Download 3.2 Complex Numbers

LESSON
3.2
Name
Complex Numbers
Class
3.2
Date
Complex Numbers
Essential Question: What is a complex number, and how can you add, subtract, and multiply
complex numbers?
Common Core Math Standards
The student is expected to:
COMMON
CORE
Resource
Locker
N-CN.A.2
Explore
Use the relation i 2 = –1 and the commutative, associative, and
distributive properties to add, subtract, and multiply complex numbers.
Also N-CN.A.1
In this lesson, you’ll learn to perform operations with complex numbers, which have a form
similar to linear binomials such as 3 + 4x and 2 − x.
Mathematical Practices
COMMON
CORE
Exploring Operations Involving
Complex Numbers
A
MP.2 Reasoning
Add the binomials 3 + 4x and 2 − x.
Group like terms.
Language Objective
Combine like terms.
Work with a partner to classify and justify the classification of real,
complex, and imaginary numbers.
B
Subtract 2 − x from 3 + 4x.
Rewrite as addition.
ENGAGE
© Houghton Mifflin Harcourt Publishing Company
(3 + 4x) (2 − x) = 6 + (−3x) + 8x
Reflect
1.
In Step A, you found that (3 + 4x) + (2 − x) = 5 + 3x. Suppose x = i (the imaginary unit).
What equation do you get? (3 + 4i) + (2 − i) = 5 + 3i
2.
In Step B, you found that (3 + 4x) + (2 − x) = 1 + 5x. Suppose x = i (the imaginary unit).
3.
In Step C, you found that (3 + 4x)(2 − x) = 6 + 5x - 4x 2. Suppose x = i (the imaginary unit).
What equation do you get? How can you further simplify the right side of this equation?
What equation do you get? (3 + 4i) - (2 − i) = 1 + 5i
(3 + 4i)(2 − i) = 6 + 5i - 4i 2; because i 2 = -1, the right side of this equation can be
simplified to 6 + 5i - 4(-1), or 10 + 5i.
Module 3
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Lesson 2
127
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A2_MNLESE385894_U2M03L2 127
operations x.
to perform
and 2 −
, you’ll learn
as 3 + 4x
In this lesson binomials such
linear
similar to
ials 3 + 4x
Add the binom
terms.
Group like

and 2 − x.
(3 + 4x)
+ (2 − x)
(
= 3+
=
like terms.
Combine
3 + 4x.
− x from
Subtract 2
addition.
Rewrite as

Group like
(3 + 4x)
− (2 − x)
terms.
like terms.
Combine
ials 3 + 4x
the binom
Multiply

HARDCOVER
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Exploring
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a form
Number
which have
Complex
x numbers,
with comple
Explore
Use FOIL.
(
(
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+ −2 +
= 3 + -2
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)
Turn to Lesson 3.2 in the
hardcover edition.
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=
and 2 − x.
) + (4x +
)
2
5
+
1
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x
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2
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Combine
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Reflect
1.
In Step A,
© Houghto
2.
Lesson 2
127
Module 3
L2 127
4_U2M03
SE38589
A2_MNLE
Lesson 3.2
+ -4x 2
5x + -4x 2
=6+
Combine like terms.
COMMON
CORE
127
)
Multiply the binomials 3 + 4x and 2 − x.
Use FOIL.
PREVIEW: LESSON
PERFORMANCE TASK
View the Engage section online. Discuss the photo
and how complex numbers can be used to generate
fractal patterns. Then preview the Lesson
Performance Task.
5
-x
x )
(
= (3 + -2 ) + (4x + x )
= ( 1 + 5x )
Combine like terms.
C
) + (4x +
+ 3x )
2
(3 + 4x) − (2 − x) = (3 + 4x) + −2 +
Group like terms.
Essential Question: What is a complex
number and how do you add, subtract,
and mutiply complex numbers?
Possible answer: A complex number has the form
a + bi where a and b are real numbers and i is the
imaginary unit. You add and subtract complex
numbers by combining like terms. You multiply
complex numbers by using the distributive
property, substituting -1 for i 2 , and combining
like terms.
(
=(
(3 + 4x) + (2 − x) = 3 +
5/22/14
10:28 AM
5/22/14 10:29 AM
Explain 1
Defining Complex Numbers
EXPLORE
A complex
number is any number that can be written in the form a + bi, where a and b are real numbers and
_
i = √-1 . For a complex number a + bi a is called the real part of the number, and b is called the imaginary part.
(Note that “imaginary part” refers to the real multiplier of i; it does not refer to the imaginary number bi.) The Venn
diagram shows some examples of complex numbers.
3 + 4i
Exploring Operations Involving
Imaginary Numbers
Complex Numbers
4 - 2i
1-i 5
Real Numbers
17
-5
π
3
√2
3.3
4
INTEGRATE TECHNOLOGY
Imaginary
Numbers
2i
-7i
i 11
Students have the option of completing the activity
either in the book or online.
6.3i
QUESTIONING STRATEGIES
Notice that the set of real numbers is a subset of the set of complex numbers. That’s because a real number a can be
written in the form a + 0i (whose imaginary part is 0). Likewise, the set of imaginary numbers is also a subset of the
set of complex numbers, because an imaginary number bi(where b ≠ 0) can be written in the form 0 + bi(whose real
part is 0).
Example 1

Do all complex numbers include an imaginary
part? Explain. No, a complex number does
not always include an imaginary part. All real
numbers are also complex numbers. The imaginary
unit i is defined as √-1 . You can use the imaginary
unit to write the square root of any negative
number.
――
Identify the real and imaginary parts of the given number. Then tell which
of the following sets the number belongs to: real numbers, imaginary
numbers, and complex numbers.
9 + 5i
The real part of 9 + 5i is 9, and the imaginary part is 5. Because both the real and imaginary parts
of 9 + 5i are nonzero, the number belongs only to the set of complex numbers.

-7i
the number belongs to these sets: imaginary numbers and complex numbers
.
Your Turn
Identify the real and imaginary parts of the given number. Then tell which of the
following sets the number belongs to: real numbers, imaginary numbers, and
complex numbers.
11
The real part of 11 is 11, and the imaginary part is 0. Because the imaginary part is 0, the
© Houghton Mifflin Harcourt Publishing Company
The real part of -7i is 0 , and the imaginary part is -7 . Because the real/imaginary part is 0,
4.
How can you tell which part of a complex
number is the real part and which is the
imaginary part? In a number of the form a + bi, the
real part is a, which does not have i as a factor. The
imaginary part is bi, where b is a nonzero real
number and i is the imaginary unit.
EXPLAIN 1
Defining Complex Numbers
number belongs to these sets: real numbers and complex numbers.
AVOID COMMON ERRORS
Module 3
128
Students may write the conjugate of a + bi as -a−bi.
Caution them to change the sign only of bi, and not
of a, when they write the complex conjugate.
Lesson 2
PROFESSIONAL DEVELOPMENT
A2_MNLESE385894_U2M03L2 128
Integrate Mathematical Practices
This lesson provides an opportunity to address Mathematical Practice MP.2,
which calls for students to translate between multiple representations and to
“reason abstractly and quantitatively.” Students explore the relationship between
operations with complex numbers and operations with binomials. They also
describe how complex-number arithmetic operations follow from operations with
rational numbers and square roots.
5/22/14 10:29 AM
QUESTIONING STRATEGIES
What are the components of a complex
number? A complex number has the form
a + bi, where a and b are real numbers. Each term of
a + bi is given a name: a is called the real part and bi
is called the imaginary part.
Complex Numbers 128
-1 + i
The real part of -1 + i is -1, and the imaginary part is 1. Because both the real and
5.
EXPLAIN 2
imaginary parts of -1 + i are nonzero, the number belongs only to the set of complex
numbers.
Adding and Subtracting Complex
Numbers
Explain 2
Adding and Subtracting Complex Numbers
To add or subtract complex numbers, add or subtract the real parts and the imaginary parts separately.
QUESTIONING STRATEGIES
Example 2
Can the sum of two imaginary numbers be
0? yes, if both the real parts and the
imaginary parts are opposites of each other

Add or subtract the complex numbers.
( -7 + 2i ) + ( 5 - 11i )
Group like terms.
A pure imaginary number has no real part.
When is the sum of two imaginary numbers a
pure imaginary number? when the real parts are
opposites, for example, (6 + 4i) + (−6 − 8i) = −4i

(−7 + 2i) + (5 − 11i) = (−7 + 5)+ (2i + ( −11i ))
Combine like terms.
= −2 + (−9i)
Write addition as subtraction.
= −2 − 9i
( 18 + 27i ) - ( 2 + 3i )
Group like terms.
(
(18 + 27i) − (2 + 3i) = 18 −
AVOID COMMON ERRORS
Some students may try to simplify a complex number
by combining the real part and the imaginary part.
Emphasize that just as unlike terms in an algebraic
expression cannot be combined, neither can the real
and imaginary parts of a complex number.
− 3i
)
Reflect
6.
Is the sum (a + bi) + (a - bi) where a and b are real numbers, a real number or an imaginary number?
Explain.
(a + bi) + (a - bi) = (a + a) + (bi + (-bi)) = 2a
Since a is a real number, 2a is a real number. So, the sum (a + bi) + (a - bi) is a
© Houghton Mifflin Harcourt Publishing Company
2, and therefore have a maximum of two real
solutions. Sometimes the two real solutions are a
double root, but many quadratic equations have zero
real solutions. It is only by introducing complex
numbers that we are able to find two solutions for all
quadratic equations.
) + ( 27i
= 16 + 24 i
Combine like terms.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Quadratic equations are equations of degree
2
real number.
Your Turn
Add or subtract the complex numbers.
7.
(17 - 6i) - (9 + 10i)
(17 - 6i) - (9 + 10i) = (17 - 9) + (-6i - 10i)
= 8 + (-16i)
= 8 - 16i
8.
(16 + 17i) + (-8 - 12i)
(16 + 17i) + (-8 - 12i) = (16 + (-8)) + (17i + (-12i))
= 8 + 5i
Module 3
129
Lesson 2
COLLABORATIVE LEARNING
A2_MNLESE385894_U2M03L2 129
Peer-to-Peer Activity
Have students work in pairs. Each student writes an addition and subtraction
problem for an imaginary number (a + bi), a pure imaginary number (bi), and a
real number (a). The partners exchange papers and solve all six problems, then
exchange papers again to check their answers.
129
Lesson 3.2
5/22/14 10:29 AM
Explain 3
Multiplying Complex Numbers
EXPLAIN 3
To multiply two complex numbers, use the distributive property to multiply each part of one number by each part
of the other. Use the fact that i 2 = -1 to simplify the result.
Example 3

Multiplying Complex Numbers
Multiply the complex numbers.
(4 + 9i)(6 − 2i)
Substitute −1 for i .
= 24 − 8i + 54i − 18(-1)
Combine like terms.
= 42 + 46i
2

QUESTIONING STRATEGIES
(4 + 9i)(6 − 2i) = 24 − 8i + 54i − 18i 2
Use the distributive property.
How is multiplying two imaginary numbers
similar to the FOIL method? The same steps
are used to multiply two imaginary numbers as are
used to multiply two binomials.
(−3 + 12i)(7 + 4i)
Use the distributive property.
(−3 + 12i)(7 + 4i) = -21 −12i + 84i + 48i 2
Substitute −1 for i 2.
= -21 − 12i + 84i + 48(−1)
Combine like terms.
= -69 + 72 i
If you use the FOIL method to multiply
two imaginary numbers, which of the
products, F, O, I, or L, are real? Which are pure
imaginary? F and L are real; O and I are pure
imaginary.
Reflect
9.
Is the product of (a + bi)(a - bi), where a and b are real numbers, a real number or an imaginary
number? Explain.
(a + bi)(a - bi) = a 2 - abi + abi - b 2i 2
AVOID COMMON ERRORS
= a 2 - b 2(-1)
= a2 + b2
Some students may try to simplify a complex number
by combining the real part and the imaginary part.
Emphasize that just as unlike terms in an algebraic
expression cannot be combined, neither can the real
and imaginary parts of a complex number.
Since a and b are real numbers, a 2 + b 2 is a real number. So the product (a + bi)(a - bi) is
a real number.
Your Turn
© Houghton Mifflin Harcourt Publishing Company
Multiply the complex numbers.
10. (6 - 5i)(3 - 10i)
(6 - 5i) (3 - 10i) = 18 - 60i - 15i + 50i 2
= 18 - 60i - 15i + 50(-1)
= -32 - 75i
11. (8 + 15i)(11 + i)
(8 + 15i)(11 + i) = 88 + 8i + 165i + 15i 2
= 88 + 8i + 165i + 15(-1)
= 73 + 173i
Module 3
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Reinforce how FOIL is used to multiply two
binomials, such as the product of 2 + 3x and 1 − 4x,
before asking students to use FOIL to multiply two
imaginary numbers, such as the product of 2 + 3i
and 1 − 4i. Show the work for the binomials side by
side with the work for the imaginary numbers. Ask
students to point out the similarities and differences
in the work steps.
Lesson 2
130
DIFFERENTIATE INSTRUCTION
A2_MNLESE385894_U2M03L2.indd 130
5/29/14 9:21 PM
Visual Learners
To help visual learners see the relationships among the various types of numbers,
show them the following diagram. Ask them to give examples of each type of number.
Rational
Real
Complex
Numbers
Irrational
Imaginary
Complex Numbers 130
Explain 4
EXPLAIN 4
Solving a Real-World Problem
Using Complex Numbers
Electrical engineers use complex numbers when analyzing electric
circuits. An electric circuit can contain three types of components:
resistors, inductors, and capacitors. As shown in the table, each type
of component has a different symbol in a circuit diagram, and each
is represented by a different type of complex number based on the
phase angle of the current passing through it.
Solving a Real-World Problem Using
Complex Numbers
QUESTIONING STRATEGIES
If two real-world quantities have values of
12 – 15i and 16 – 20i, do you know which
quantity is larger? Explain. No; you can compare the
real part only to the real part, and the imaginary
part only to the imaginary part.
Circuit
Component
Phase
Angle
Representation as a Complex Number
Resistor
0°
Inductor
90°
An imaginary number bi where b > 0
-90°
An imaginary number bi where b < 0
Capacitor
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Critical Thinking
MP.3 Ask students to compare the product of
A real number a
A diagram of an alternating current (AC) electric circuit is shown along with
the impedance (measured in ohms, Ω) of each component in the circuit. An AC
power source, which is shown on the left in the diagram and labeled 120 V (for
volts), causes electrons to flow through the circuit. Impedance is a measure of each
component’s opposition to the electron flow.
Example 4

© Houghton Mifflin Harcourt Publishing Company
2 + 3i and 2 – 3i to the product of 3 + 2i and 3 – 2i.
Have them explain the relationship between the
products. Both have a value of 13. The first product
is (2 + 3i)(2 – 3i) = 2 2 – 6i + 6i – 9i 2 = 4 + 9 = 13.
The second has the same middle terms and has a
value of 9 + 4 = 13.
Symbol in Circuit
Diagram
4Ω
3Ω
120 V
5Ω
Use the diagram of the electric circuit to answer the following questions.
The total impedance in the circuit is the sum of the impedances for the individual
components. What is the total impedance for the given circuit?
Write the impedance for each component as a complex number.
• Impedance for the resistor: 4
• Impedance for the inductor: 3i
• Impedance for the capacitor: -5i
Then find the sum of the impedances.
Total impedance = 4 + 3i + (−5i) = 4 − 2i

Ohm’s law for AC electric circuits says that the voltage V (measured in volts) is the product
of the current I (measured in amps) and the impedance Z (measured in ohms):
V = I ∙ Z. For the given circuit, the current I is 24 + 12i amps. What is the voltage
V for each component in the circuit?
Use Ohm’s law, V = I ∙ Z, to find the voltage for each component. Remember that Z is the
impedance from Part A.
Module 3
A2_MNLESE385894_U2M03L2.indd 131
131
Lesson 3.2
131
Lesson 2
19/03/14 3:00 PM
( )
( )
( )
= I ∙ Z = ( 24 + 12i)
4
= 96 + 48 i
Voltage for the inductor = I ∙ Z = (24 + 12i)
3i
= -36 + 72 i
Voltage for the resistor
Voltage for the capacitor = I ∙ Z = (24 + 12i) -5i
CONNECT VOCABULARY
Have students complete a Venn diagram in which
one circle contains real numbers and the other circle
imaginary numbers. Emphasize that the overlap
shows complex numbers. Have students write three
examples of each kind of number in their diagrams.
= 60 − 120i
Reflect
12. Find the sum of the voltages for the three components in Part B. What do you notice?
Sum of voltages = (96 + 48i) + (-36 + 72i) + (60 - 120i)
⌉ ⌈
⌉
=⌈
⌊96 + (-36) + 60⌋ + ⌊48i + 72i + (-120i)⌋
ELABORATE
= 120 + 0i = 120
The sum of the voltages equals the voltage supplied by the power source.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Relate the real and imaginary parts of a
Your Turn
13. Suppose the circuit analyzed in Example 4 has a second resistor with an impedance of 2 Ω added to it.
Find the total impedance. Given that the circuit now has a current of 18 + 6i amps, also find the voltage
for each component in the circuit.
complex number to the real (horizontal) and
imaginary (vertical) axes. Students should realize that
points on the horizontal axis represent real numbers,
points on the vertical axis represent pure imaginary
numbers, and points in the quadrants represent
complex numbers.
Total impedance = 2 + 4 + 3i + (-5i) = 6 - 2i
Voltage for the first resistor = I ∙ Z = (18 + 6i)(4) = 72 + 24i
Voltage for the second resistor = I ∙ Z = (18 + 6i)(2) = 36 + 12i
Voltage for the inductor = I ∙ Z = (18 + 6i)(3i) = -18 + 54i
Voltage for the capacitor = I ∙ Z = (18 + 6i)(-5i) = 30 - 90i
14. What kind of number is the sum, difference, or product of two complex numbers?
The sum, difference, or product of two complex numbers is always a complex number.
15. When is the sum of two complex numbers a real number? When is the sum of two complex numbers an
imaginary number?
The sum of two complex numbers is a real number when the imaginary parts of the
numbers are additive inverses or both 0. The sum of two complex numbers is an imaginary
number when the real parts of the numbers are additive inverses or both 0.
© Houghton Mifflin Harcourt Publishing Company
Elaborate
QUESTIONING STRATEGIES
If you multiply a nonzero real number and an
imaginary number, is the product real or
imaginary? Why? Imaginary; c(bi)= cbi, which is
imaginary since cb is real and neither b nor c
equals 0.
SUMMARIZE THE LESSON
Module 3
132
Lesson 2
LANGUAGE SUPPORT
A2_MNLESE385894_U2M03L2 132
Connect Vocabulary
5/22/14 10:29 AM
How do you add, subtract, and multiply
complex numbers? To add or subtract
complex numbers, add or subtract their real parts
and their imaginary parts separately. To multiply
complex numbers, use the distributive property or
the FOIL method.
Provide pairs of students with 6 to 8 “number cards” or index cards on which are
different complex numbers, imaginary numbers, and real numbers. Ask them to
sort the cards into those categories. Pairs must agree on the classifications and
justify their decisions by writing a short explanation on each card. Students then
identify the real and imaginary parts of the complex numbers by labeling.
Complex Numbers 132
16. Discussion What are the similarities and differences between multiplying two complex numbers and
multiplying two binomial linear expressions in the same variable?
The distributive property is used to multiply both complex numbers and binomial linear
EVALUATE
expressions. When two binomial linear expressions in the same variable are multiplied, the
result is a trinomial quadratic expression. When two complex numbers are multiplied, the
result is another complex number.
17. Essential Question Check-In How do you add and subtract complex numbers?
To add or subtract complex numbers, combine like terms.
ASSIGNMENT GUIDE
Concepts and Skills
Practice
Explore
Exploring Operations Involving
Imaginary Numbers
Exercise 1
Example 1
Defining Complex Numbers
Exercises 2–5
Example 2
Adding and Subtracting Complex
Numbers
Exercises 6–9
Example 3
Multiplying Complex Numbers
Exercises 10–20
Evaluate: Homework and Practice
1.
Find the sum of the binomials 3 + 2x and 4 - 5x. Explain how you can use the
result to find the sum of the complex numbers 3 + 2i and 4 - 5i.
• Online Homework
• Hints and Help
• Extra Practice
(3 + 2x) + (4 - 5x) = (3 + 4) + (2x - 5x) = 7 - 3x
Replacing x with the imaginary unit i gives this result: (3 + 2i) + (4 - 5i) = 7 - 3i.
2.
Find the product of the binomials 1 - 3x and 2 + x. Explain how you can use the
result to find the product of the complex numbers 1 - 3i and 2 + i.
(1 - 3x)(2 + x) = 2 - 6x + x - 3x 2 = 2 - 5x - 3x 2
Replacing x with the imaginary unit i gives this result: (1 - 3i)(2 + i) = 2 - 5i-3i 2.
Because i 2 = -1, the result can be further simplified as follows:
(1 - 3i)(2 + i) = 2 - 5i - 3i 2 = 2 - 5i - 3(-1) = 5 - 5i
© Houghton Mifflin Harcourt Publishing Company
Identify the real and imaginary parts of the given number. Then tell which of the following
sets the number belongs to: real numbers, imaginary numbers, and complex numbers.
3.
5+i
The real part is 5, and the imaginary part is 1. Because both the real and imaginary parts
are nonzero, the number belongs only to the set of complex numbers.
4.
7 - 6i
The real part is 7, and the imaginary part is -6. Because both the real and imaginary parts
are nonzero, the number belongs only to the set of complex numbers.
Module 3
Exercise
A2_MNLESE385894_U2M03L2 133
COMMON
CORE
Mathematical Practices
1 Recall of Information
MP.2 Reasoning
15–18
2 Skills/Concepts
MP.2 Reasoning
19–22
2 Skills/Concepts
MP.4 Modeling
23
2 Skills/Concepts
MP.2 Reasoning
3 Strategic Thinking
MP.2 Reasoning
2 Skills/Concepts
MP.3 Logic
26
Lesson 3.2
Depth of Knowledge (D.O.K.)
1–14
24–25
133
Lesson 2
133
5/22/14 10:29 AM
5.
25
QUESTIONING STRATEGIES
The real part is 25, and the imaginary part is 0. Because the imaginary part is 0, the
number belongs to these sets: real numbers and complex numbers.
6.
_
i√ 21
What is a complex conjugate? Two complex
numbers of the form a + bi and a − bi are
complex conjugates. The product of complex
conjugates is always a real number.
―
The real part is 0, and the imaginary part is √21 . Because the real part is 0, the number
belongs to these sets: imaginary numbers and complex numbers.
Add.
7.
(3 + 4i) + (7 + 11i)
8.
Just as every real number corresponds to a point on
the real number line, every complex number
corresponds to a point in the complex plane. The
complex plane has a horizontal axis called the real
axis and a vertical axis called the imaginary axis. Ask
students whether they can graph the line y = 3x + 4
on the complex plane.
= 8 - 2i
= 10 + 15i
(-1 - i) + (-10 + 3i)
CRITICAL THINKING
(2 + 3i) + (6 - 5i) = (2 + 6) + (3i + 5i)
(3 + 4i) + (7 + 11i) = (3 + 7) + (4i + 11i)
9.
(2 + 3i) + (6 - 5i)
10. (-9 - 7i) + (6 + 5i)
(-1 - i) + (-10 + 3i) = (-1 - 10) + (-i + 3i) (-9 - 7i) + (6 + 5i) = (-9 + 6) + (-7i + 5i)
= -3 - 2i
= -11 + 2i
Subtract.
11. (2 + 3i) - (7 + 6i)
12. (4 + 5i) - (14 - i)
(2 + 3i) - (7 + 6i) = (2 - 7) + (3i - 6i)
(4 + 5i) - (14 - i) = (4 - 14) + (5i + i)
= -5 - 3i
= -10 + 6i
13. (-8 - 3i) - (-9 - 5i)
14. (5 + 2i) - (5 - 2i)
(5 + 2i) - (5 - 2i) = (5 - 5) + (2i + 2i)
(-8 - 3i) - (-9 - 5i) = (-8 + 9) + (-3i + 5i)
= 4i
= 1 + 2i
15. (2 + 3i)(3 + 5i)
16. (7 + i)(6 - 9i)
(2 + 3i)(3 + 5i) = 6 + 10i + 9i + 15i 2
(7 + i)(6 - 9i) = 42 - 63i + 6i - 9i 2
= 6 + 10i + 9i + 15(-1)
= -9 + 19i
17. (-4 + 11i)(-5 - 8i)
(-4 + 11i)(-5 - 8i) = 20 + 32i - 55i - 88i
= 42 - 63i + 6i - 9(-1)
= 51 - 57i
18. (4 - i)(4 + i)
2
= 20 + 32i - 55i - 88(-1)
= 108 - 23i
Module 3
A2_MNLESE385894_U2M03L2 134
134
© Houghton Mifflin Harcourt Publishing Company
Multiply.
(4 - i)(4 + i) = 16 + 4i - 4i - i 2
= 16 + 4i - 4i -(-1)
= 17
Lesson 2
5/22/14 10:29 AM
Complex Numbers 134
AVOID COMMON ERRORS
Use the diagram of the electric circuit and the given current to find the total
impedance for the circuit and the voltage for each component.
Watch for students who simplify a complex number
by combining the real part and the imaginary part.
Clarify that just as unlike terms in an algebraic
expression cannot be combined, real and imaginary
parts of a complex number cannot be combined.
19.
20.
1Ω
120 V
4Ω
3Ω
120 V
3Ω
The circuit has a current of 12 + 36i amps.
The circuit has a current of 19.2 - 14.4i amps.
Total impedance = 4 + 3i
Total impedance = 1 - 3i
Voltage for the resistor = I ∙ Z
Voltage for the resistor = I ∙ Z
= (19.2 - 14.4i)(4)
= (12 + 36i)(1)
= 76.8 - 57.6i
= 12 + 36i
Voltage for the inductor = I ∙ Z
Voltage for the capacitor = I ∙ Z
= (19.2 - 14.4i)(-3i)
= (12 + 36i)(-3i)
= 43.2 + 57.6i
= 108 - 36i
6Ω
21.
2Ω
120 V
4Ω
The circuit has a current of 7.2 + 9.6i amps.
The circuit has a current of 16.8 + 2.4i amps.
Total impedance = 6 + 2i + (-10i) = 6 - 8i
Total impedance = 7 + 3i + (-4i) = 7 - i
Voltage for the resistor = I ∙ Z
Voltage for the resistor = I ∙ Z
© Houghton Mifflin Harcourt Publishing Company
= (7.2 + 9.6i)(6)
= (16.8 + 2.4i)(7)
= 43.2 + 57.6i
= 117.6 + 16.8i
Voltage for the inductor = I ∙ Z
Voltage for the inductor = I ∙ Z
= (7.2 + 9.6i)(2i)
= (16.8 + 2.4i)(3i)
= -19.2 + 14.4i
= -7.2 + 50.4i
Voltage for the capacitor = I ∙ Z
Voltage for the capacitor = I ∙ Z
= (7.2 + 9.6i)(-10i)
= (16.8 + 2.4i)(-4i)
= 96 - 72i
Module 3
A2_MNLESE385894_U2M03L2 135
Lesson 3.2
3Ω
120 V
10 Ω
135
7Ω
22.
= 9.6 - 67.2i
135
Lesson 2
5/22/14 10:29 AM
23. Match each product on the right with the corresponding expression on the left.
B
−16 + 30i
A. (3 − 5i)(3 + 5i)
B. (3 + 5i)(3 + 5i)
D
−34
C. (−3 − 5i)(3 + 5i)
A
34
D. (3 − 5i)(−3 − 5i)
C
16 − 30i
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Have students look for a pattern in powers of
the imaginary unit i, going beyond i2, the highest
power used in the lesson. Students will need to
deduce that all the higher powers can be simplified
by repeatedly dividing out i2.
A. (3 - 5i)(3 + 5i) = 9 + 15i - 15i - 25i 2
= 9 + 15i - 15i - 25(-1)
= 34
B. (3 + 5i)(3 + 5i) = 9 + 15i + 15i + 25i 2
= 9 + 15i + 15i + 25(-1)
= -16 + 30i
C. (-3 - 5i)(3 + 5i) = -9 - 15i - 15i - 25i 2
= -9 - 15i - 15i - 25(-1)
= 16 - 30i
D. (3 - 5i)(-3 - 5i) = -9 - 15i + 15i + 25i 2
= -9 - 15i + 15i + 25(-1)
= -34
H.O.T. Focus on Higher Order Thinking
24. Explain the Error While attempting to multiply the expression (2 - 3i)(3 + 2i),
a student made a mistake. Explain and correct the error.
(2 - 3i)(3 + 2i) = 6 - 9i + 4i - 6i 2
= 6 - 9(-1) + 4(-1) - 6(1)
=5
© Houghton Mifflin Harcourt Publishing Company
=6+9-4- 6
_
The student incorrectly defined i as being equal to −1 instead of √ -1 . The
student should have written the product as 6 − 9i + 4i − 6(−1) = 12 − 5i.
Module 3
A2_MNLESE385894_U2M03L2.indd 136
136
Lesson 2
23/05/14 3:24 PM
Complex Numbers 136
_
_
_
_
25. Critical Thinking Show that √ 3 + i√3 and -√ 3 - i√3 are the square roots of 6i.
PEERTOPEER DISCUSSION
―
―
―
―
Show that the square of each number is 6i.
( √3 + i √3)( √3 + i √3) = 3 + 3i + 3i + 3i 2
Ask students to work with a partner to find examples
of each of the following terms: a complex number, a
real number, and an imaginary number. complex:
a + bi; real: a; imaginary: bi
= 3 + 3i + 3i + 3(-1)
= 6i
(- √―3 - i √―3)(- √―3 - i √―3) = 3 + 3i + 3i + 3i 2
= 3 + 3i + 3i + 3(-1)
= 6i
JOURNAL
26. Justify Reasoning What type of number is the product of two complex numbers
that differ only in the sign of their imaginary parts? Prove your conjecture.
Have students write about how complex numbers can
be applied to the real world. How might complex
numbers help describe how electric circuits operate?
The product of two complex numbers that differ only in the sign of their imaginary parts is
a real number. Proof: Let the complex numbers be a + bi and a - bi where a and b are real
and b ≠ 0. Multiplying the numbers gives the following result:
(a + bi)(a - bi) = a 2 + abi - abi - b 2i 2
= a 2 - b 2(-1)
= a2 + b2
© Houghton Mifflin Harcourt Publishing Company
Since the set of real numbers is closed under all operations, a 2 + b 2 is a real number.
Module 3
A2_MNLESE385894_U2M03L2.indd 137
137
Lesson 3.2
137
Lesson 2
19/03/14 2:59 PM
Lesson Performance Task
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Patterns
MP.8 Students should be able to recognize when
Just as real numbers can be graphed on a real number line, complex
numbers can be graphed on a complex plane, which has a horizontal
real axis and a vertical imaginary axis. When a set that involves complex
numbers is graphed on a complex plane, the result can be an elaborate
self-similar figure called a fractal. Such a set is called a Julia set.
iterations for the Julia set result in a constant or two
distinct values which switch back and forth. Students
can try different values of Z 0 and c to explore this
phenomenon, such as Z 0 = i – 1, and c = i + 1, which
returns a constant value Z n = –i +1.
Consider Julia sets having the quadratic recursive rule
2
ƒ(n + 1) = (ƒ(n)) + c for some complex number ƒ(0) and some
complex constant c. For a given value of c, a complex number
ƒ(0) either belongs or doesn’t belong to the “’filled-in” Julia set
corresponding to c depending on what happens with the sequence of
numbers generated by the recursive rule.
a. Letting c = i, generate the first few numbers in the sequence defined by ƒ(0) = 1 and
2
ƒ(n + 1) = (ƒ(n)) + i . Record your results in the table.
AVOID COMMON ERRORS
f(n + 1) =( f(n)) + i
2
f (n)
n
0
f (0) = 1
2
f (1) = (f (0)) + i = (1) + i = 1 + i
1
f (1) = 1 + i
f (2) = (f (1)) + i = (1 + i) + i = 3i
2
f (2) = 3i
f (3) = (f (2)) + i = 3i
3
f (3) = -9 + i
f (4) = (f (3)) + i =
Students may have difficulty with the notation
2
f(n + 1) = (f(n)) + c. Have students read this aloud
as: “f(n + 1) equals the quantity f(n) squared plus c.”
Note that the nested parentheses indicate that the
entire quantity f(n) is squared.
2
2
2
(
2
2
2
-9 + i
-9 + i
) +i=
2
80 -17i
_
b. The magnitude of a complex number a + bi is the real number √a 2 + b 2 . In the complex plane, the
magnitude of a complex number is the number’s distance from the origin. If the magnitudes of the
numbers in the sequence generated by a Julia set’s recursive rule, where ƒ(0) is the starting value,
remain bounded, then ƒ(0) belongs to the “filled-in” Julia set. If the magnitudes increase without
bound, then ƒ(0) doesn’t belong to the “filled-in” Julia set. Based on your completed table for ƒ(0) = 1,
would you say that the number belongs to the “filled-in” Julia set corresponding to c = i? Explain.
c. Would you say that ƒ(0) = i belongs to the “filled-in” Julia set corresponding to c = i? Explain.
__
_
b. The magnitude of f(0) = 1 is √1 2 + 02 = √1 = 1. The magnitude of f(1) = 1 + i is
__
__
_
_
√1 2 + 12 = √2 . The magnitude of f(2) = 3i is √0 2 + 32 = √9 = 3. The magnitude
__
_
of f(3) = -9 + i is √(-9 2) + 12 = √82 . The magnitude of f(4) = 80 - 17i
__
_
is √80 2 + ( -17)2 = √6689 . The magnitudes appear to be increasing without bound,
so f(0) = 1 does not belong to the “filled-in” Julia set.
c. For f(0) = i , the sequence of numbers generated by the recursive rule is f (0) = i,
f (1) = -1 + i, f (2) = -i, f (3) = -1 + i, f (3) = -1 + i, f (4) = -i and so on. Since the
© Houghton Mifflin Harcourt Publishing Company ⋅ Image Credits: ©koi88/
Shutterstock
(
) +i=
_
magnitudes of the numbers never exceed √2 , f(0) = i belongs to the “filled-in” Julia set.
Module 3
Lesson 2
138
EXTENSION ACTIVITY
A2_MNLESE385894_U2M03L2 138
Students can use the Julia set to generate a visual pattern. Each set of ordered
pairs (a, b) in the grid represents a complex number a + bi. Use these complex
numbers to calculate up to three iterations for the Julia set, using c = 0. Then find
the absolute value of the result, which is √a 2 + b 2 .
5/22/14 10:29 AM
―――
If |Z 1|>2, stop, and color that square red.
If |Z 2|>2, stop, and color that square green.
If |Z 3|>2, stop, and color that square blue.
Have students experiment with larger grids and with c ≠ 0.
Scoring Rubric
2 points: Student correctly solves the problem and explains his/her reasoning.
1 point: Student shows good understanding of the problem but does not fully
solve or explain his/her reasoning.
0 points: Student does not demonstrate understanding of the problem.
Complex Numbers 138