C. Multiplying Complex Numbers; Powers of i
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As before, computations with complex numbers are easily checked using a graphing
calculator. The results from Example 5(a), 5(c), and 5(d) are verified in Figure 3.5. For
Example 5(b), the coefficients are irrational numbers that run off of the screen (Figure 3.6).
To check this answer, we could compare the decimal forms of Ϫ312 and 216 to the
given decimal numbers, but instead we chose to simply check the product using division
(to check the result of a multiplication, divide by one of the factors). Dividing the
result by the factor 1Ϫ6, gives 2 ϩ 1.732050808i, which we easily recognize as the
other original factor 2 ϩ i 13 ✓.
Figure 3.5
Figure 3.6
Recall that expressions 2x ϩ 5 and 2x Ϫ 5 are called binomial conjugates. In the
same way, a ϩ bi and a Ϫ bi are called complex conjugates. Note from Example 5(d)
that the product of the complex number a ϩ bi with its complex conjugate a Ϫ bi is a
real number. This relationship is useful when rationalizing expressions with a complex
number in the denominator, and we generalize the result as follows:
Product of Complex Conjugates
WORTHY OF NOTE
Notice that the product of a
complex number and its conjugate
also gives us a method for factoring
the sum of two squares using
complex numbers! For the
expression x2 ϩ 4, the factored
form would be 1x ϩ 2i21x Ϫ 2i2. For
more on this idea, see Exercise 79.
EXAMPLE 6
ᮣ
For a complex number a ϩ bi and its conjugate a Ϫ bi,
their product 1a ϩ bi2 1a Ϫ bi2 is the real number a2 ϩ b2;
1a ϩ bi21a Ϫ bi2 ϭ a2 ϩ b2
Showing that 1a ϩ bi21a Ϫ bi2 ϭ a2 ϩ b2 is left as an exercise (see Exercise 79),
but from here on, when asked to compute the product of complex conjugates, simply
refer to the formula as illustrated here: 1Ϫ3 ϩ 5i2 1Ϫ3 Ϫ 5i2 ϭ 1Ϫ32 2 ϩ 52 or 34. See
Exercises 45 through 48.
These operations on complex numbers enable us to verify complex solutions by
substitution, in the same way we verify solutions for real numbers. In Example 2 we
stated that x ϭ Ϫ3 ϩ 2i was one solution to x2 ϩ 6x ϩ 13 ϭ 0. This is verified here.
Checking a Complex Root by Substitution
Verify that x ϭ Ϫ3 ϩ 2i is a solution to x2 ϩ 6x ϩ 13 ϭ 0.
Solution
ᮣ
x2 ϩ 6x ϩ 13 ϭ 0 original equation
1Ϫ3 ϩ 2i2 ϩ 61Ϫ3 ϩ 2i2 ϩ 13 ϭ 0 substitute Ϫ3 ϩ 2i for x
1Ϫ32 2 ϩ 21Ϫ32 12i2 ϩ 12i2 2 Ϫ 18 ϩ 12i ϩ 13 ϭ 0 square and distribute
9 Ϫ 12i ϩ 4i2 ϩ 12i Ϫ 5 ϭ 0 simplify
9 ϩ 1Ϫ42 Ϫ 5 ϭ 0 combine terms 112i Ϫ 12i ϭ 0; i 2 ϭ Ϫ12
0 ϭ 0✓
2
A calculator verification is also shown.
Now try Exercises 49 through 56
ᮣ
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Section 3.1 Complex Numbers
EXAMPLE 7
ᮣ
287
Checking a Complex Root by Substitution
Show that x ϭ 2 Ϫ i 13 is a solution of x2 Ϫ 4x ϭ Ϫ7.
Solution
ᮣ
x2 Ϫ 4x ϭ Ϫ7
12 Ϫ i 132 2 Ϫ 412 Ϫ i 132 ϭ Ϫ7
4 Ϫ 4i 13 ϩ 1i 132 2 Ϫ 8 ϩ 4i 13 ϭ Ϫ7
4 Ϫ 4i 13 Ϫ 3 Ϫ 8 ϩ 4i 13 ϭ Ϫ7
Ϫ7 ϭ Ϫ7✓
original equation
substitute 2 Ϫ i 13 for x
square and distribute
1i 132 2 ϭ Ϫ3
solution checks
A calculator verification is also shown.
Now try Exercises 57 through 60
ᮣ
The imaginary unit i has another interesting and useful property. Since i ϭ 1Ϫ1
and i 2 ϭ Ϫ1, it follows that i 3 ϭ i 2 # i ϭ 1Ϫ12i ϭ Ϫi and i 4 ϭ 1i 2 2 2 ϭ 1.We can now
simplify any higher power of i by rewriting the expression in terms of i 4, since
1i 4 2 n ϭ 1 for any natural number n.
iϭi
i ϭ Ϫ1
i 3 ϭ Ϫi
i4 ϭ 1
i5 ϭ i4 # i
i ϭ i4 # i2
i7 ϭ i4 # i3
i 8 ϭ 1i 4 2 2
2
6
ϭi
ϭ Ϫ1
ϭ Ϫi
ϭ1
Notice the powers of i “cycle through” the four values i, Ϫ1, Ϫi and 1. In more advanced classes, powers of complex numbers play an important role, and next we learn
to reduce higher powers using the power property of exponents and i 4 ϭ 1. Essentially, we divide the exponent on i by 4, then use the remainder to compute the value of
the expression. For i 35, 35 Ϭ 4 ϭ 8 remainder 3, showing i 35 ϭ 1i 4 2 8 # i 3 ϭ Ϫi.
EXAMPLE 8
ᮣ
Simplifying Higher Powers of i
Simplify:
a. i22
Solution
ᮣ
b. i28
c. i57
a. 22 Ϭ 4 ϭ 5 remainder 2
i22 ϭ 1i 4 2 5 # 1i2 2
ϭ 112 5 1Ϫ12
ϭ Ϫ1
c. 57 Ϭ 4 ϭ 14 remainder 1
i57 ϭ 1i 4 2 14 # i
ϭ 112 14i
ϭi
d. i75
b. 28 Ϭ 4 ϭ 7 remainder 0
i28 ϭ 1i 4 2 7
ϭ 112 7
ϭ1
d. 75 Ϭ 4 ϭ 18 remainder 3
i75 ϭ 1i 4 2 18 # 1i3 2
ϭ 112 18 1Ϫi2
ϭ Ϫi
Now try Exercises 61 and 62 ᮣ
C. You’ve just seen how
we can multiply complex
numbers and find powers of i
While powers of i can likewise be checked on a graphing calculator, the result must be interpreted carefully.
For instance, while we know i 22 ϭ Ϫ1, the calculator
returns an answer of Ϫ1 Ϫ 2E Ϫ 13i (Figure 3.7). To
interpret this result correctly, we identify the real
number part as a ϭ Ϫ1, and the imaginary part as
b ϭ 0, which due to limitations in the technology
is approximated by Ϫ2E Ϫ 13 ϭ Ϫ2 ϫ 10Ϫ13 (an
extremely small number).
Figure 3.7
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D. Division of Complex Numbers
3Ϫi
actually have a radical in the denominator.
2ϩi
To divide complex numbers, we simply apply our earlier method of rationalizing
denominators (Section R.6), but this time using a complex conjugate.
Since i ϭ 1Ϫ1, expressions like
EXAMPLE 9
ᮣ
Dividing Complex Numbers
Divide and write each result in a ϩ bi form.
2
3Ϫi
6 ϩ 1Ϫ36
a.
b.
c.
5Ϫi
2ϩi
3 ϩ 1Ϫ9
Solution
ᮣ
2
2 #5ϩi
ϭ
5Ϫi
5Ϫi 5ϩi
215 ϩ i2
ϭ 2
5 ϩ 12
10 ϩ 2i
ϭ
26
10
2
ϭ
ϩ i
26
26
1
5
ϩ i
ϭ
13
13
6 ϩ i 136
6 ϩ 1Ϫ36
ϭ
c.
3 ϩ 1Ϫ9
3 ϩ i 19
6 ϩ 6i
ϭ
3 ϩ 3i
a.
b.
3Ϫi
3Ϫi#2Ϫi
ϭ
2ϩi
2ϩi 2Ϫi
6 Ϫ 3i Ϫ 2i ϩ i2
ϭ
22 ϩ 12
6 Ϫ 5i ϩ 1Ϫ12
ϭ
5
5
5i
5 Ϫ 5i
ϭ Ϫ
ϭ
5
5
5
ϭ1Ϫi
convert to i notation
simplify
The expression can be further simplified by reducing common factors.
ϭ
611 ϩ i2
311 ϩ i2
ϭ 2 ϩ 0i
factor and reduce
Now try Exercises 63 through 68
ᮣ
As mentioned, operations on complex numbers can be checked using inverse
operations, just as we do for real numbers. To check the division from Example 9b,
we multiply 1 Ϫ i by the divisor 2 ϩ i:
11 Ϫ i212 ϩ i2 ϭ 2 ϩ i Ϫ 2i Ϫ i2
ϭ 2 Ϫ i Ϫ 1Ϫ12
ϭ2Ϫiϩ1
ϭ 3 Ϫ i✓
D. You’ve just seen how
we can divide complex
numbers
Several checks are asked for in the exercises. A calculator check is shown for Example 9(a) in Figure 3.8,
where we note that converting the coefficients to
rational numbers (where possible): MATH 1: ᭤Frac,
makes the result easier to understand. As you read and
5
1
ϩ i, which
interpret this result, note the intent is
13 13
1
5
ϩ
must not be confused with
. The latter is an
13 13i
entirely different (and incorrect) number.
Figure 3.8
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Section 3.1 Complex Numbers
289
3.1 EXERCISES
ᮣ
CONCEPTS AND VOCABULARY
Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary.
1. Given the complex number 3 ϩ 2i, its complex
conjugate is
.
4 ϩ 6i12
is written in the standard
2
form a ϩ bi, then a ϭ
and b ϭ
.
3. If the expression
5. Discuss/Explain which is correct:
a. 1Ϫ4 # 1Ϫ9 ϭ 11Ϫ42 1Ϫ92 ϭ 136 ϭ 6
b. 1Ϫ4 # 1Ϫ9 ϭ 2i # 3i ϭ 6i 2 ϭ Ϫ6
ᮣ
2. The product 13 ϩ 2i2 13 Ϫ 2i2 gives the real
number
.
4. For i ϭ 1Ϫ1, i 2 ϭ
, i4 ϭ
, i6 ϭ
, and
8
3
5
7
i ϭ
,i ϭ
,i ϭ
,i ϭ
, and i 9 ϭ .
6. Compare/Contrast the product 11 ϩ 12211 Ϫ 132
with the product 11 ϩ i12211 Ϫ i132. What is the
same? What is different?
DEVELOPING YOUR SKILLS
Simplify each radical (if possible). If imaginary, rewrite
in terms of i and simplify.
7. a. 1Ϫ144
c. 127
b. 1Ϫ49
d. 172
8. a. 1Ϫ100
c. 164
Write each complex number in the standard form
a ؉ bi and clearly identify the values of a and b.
17. a. 5
b. 3i
18. a. Ϫ2
b. Ϫ4i
b. 1Ϫ169
d. 198
19. a. 2 1Ϫ81
b.
1Ϫ32
8
9. a. Ϫ 1Ϫ18
c. 3 1Ϫ25
b. Ϫ 1Ϫ50
d. 2 1Ϫ9
20. a. Ϫ31Ϫ36
b.
1Ϫ75
15
10. a. Ϫ 1Ϫ32
c. 3 1Ϫ144
b. Ϫ 1Ϫ75
d. 2 1Ϫ81
21. a. 4 ϩ 1Ϫ50
b. Ϫ5 ϩ 1Ϫ27
11. a. 1Ϫ19
Ϫ12
c.
A 25
b. 1Ϫ31
Ϫ9
d.
A 32
22. a. Ϫ2 ϩ 1Ϫ48
b. 7 ϩ 1Ϫ75
23. a.
14 ϩ 1Ϫ98
8
b.
5 ϩ 1Ϫ250
10
12. a. 1Ϫ17
Ϫ45
c.
A 36
b. 1Ϫ53
Ϫ49
d.
A 75
24. a.
21 ϩ 1Ϫ63
12
b.
8 ϩ 1Ϫ27
6
Simplify each expression, writing the result in terms of i.
13. a.
14. a.
15. a.
16. a.
2 ϩ 1Ϫ4
2
16 Ϫ 1Ϫ8
2
8 ϩ 1Ϫ16
2
6 Ϫ 1Ϫ72
4
b.
b.
b.
b.
6 ϩ 1Ϫ27
3
4 ϩ 31Ϫ20
2
10 Ϫ 1Ϫ50
5
12 ϩ 1Ϫ200
8
Perform the addition or subtraction. Write the result in
a ؉ bi form. Check your answers using a calculator.
25. a. 112 Ϫ 1Ϫ42 ϩ 17 ϩ 1Ϫ92
b. 13 ϩ 1Ϫ252 ϩ 1Ϫ1 Ϫ 1Ϫ812
c. 111 ϩ 1Ϫ1082 Ϫ 12 Ϫ 1Ϫ482
26. a. 1Ϫ7 Ϫ 1Ϫ722 ϩ 18 ϩ 1Ϫ502
b. 1 13 ϩ 1Ϫ22 Ϫ 1 112 ϩ 1Ϫ82
c. 1 120 Ϫ 1Ϫ32 ϩ 1 15 Ϫ 1Ϫ122
27. a. 12 ϩ 3i2 ϩ 1Ϫ5 Ϫ i2
b. 15 Ϫ 2i2 ϩ 13 ϩ 2i2
c. 16 Ϫ 5i2 Ϫ 14 ϩ 3i2