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1
Equations and
Inequalities
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1
1.3
Complex Numbers
• Basic Concepts of Complex Numbers
• Operations on Complex Numbers
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21.3 - 2
Basic Concepts of Complex
Numbers
There is no real number solution of the
equation
2
x  1.
since no real number, when squared, gives 1.
To extend the real number system to include
solutions of equations of this type, the
number i is defined to have the following
property.
2
i  1, and therefore, i  1.
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Basic Concepts of Complex
Numbers
If a and b are real numbers, then any
number of the form a + bi is a complex
number.
In the complex number a + bi , a is the real
part and b is the imaginary part.
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Basic Concepts of Complex
Numbers
Two complex numbers a + bi and c + di are
equal provided that their real parts are equal
and their imaginary parts are equal; that is
a  bi  c  di if and only if a  c and b  d .
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Basic Concepts of Complex
Numbers
For complex number a + bi, if b = 0, then
a + bi = a.
Thus, the set of real numbers is a subset of
the set of complex numbers.
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Basic Concepts of Complex
Numbers
If a = 0 and b ≠ 0, the complex number is said
to be a pure imaginary number.
A pure imaginary number, or a number like
7 + 2i with a ≠ 0 and b ≠ 0, is a nonreal
complex number.
A complex number written in the form a + bi (or
a + ib) is in standard form.
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THE EXPRESSION
a
If a  0, then a  i a.
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Example 1
WRITING
a AS i a
Write as the product of a real number and i,
using the definition of a.
(a)
16
Solution
16  i 16  4i
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Example 1
WRITING
a AS i a
Write as the product of a real number and i,
using the definition of a.
(b)
70
Solution
70  i 70
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Example 1
WRITING
a AS i a
Write as the product of a real number and i,
using the definition of a.
(c)
 48
Solution
48  i 48  i 16 3  4i 3
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Product rule
for radicals
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Operations on Complex Numbers
Products or quotients with negative
radicands are simplified by first rewriting
a as i a for a positive number a.
Then the properties of real numbers and the
2
fact that i  1 are applied
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Operations on Complex Numbers
Caution When working with
negative radicands, use the definition
a  i a
before using any of the other rules for
radicals.
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Operations on Complex Numbers
Caution In particular, the rule
c
d  cd
is valid only when c and d are not both
negative.
4 9  2i 3i  6i 2  6 is correct,
while
4
9  ( 4)( 9)  36  6 is incorrect.
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FINDING PRODUCTS AND
QUOTIENTS INVOLVING a
Example 2
Multiply or divide, as indicated. Simplify
each answer.
(a)
7
7
Solution
7
7  i 7 i 7
First write all square
roots in terms of i.
i
2
 7
2
 1 7
 7
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i 2 = −1
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Example 2
FINDING PRODUCTS AND
QUOTIENTS INVOLVING a
Multiply or divide, as indicated. Simplify
each answer.
(b) 6 10
Solution
6
10  i 6 i 10
i
 1
 1
 2
2
60
4 15
2 15
15
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Example 2
FINDING PRODUCTS AND
QUOTIENTS INVOLVING a
Multiply or divide, as indicated. Simplify
each answer.
20
(c)
2
Solution
20 i 20
20


 10
2
2
i 2
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Quotient
rule for
radicals
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Example 2
FINDING PRODUCTS AND
QUOTIENTS INVOLVING a
Multiply or divide, as indicated. Simplify
each answer.
48
(d)
24
Solution
48 i 48
48

i
i 2
24
24 i 24
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Example 3
SIMPLIFYING A QUOTIENT
INVOLVING a
8  128
Write
4
Solution
in standard form a + bi.
8  128 8  64 2

4
4
8  8i 2

4
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64  8i
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SIMPLIFYING A QUOTIENT
INVOLVING a
Example 3
8  128
Write
4
Solution
Be sure to
factor before
simplifying
in standard form a + bi.
8  8i 2

4


4 2  2i 2

4
 2  2i 2
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Factor.
Lowest terms
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Addition and Subtraction of
Complex Numbers
For complex numbers a + bi and c + di,
(a  bi )  (c  di )  (a  c )  (b  d )i
and (a  bi )  (c  di )  (a  c )  (b  d )i .
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Example 4
ADDING AND SUBTRACTING
COMPLEX NUMBERS
Find each sum or difference.
(a) (3  4i )  ( 2  6i )
Solution
Add real
parts.
Add
imaginary
parts.
(3  4i )  ( 2  6i )  3  ( 2)    4  6 i
Commutative, associative,
distributive properties
 1  2i
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Example 4
ADDING AND SUBTRACTING
COMPLEX NUMBERS
Find each sum or difference.
(b) ( 4  3i )  (6  7i )
Solution
( 4  3i )  (6  7i )  ( 4  6)   3  ( 7) i
 10  10i
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Multiplication of Complex Numbers
The product of two complex numbers is
found by multiplying as though the numbers
were binomials and using the fact that
i2 = –1, as follows.
FOIL
(a  bi )(c  di )  ac  adi  bic  bidi
 ac  adi  bci  bdi
Distributive property;
i 2 = –1
2
 ac  (ad  bc )i  bd ( 1)
 (ac  bd )  (ad  bc )i
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Multiplication of Complex
Numbers
For complex numbers a + bi and c + di,
(a  bi )(c  di )  (ac  bd )  (ad  bc )i .
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Example 5
MULTIPLYING COMPLEX
NUMBERS
Find each product.
(a) (2  3i )(3  4i )
Solution
(2  3i )(3  4i )  2(3)  2(4i )  3i (3)  3i (4i )
FOIL
 6  8i  9i  12i
2
 6  i  12( 1)
Combine like terms;
i2 = −1
 18  i
Standard form
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Multiply.
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MULTIPLYING COMPLEX
NUMBERS
Example 5
Find each product.
(b) (4  3i )2
Remember to add twice the
product of the two terms.
Solution
2
2
2
(4  3i )  4  2(4)(3i )  (3i ) Square of a binomial
 16  24i  9i
2
Multiply.
 16  24i  9( 1)
i 2 = −1
 7  24i
Standard form
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Example 5
MULTIPLYING COMPLEX
NUMBERS
Find each product.
(c) (6  5i )(6  5i )
Solution
(6  5i )(6  5i )  6  (5i )
2
2
Product of the sum
and difference of
two terms
 36  25( 1)
i 2 = −1
 36  25
Multiply.
 61, or 61  0i
Standard form
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Example 5(c) showed that
(6  5i )(6  5i )  61
The numbers 6 + 5i and 6 – 5i differ only in the
sign of their imaginary parts and are called
complex conjugates. The product of a
complex number and its conjugate is
always a real number. This product is the
sum of the squares of real and imaginary
parts.
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Property of Complex
Conjugates
For real numbers a and b,
(a  bi )(a  bi )  a 2  b 2.
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Example 6
DIVIDING COMPLEX NUMBERS
Write each quotient in standard form a + bi.
3  2i
(a)
5i
Multiply by the
Solution
complex conjugate of
3  2i (3  2i )(5  i )
the denominator in

both the numerator
5i
(5  i )(5  i )
and the denominator.
15  3i  10i  2i 2

2
25  i
Multiply.
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Example 6
DIVIDING COMPLEX NUMBERS
Write each quotient in standard form a + bi.
3  2i
(a)
5i
Solution
13  13i

Combine like terms; i 2 = −1
26
13 13i


26 26
a  bi a bi
 
c
c c
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Example 6
DIVIDING COMPLEX NUMBERS
Write each quotient in standard form a + bi.
3  2i
(a)
5i
Solution
1 1
Write in lowest terms and standard
  i
form.
2 2
Check
1 1 
  i  (5  i )  3  2i
2 2
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DIVIDING COMPLEX NUMBERS
Example 6
Write each quotient in standard form a + bi.
3
(b)
i
Solution
3 3( i )

i
i ( i )
3i
 2
i
– i is the conjugate of i.
Multiply.
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Example 6
DIVIDING COMPLEX NUMBERS
Write each quotient in standard form a + bi.
3
(b)
i
Solution
3i

1
i 2 = −1(−1) = 1
 3i , or 0  3i
Standard form
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Simplifying Powers of i
Powers of i can be simplified using the facts
i 2  1 and i 4  ( i 2 )2  ( 1)2  1.
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Powers of i
i i
i i
i  1
i  1
i  i
i  i
i 1
i  1 and so on.
1
2
3
4
5
6
7
8
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Powers of i cycle through the same four
outcomes (i, –1, –i, and 1) since i4 has the
same multiplicative property as 1. Also,
any power of i with an exponent that is a
multiple of 4 has value 1. As with real
numbers, i0 = 1.
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SIMPLIFYING POWERS OF i
Example 7
Simplify each power of i.
(a) i 15
Solution
Since i 4 = 1, write the given power as a
product involving i 4.
i
15
i
12
 
i  i
3
4
3
i 3  13 ( i )  i
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Example 7
SIMPLIFYING POWERS OF i
Simplify each power of i.
(b) i 3
Solution
3
Multiply i by 1 in the form of i 4 to create the
least positive exponent for i.
i 3  i 3 1  i 3 i 4  i
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