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Section 7.7
• Previously, when we encountered square
•
•
•
roots of negative numbers in solving
equations, we would say “no real solution”
or “not a real number”.
That’s still true.
However, we will now introduce a new set
of numbers.
Imaginary numbers which includes the
imaginary unit i.
Real numbers and imaginary numbers are both
subsets of a new set of numbers.
Complex numbers
Every complex number can be written as a sum
of a real number and an imaginary number.
The imaginary unit i is the number whose
square is –1.
i 2  1
i  1
We can write the square root of a negative
number in terms of i.
Example
Write the following with the i notation.
 25 
25   1  5 i
 32 
32   1  16  2   1  4 2  i  4 i 2


 121   121   1   11i
Complex numbers can be written in the form
a + bi (called standard form), with both a
and b as real numbers.
a is a real number and bi would be an
imaginary number.
If b = 0, a + bi is a real number.
If a = 0, a + bi is an imaginary number.
Example
Write each of the following in the form of a
complex number in standard form a + bi.
6 = 6 + 0i
8i = 0 + 8i
 24 
4  6  1  2 i 6  0  2 i 6
6   25  6  25   1  6 + 5i
Adding and subtracting complex numbers is
accomplished by combining their
corresponding components.
(a + bi) + (c + di) = (a + c) + (b + d)i
(a + bi) – (c + di) = (a – c) + (b – d)i
Example
Add or subtract the following complex numbers.
Write the answer in standard form a + bi.
(4 + 6i) + (3 – 2i) = (4 + 3) + (6 – 2)i = 7 + 4i
(8 + 2i) – (4i) = (8 – 0) + (2 – 4)i = 8 – 2i
The technique for multiplying complex
numbers varies depending on whether the
numbers are written as single term (either
the real or imaginary component is missing)
or two terms.
Note that the product rule for radicals does
NOT apply for imaginary numbers.
2
20i
 20(1)   20
 16   25  4i  5i 
 16   25 
16  25 
400  20
Example
Multiply the following complex numbers.
8i • 7i
56i2
56(-1)
-56
Example
Multiply the following complex numbers.
Write the answer in standard form a + bi.
5i(4 – 7i)
20i – 35i2
20i – 35(-1)
20i + 35
35 + 20i
Example
Multiply the following complex numbers.
Write the answer in standard form a + bi.
(6 – 3i)(7 + 4i)
42 + 24i – 21i – 12i2
42 + 3i – 12(-1)
42 + 3i + 12
54 + 3i
In the previous chapter, when trying to
rationalize the denominator of a rational
expression containing radicals, we used the
conjugate of the denominator.
Similarly, to divide complex numbers, we
need to use the complex conjugate of the
number we are dividing by.
The conjugate of a + bi is a – bi.
The conjugate of a – bi is a + bi.
The product of (a + bi) and (a – bi) is
(a + bi)(a – bi)
a2 – abi + abi – b2i2
a2 – b2(-1)
a2 + b2, which is a real number.
Example
Divide the following complex numbers. Write
the answer in standard form.
24  18i  8i  6i 2
6  2i 6  2i 4  3i




2
4  3i 4  3i 4  3i 16  12i  12i  9i
18 26
24  26i  6(1) 18  26i

 i

25
25 25
16  9(1)
Example
Divide the following complex numbers.
5  6i
 30i
 30i
5
5
 30i





i

2
6i  6i  36i
36
6i
6
 36(1)
• So far, we have looked at only two powers of i,
i and i2
• There is an interesting pattern within the
powers of i.
i  1
i  i  i  (1)i  i
i 2  1
i 6  i 4  i 2  (1)(1)  1
i  i  i  (1)i  i
4
2
2
i  i  i  (1)(1)  1
i  i  i  (1)(i)  i
8
4
4
i  i  i  (1)(1)  1
3
5
2
7
4
4
3
The powers recycle through each multiple of 4.
i
4k
1
Example
Simplify each of the following powers.
i  i i  i
53
i
17
52
413
 i  1i  i
1
1
1 1 i i
1
  
 2 
 17  16 
i
i i 1 i i i  i  i
i
i
 i

1
 (1)