Section 1.6
The Complex Number System
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Topics
o The imaginary unit i and its properties
o The algebra of complex numbers
o Roots and complex numbers
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The Imaginary Unit i and its Properties
The Imaginary Unit i
The imaginary unit i is defined as i  1. In other
2
i
words, i has the property that its square is −1:  1.
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The Imaginary Unit i and its Properties
Square Roots of Negative Numbers
If a is a positive real number, a  i a . Note that by
this definition, and by a logical extension of
2
2
2
exponentiation, i a  i
a  a.
 
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 
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Example 1
a.
b.
16  i 16  i  4   4i.
As is customary, we write a constant such as 4
before letters in algebraic expressions, even if, as in
this case, the letter is not a variable. Remember that
i has a fixed meaning: i is the square root of −1.
 
8  i 8  i 2 2  2i 2.
As is customary, again, we write the
radical factor
2
last. You should verify that  2i 2  is indeed −8.
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Example 1 (cont.)
4
2 2
and
i  i i   1 1  1.
c. i  i i   1 i   i ,
The simple fact that i2 = −1 allows us, by our
extension of exponentiation, to determine in for any
natural number n.
3
2
d.  i    1 i  i 2  1.
2
2 2
This observation shows that −i also has the property
that its square is −1.
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The Imaginary Unit i and its Properties
Complex Numbers
For any two real numbers a and b, the sum a + bi is a
complex number. The collection = {a + bi|a and b are
both real} is called the set of complex numbers and is
another example of a field. The number a is called the
real part of a + bi, and the number b is called the
imaginary part. If the imaginary part of a given
complex number is 0, the number is simply a real
number. If the real part of a given complex number is 0,
the number is a pure imaginary number.
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The Algebra of Complex Numbers
Simplifying Complex Expressions
Step 1: Add, subtract, or multiply the complex
numbers, as required, by treating every
complex number a + bi as a polynomial
expression. Remember, though, that i is not
actually a variable. Treating a + bi as a binomial
in i is just a handy device.
Step 2: Complete the simplification by using the fact
that i2 = −1.
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Example 2
Simplify the following complex number expressions.
a.  4  3i    5  7i 
b.  2  3i    3  3i 
c.  3  2i  2  3i 
d.  2  3i 
2
Solutions:
a.  4  3i    5  7i 
  4  5   3  7 i
 1  10i
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Treating the two complex numbers
as polynomials in i, we combine
the real parts and then the
imaginary parts.
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Example 2 (cont.)
b.  2  3i    3  3i 
  2  3i      3  3i 
We begin by distributing the
minus sign over the two terms
of the second complex number,
and then combine as in part a.
  2  3   3  3 i
 1  0i
1
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Example 2 (cont.)
c.  3  2i  2  3i 
 6  9i  4i  6i 2
 6   9  4  i  6  1
The product of two complex numbers
leads to four products via the
distributive property, as illustrated
here. After multiplying, we combine
the two terms containing i, and
rewrite i2 as −1.
 6  5i  6
 12  5i
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Example 2 (cont.)
d.  2  3i    2  3i  2  3i 
2
 4  6i  6i  9i 2
 4  12i  9  1
Squaring this complex number
also leads to four products,
which we simplify as in part c.
Remember that a complex
number is not simplified until it
has the form a + bi.
 5  12i
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Example 3
Simplify the quotients.
2  3i
a.
3i
b.  4  3i 
1
1
c.
i
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Example 3(cont.)
Solutions:
2  3i  2  3i   3  i 

a.
 


3i
3 i
3i
2  3i  3  i 


 3  i  3  i 
6  2i  9i  3i 2

9  3i  3i  i 2
3  11i
3 11

  i
10
10 10
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We multiply the top and bottom
of the fraction by 3 + i, which is
the complex conjugate of the
denominator. The rest of the
simplification involves multiplying
complex numbers as in the last
example.
We would often leave the answer
3  11i
in the form
unless it is
10
necessary to identify the real and
imaginary parts.
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Example 3(cont.)
b.  4  3i 
1
1

4  3i
 1   4  3i 

 4  3i   4  3i 
4  3i

 4  3i  4  3i 
In this example, we simplify
the reciprocal of the complex
number 4 − 3i. After writing
the original expression as a
fraction, we multiply the top
and bottom by the complex
conjugate of the denominator
and proceed as in part a.
4  3i

16  9i 2
4  3i
4 3

  i
25
25 25
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Example 3(cont.)
1  1   i 
c.
  
 i   i 
i
i
 2
i
This problem illustrates the process of
writing the reciprocal of the
imaginary unit as a complex number.
Note that with this as a starting point,
we could now calculate i –2, i –3, …
i

1
 i
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Example 4
Simplify the following expressions.

a. 2  3
b.

2
4
4
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Example 4 (cont.)
Solutions:

 
2

a. 2  3  2  3 2  3

 4  4 3  3 3
 
 4  4i 3  i 3
 4  4i 3  3
2
Each 3 is converted
to i 3 before carrying
out the associated
multiplications. Note that
incorrect use of one of the
properties of radicals
would have led to adding 3
instead of subtracting 3.
 1  4i 3
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Example 4 (cont.)
b.
4
2

4 2i
1

i
 i
1
We have already simplified
in Example 3c, so
i
we quickly obtain the correct answer of −i. If we
had incorrectly rewritten the original fraction as
4
, we would have obtained 1 or i as the
4
final answer.
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