 2i
= 3 + 2i
 F.O.I.L.:
 (3 + 2i)(1 + 4i) = 3 + 12i + 2i + 8i2
= 3 + 14i + 8(-1)
= -5 + 14i

Multiplying can be pretty
“complex”
217  5i 
217  5i   217  25i   34 10i
a)
b)
3  i 5  4i 
3  i 5  4i   15  12i  5i  4i
 15  12i  5i  4
 19  7i
Example 2. Multiply the following
complex numbers.
2
 i1 = √-1
 i5 = √-1
 i9
 i2 = -1
 i6
 i10
 i3 = - √-1
 i7
 i11
 i4 = -1(-1)
 i8
 i12
What do we notice about a pattern of multiplication?
What is this “i” business?

Perform the appropriate operation…please

Simplify
6  2i 2  3i 
2
5 3
Warm - Up
 Assignment
◦p. 258
◦# 28, 30, 56 – 62, 66 – 69
 Notebook
Quiz next week
 Test Corrections Thursday
Announcement

Changing the square root of a negative
number to a complex number
 3  3(1)  3 1  3i

Principal square root of a negative
number
 a  ai
Complex Solutions of Quadratic
Equations

Solve
3x2
– 2x + 5 = 0
1.Substitute into the
quadratic equation
2.Simplify
3.Combine like terms
4.Convert square root of
negative numbers into
complex numbers
5.Simplify
 b  b 2  4ac
x
2a
 (2)  (2) 2  4(3)(5)

2(3)
2  4  60

6
2   56

6
2  2 14i

6
1
14
 
i
3
3
Complex solution of a quadratic
equations

Solve 16x2 – 4x + 3 = 0
 b  b 2  4ac
x
2a
 (4)  (4) 2  4(16)(3)

2(16)
4  16  192
32
4   176

32
4  4 11i

32
1
11
 
i
8
8

Complex solution of a quadratic
equations
6  2
 6i  2i
 12i
2
 12 (1)
1.Convert the negative
square roots into
square roots times i
2. Combine like terms
3.If necessary, change
i squared into -1
4.Simplify
  12
Perform the operation and write
the result in standard form
 3  12
 48   27 
 3i 12i
  48   27
 36i
2
 48i  27i
 6(1)
 6
 4 3i  3 3i
 3i
Writing complex numbers in
standard form
 Can
we have an imaginary
number in the denominator?
◦ Can we have a square root in the
denominator?
◦ Is i a square root?
Dividing imaginary numbers and
creating conjugates

Complex numbers of the forms a + bi and
a – bi are called complex conjugates.
(a + bi)( a – bi) = a2 – (bi)2 = a2 + b2

If there is a complex number in the
denominator…
2
2 1  i 21  i


 2 2 
1  i 1 i 1  i 1  1
Complex Conjugates
2
1

i
21  i
2
 1 i
2i
4  3i
2i
2  i 4  3i


4  3i 4  3i 4  3i
8  6i  4i  3i 2

4 2  32
8  3  6i  4i

16  9
5  10i

25
1 2
  i
5 5
Example 3. Find the quotient of
the following complex numbers.