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C. Multiplying Complex Numbers; Powers of i

C. Multiplying Complex Numbers; Powers of i

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CHAPTER 3 Quadratic Functions and Operations on Functions



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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CHAPTER 3 Quadratic Functions and Operations on Functions



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



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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



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