1.3
Complex Numbers
Basic Concepts of Complex Numbers
Operations on Complex Numbers
1.3 - 1
Basic Concepts of Complex
Numbers
There are no real numbers for the solution of
the equation
2
x = −1.
To extend the real number system to include
such numbers as,
−1,
the number i is defined to have the following
property;
2
i = −1.
1.3 - 2
Basic Concepts of Complex
Numbers
So… i=
−1
The number i is called the imaginary unit.
Numbers of the form a + bi, where a and b
are real numbers are called complex
numbers.
In this complex number, a is the real part
and b is the imaginary part.
1.3 - 3
Complex
numbers
a + bi,
a and b
real
Nonreal
complex
numbers
a + bi,
b≠0
Real
numbers
a + bi,
b=0
Irrational
numbers
Integers
Rational
numbers
Nonintegers
Basic Concepts of Complex
Numbers
Two complex numbers are equal provided
that their real parts are equal and their
imaginary parts are equal;
a + bi =c + di if and only if a = c and b = d
1.3 - 5
Basic Concepts of Complex
Numbers
If a = 0 and b ≠ 0, the complex number is pure
imaginary.
A pure imaginary number or a number, like 7 +
2i with a ≠ 0 and b ≠ 0, is a nonreal complex
number.
The form a + bi (or a + ib) is called standard
form.
1.3 - 6
THE EXPRESSION
If a > 0, then
−a
−a =
i a.
1.3 - 7
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
1.3 - 8
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
1.3 - 9
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
Product rule
for radicals
1.3 - 10
Operations on Complex Numbers
Products or quotients with negative
radicands are simplified by first rewriting
−a as i a for a positive number.
Then the properties of real numbers are
applied, together with the fact that
i 2 = −1.
1.3 - 11
Example 2
FINDING PRODUCTS AND
QUOTIENTS INVOLVING
NEGATIVE RADICALS
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  7
2
= −17
= −7
2
i 2 = −1
1.3 - 12
Example 2
FINDING PRODUCTS AND
QUOTIENTS INVOLVING
NEGATIVE RADICALS
Multiply or divide, as indicated. Simplify
each answer.
b.
−6  −10
Solution:
i 6 i 10
−6  −10 =
= i  60
2
= −12 15
= −2 15
= −1 4 15
1.3 - 13
Example 2
FINDING PRODUCTS AND
QUOTIENTS INVOLVING
NEGATIVE RADICALS
Multiply or divide, as indicated. Simplify
each answer.
−20
c.
−2
Solution:
−20 i 20
= =
−2
i 2
20
=
2
10
Quotient
rule for
radicals
1.3 - 14
Example 3
SIMPLIFYING A QUOTIENT
INVOLVING A NEGATIVE
RADICAND
Write −8 + −128
4
Solution:
in standard form a + bi.
−8 + −128 −8 + −64 2
=
4
4
−8 + 8i 2
=
4
−64 =
8i
1.3 - 15
SIMPLIFYING A QUOTIENT
INVOLVING A NEGATIVE
RADICAND
Example 3
Write −8 + −128
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
−64 =
8i
)
Factor.
Lowest terms
1.3 - 16
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 .
1.3 - 17
Example 4
ADDING AND SUBTRACTING
COMPLEX NUMBERS
Find each sum or difference.
a. (3 − 4i ) + ( −2 + 6i )
Solution:
Add real
parts.
(3 − 4i ) =
+ ( −2 + 6i )
Add
imaginary
parts.
[3 + ( −2)] + [ − 4 + 6] i
Commutative, associative,
distributive properties
= 1 + 2i
1.3 - 18
Example 4
ADDING AND SUBTRACTING
COMPLEX NUMBERS
Find each sum or difference.
b. ( −9 + 7i ) + (3 − 15i )
Solution:
( −9 + 7i ) + (3 − 15i ) =− 6 − 8i
1.3 - 19
Example 4
ADDING AND SUBTRACTING
COMPLEX NUMBERS
Find each sum or difference.
c. ( − 4 + 3i ) − (6 − 7i )
Solution:
( −4 + 3i ) − (6 − 7i ) =−
( 4 − 6) + [3 − ( −7)] i
=
−10 + 10i
1.3 - 20
Example 4
ADDING AND SUBTRACTING
COMPLEX NUMBERS
Find each sum or difference.
d. (12 − 5i ) − (8 − 3i )
Solution:
(12 − 5i ) − (8 − 3i ) =4 − 2i
1.3 - 21
Multiplication of Complex
Numbers
For complex numbers a + bi and c + di,
(a + bi )(c + di ) = (ac − bd ) + (ad + bc )i .
1.3 - 22
MULTIPLYING COMPLEX
NUMBERS
Example 5
Find each product.
a. (2 − 3i )(3 + 4i )
Solution:
(2 − 3i )(3 + 4i ) = 2(3) + 2(4i ) − 3i (3) − 3i (4i )
=6 + 8i − 9i − 12i
= 6 − i − 12( −1)
FOIL
2
i2 = −1
= 18 − i
1.3 - 23
Example 5
MULTIPLYING COMPLEX
NUMBERS
Find each product.
2
(4
+
3
i
)
b.
Solution:
(4 + 3i ) =4 + 2(4)(3i ) + (3i )
2
2
=16 + 24i + 9i
2
2
= 16 + 24i + 9( −1)
Square of a binomial
Remember to add twice the
product of the two terms.
i 2 = −1
= 7 + 24i
1.3 - 24
MULTIPLYING COMPLEX
NUMBERS
Example 5
Find each product.
c. (6 + 5i )(6 − 5i )
Solution:
(6 + 5i )(6 − 5i ) =62 − (5i )2
= 36 − 25( −1)
Product of the sum
and difference of
two terms
i 2 = −1
= 36 + 25
= 61, or 61 + 0i
Standard form
1.3 - 25
Simplifying Powers of i
Powers of i can be simplified using the facts
(i 2 )2 =
( −1)2 =
1
i2 =
−1 and i 4 =
1.3 - 26
SIMPLIFYING POWERS OF i
Example 6
Simplify each power of i.
a. i 15
Solution:
Since i 2 = –1 and i 4 = 1, write the given
power as a product involving i 2 or i 4. For
example, i 3 = i 2  i = ( −1)  i = −i .
Alternatively, using i4 and i3 to rewrite i15 gives
i
15
( )
=i  i =i
12
3
4
3
 i =1 ( −i ) =−i
3
3
1.3 - 27
Powers of i
i =i
i =i
i =i
i = −1
i = −1
i
10
i = −i
i = −i
i = −i
i =1
i =1
1
2
3
4
5
6
7
8
9
= −1
11
i
12
= 1, and so on.
1.3 - 28
Ex 5c. showed that…
(6 + 5i )(6 − 5i ) =
61
The numbers 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 squares
of real and imaginary parts.
1.3 - 29
Property of Complex
Conjugates
For real numbers a and b,
(a + bi )(a − bi ) =a 2 + b 2 .
1.3 - 30
Example 7
DIVIDING COMPLEX NUMBERS
Write each quotient in standard form a + bi.
a. 3 + 2i
5−i
Solution:
3 + 2i (3 + 2i )(5 + i )
=
5−i
(5 − i )(5 + i )
15 + 3i + 10i + 2i 2
=
2
25 − i
Multiply by the
complex conjugate of
the denominator in
both the numerator
and the denominator.
Multiply.
1.3 - 31
Example 7
DIVIDING COMPLEX NUMBERS
Write each quotient in standard form a + bi.
a. 3 + 2i
5−i
Solution:
2
15 + 3i + 10i + 2i
=
2
25 − i
13 + 13i
=
26
Multiply.
i 2 = −1
1.3 - 32
Example 7
DIVIDING COMPLEX NUMBERS
Write each quotient in standard form a + bi.
a. 3 + 2i
5−i
Solution:
13 + 13i
=
26
13 13i
=
+
26 26
i 2 = −1
a + bi a bi
=
+
c
c c
1.3 - 33
Example 7
DIVIDING COMPLEX NUMBERS
Write each quotient in standard form a + bi.
a. 3 + 2i
5−i
Solution:
13 13i
=
+
26 26
1 1
=
+ i
2 2
a + bi a bi
=
+
c
c c
Lowest terms;
standard form
1.3 - 34