Complex numbers - MGMP Matematika Satap Malang

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Transcript Complex numbers - MGMP Matematika Satap Malang

By: Jeffrey Bivin
Lake Zurich High School
[email protected]
Last Updated: October 30, 2006
The Imaginary Unit
1  i
x  5
x   5
2
x   1 5
x  i 5
Jeff Bivin -- LZHS
x  9
x   9
2
x   1 9
x   3i
More about Imaginary Numbers
i
i 
2

1
1

2
i 5  i 4  i  (1)(i)  i
 1
i  i  i  (1)(1)  1
6
4
2
i 3  i 2  i  1 i   i
i 7  i 4  i 3  (1)(i)   i
i 4  i 2  i 2  (1)(1)  1
i8  i 4  i 4  (1)(1)  1
i9
i10
i11
i12
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



i 4  i 4  i  (1)(1)i  i
i 4  i 4  i 2  (1)(1)(1)  1
i 4  i 4  i 3  (1)(1)(i)   i
i 4  i 4  i 4  (1)(1)(1)  1
This is fun!
i 23  i 4  i 4  i 4  i 4  i 4  i 3  (1)(1)(1)(1)(1)(i)   i
i
23
i
i
41
74
122
i
Jeff Bivin -- LZHS
  i  (1) (i)   i
 i  i  (1) (i)  i
 i  i  (1) (1)   1
 i  i  (1) (1)   1
 i
4 5 3
5
4 10
10
4 18 2
18
4 30 2
30
Complex Numbers
For any real numbers a and b, the number
a + bi
is a complex number.
Examples:
1 i 3
3  2i
2  5i
Jeff Bivin -- LZHS
 7i
2 i 5
i
3i
2
i 2
7  2i
5  2i 3
Jeff Bivin -- LZHS
3 5
3
7
2  5 11

1 4  15
, ,
2 7 11
Addition & Subtraction
Example 1
(3  2i)  (5  7i)  8  9i
Example 2
(5  3i)  (2  9i)  3  6i
Example 3
(8  6i)  (2  5i)  10  i
Jeff Bivin -- LZHS
Multiplication
Example 1
(3  2i) (4  5i) 



12  15i  8i  10i 2
12  15i  8i  10(1)
12  15i  8i  10
2  23i
Example 2
(2  5i) (3  7i)  6  14i  15i  35i 2
 6  14i  15i  35(1)
 6  14i  15i  35
 41  i
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Graphing Complex Numbers
3  2i
 4  5i
 3  2i
4  3i
3i
2
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Solve using the Quadratic Formula
x2 + 6x + 13 = 0
a=1
b=6
c = 13
x 
b  b 2  4 ac
2a
x 
6 
6 2  4113
21
x 
6 
36  52
2
x 
6  16
2
2( 3  2i )
6  4i
2
2
x 
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
  3  2i
Solve using the Quadratic Formula
3x2 + 7x + 5 = 0
a=3
b=7
c=5
x 
b  b 2  4 ac
2a
x 
7 
7 2  435
23
x 
7 
49  60
6
x 
7  11
6
7  i 11
6
x 
Jeff Bivin -- LZHS
Solve using the Quadratic Formula
3x2 - 2x + 5 = 0
a=3
b = -2
c=5
x 
b  b 2  4 ac
2a
x 
2
( 2 ) 2  435
23
x 
2
4  60
6
x 
2  56
6
2  2i 14
6
x 
Jeff Bivin -- LZHS

2 (1  i 14 )
6

1  i 14
3