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CURRICULUM
AMES: Matrix Multiplication
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Lesson Information
Teacher Notes:
Contents
•
•
•
•
•
Matrix multiplication
Commutativity in matrix multiplication
Distributivity in matrix multiplication
Associativity in matrix multiplication
Identity Matrix
Standards
• CCSS-M: N-VM.C.8
Add, subtract, and multiply matrices of appropriate dimensions.
• CCSS-M: N-VM.C.9
Understand that, unlike multiplication of numbers, matrix
multiplication for square matrices is not a commutative
operation, but still satisfies the associative and distributive
properties.
• CCSS-M: N-VM.C.10
Understand that the zero and identity matrices play a role in matrix addition and multiplication similar to the role of 0 and 1
in the real numbers. The determinant of a square matrix is
nonzero if and only if the matrix has a multiplicative inverse.
Objectives
•
•
This material is based upon work
supported by the U.S. Department
of Homeland Security under
Grant Award Number, 2013-PD127-000001, Modification #2.
•
Copyright © 2015 Cyber Innovation Center
Introduce students to matrix multiplication
procedures, as well as helping students
determine when matrices are able to be
multiplied.
Students discover whether matrix
multiplication is commutative, distributive,
or associative.
Introduce students to the identity matrix for
multiplication.
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AMES: Matrix Multiplication
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MATRIX MULTIPLICATION
Teacher Notes:
As seen in the previous lesson, rows and columns are important when dealing with matrices.
Multiplication of matrices is another example of that importance. Here’s the main idea with
multiplying - think row times column, like this:
Row1Column1 Row1Column2 
 Row 1

 × Column 1 Column 2  = 

Row2Column1 Row2Column2 
Row 2 
Is the pattern apparent? Look closely at the numbers that are part of each position. If the
name of the resulting matrix is A, then A11 is synonymous with the row1column1 spot. The
numbers of the row and column match the subscript that indicates position.
Multiple Rows and Columns
Thinking more in depth, you may realize that rows and columns could have multiple
numbers. So, how do you multiply a string of numbers against another string of numbers?
Take the first number in Row 1, multiply it by the first number in Column 1, add it to
the second number in the row times the second number in the column, and continue the
process for that row and column. This sum will be in the position of the Row # and Column
# in the resulting matrix (e.g. A11). Repeat the steps for each position. The visual below may
assist in comprehending.
a b   e f  ae+bg af+bh 

=
×

 c d  g h  ce+dg cf+dh 
Exercise 1 Solution:
Row 1 × Column 1 =
2(0) + 3(-3) + 4(2) + 1(2) +
2(0) + 3(2) + 4(-3) = -5
Row 2 × Column 1 =
7(0) - 1(-3) + 3(2) +0(2) +
5(0) - 2(2) + 0(-3) = 5
 Row1× Column1

=
Row2 × Column1
 −5
 
5
Let’s try one with numbers.
Exercise 1
0
 
 −3
2
2 3 4 1 2 3 4 
 
LetY = 
 and Z =  2  . What is Z × Y ?
7 −1 3 0 5 −2 0 
0
 
2
 −3
 
*Special note: The final
answer is a 2 × 1 matrix,
but it could also be
called a vector! A vector
transformed into a matrix
is a column matrix
consisting of the vector
components.
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Teacher Notes:
Ability to Multiply Matrices
Exercise 2 Solution:
To begin, look at row 1
times column 1.
Row 1 is 1 and column 1
12 
 
is  7  .
2
 
Usually, when we multiply
matrices, the first number
in the row goes with the
first number in the column,
the second number in the
row goes with the second
number in the column,
and so on and so forth.
However, in this problem,
row 1 only has a single
number to correspond
with 3 numbers in column
1; therefore, these two
matrices cannot be
multiplied.
Exercise 2
 1  12 
   
What is 17  ×  7  ?
9 2
   
In Exercise 2, a 3×1 could not be multiplied by another 3×1. While in Exercise 1, a 1×7
matrix is multiplied against a 7×2 matrix. For the arbitrary example, a 2×2 times 2×2 is
shown. Is there a common thread between these?
3 × 1× 3 × 1
Can NOT multiply
1× 7 × 7 × 2
Can multiply
Produces a 1× 2 matrix
2×2×2×2
Can multiply
Produces a 2 × 2 matrix
The column dimension of the first matrix and the row dimension of the second matrix
must be the same in order to multiply. Also, the row dimension of the first matrix by the
column dimension of the second matrix is the dimension of the matrix produced when
multiplying the two matrices.
Multiplication Properties
Exercise 3 Solutions:
18 −10 
A× B = 

 −1 58 
 32 −22 
B× A = 

 −17 44 
In Matrix Addition and Subtraction, the commutative, distributive, and associative
properties remained as they would in the addition and subtraction of numbers. Is the same
true for multiplication?
Exercise 3
 7 4
2 −6 
Let A = 
 and B = 
 . Multiply A × B , then multiply B × A .
 −3 5 
1 8 
The two resulting matrices
are different, so we learn
that matrix multiplication
is not commutative ( A × B
does not always equal
B × A ).
For full solution, see page
5.
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AMES: Matrix Multiplication
Teacher Notes:
Exercise 4 Solution:
636 −53 
(C × D) × F = 

564 −47 
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Exercise 4
5 
10 1 6 
 
=
F 12 −1 . Compute (C × D) × F and C × (D × F ) .
Let C = 
 , D = 0  , and
 9 2 4
 12 
 
636 −53 
C × (D × F ) = 

564 −47 
The final answer for each
computation was the same;
matrix multiplication is
associative.
Exercise 5
12 −10 
1 5 
5 X + 5Y ?
If X = 
 and Y = 
 , does 5( X + Y ) =
2 
3
0 −1
For full solution, see page
5.
Exercise 5 Solution:
65 −25
5( X + Y ) = 

15 5 
65 −25
5 X + 5Y = 

15 5 
The third property,
distribution, holds when
dealing with matrices.
Let’s look at an interesting case for matrix multiplication in the next exercise, and evaluate
what we find.
Exercise 6
 5 −18 
1 0 
If M = 
 and N = 
 , then what is M × N ?
22 3 
0 1 
For full solution, see page
5.
Exercise 6 Solution:
 5 −18 
M×N = 

22 3 
For full solution, see page
5.
Exercise 7 Solution:
C × D = 2 12 4 
For full solution, see page
5.
Check out the exercise above. Matrix M is exactly the same as the final answer. This is
because N is a special kind of matrix called an identity matrix – a square matrix consisting
of only ones in the diagonal and zeros in every other position. This type of matrix has
several special properties (which will be covered more in a following lesson), but as seen in
the exercise, multiplying any matrix by the identity results in the original matrix. Let’s look
at another example.
Exercise 7
1 0 0 


Let C = 2 12 4  and D = 0 1 0  . Find C × D .
0 0 1 


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AMES: Matrix Multiplication
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Exercise 3 Full Solution:
 7(2) + 4(1) 7(−6) + 4(8)  18 −10 
A× B = 

 =
 −3(2) + 5(1) −3(−6) + 5(8)  −1 58 
2(7) + −6(−3) 2(4) + −6(5)  32 −22 
B× A = 

 =
1(4) + 8(5)   −17 44 
 1(7) + 8(−3)
The two resulting matrices are different, so we learn that matrix multiplication is not commutative (A×B does not always
equal B×A).
Exercise 4 Full Solution:
10(5) + 1(0) + 6( 12 )  53 
1. C × D = 
= 
1 
 9(5) + 2(0) + 4( 2 )   47 
2.
 53 
 53(12) 53(−1)  636 −53 
(C × D) × F =   × 12 −1 = 
 =



 47 
 47(12) 47(−1) 564 −47 
 5(12) 5(−1)  60 −5 

 

D × F =  0(12) 0(−1)  =  0 0 
 12 (12) 12 (−1)  6 − 12 

 

60 −5 
1
10 1 6  
 10(60) + 1(0) + 6(6) 10(−5) + 1(0) + 6(− 2 ) 636 −53 
C × (D × F ) = 
=
×
0
0
=

 
 
1 
 9 2 4   6 − 1   9(60) + 2(0) + 4(6) 9(−5) + 2(00) + 4(− 2 )  564 −47 
2

The final answer for each computation was the same; matrix multiplication is associative.
Exercise 5 Full Solution:
13 −5 65 −25
12 + 1 −10 + 5
5( X + Y ) = 5 
 = 5
 =

 3 + 0 2 + −1 
 3 1  15 5 
12 −10  1 5  60 −50  5 25  65 −25
=
+5
5 X + 5Y = 5 
=
+
2  0 −1 15 10  0 −5 15 5 
3
The third property, distribution, holds when dealing with matrices.
Exercise 6 Full Solution:
 5(1) − 18(0) 5(0) − 18(1)   5 −18 
M×N = 

 =
22(1) + 3(0) 22(0) + 3(1) 22 3 
Exercise 7 Full Solution:
C ×=
D 2(1) + 12(0) + 4(0) 2(0) + 12(1) + 4(0) 2(0) + 12(0) + 4(1) = 2 12 4 
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AMES: Matrix Multiplication
Teacher Notes:
Exercise 1 Solution:
15 7 
A× B = 

 48 30 
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Matrix Multiplication Practice Exercises
Exercise 1
1 3 
6 4 
Given Matrices A = 
 and B = 
 , find A × B and B × A .
7 2 
3 1 
34 26 
B× A = 

10 11 
Exercise 2
Exercise 2 Solution:
 12 
C × D =  −30 
 −11


 −1 8 
 −4 


Given Matrices C =  7 −2  and D =   , find C × D .
1
5 9


Exercise 3
Exercise 3 Solution:
E × F = 1
 −2 
 
1


Given E = 7 3 4 1 and F =   , find E × F .
5
 
 −8 
Exercise 4
Exercise 4 Solution:
Not possible.
 −7 4 
1 3 


Given G = 
 and H =  5 2  , find G × H .
2
4
−


 1 −9 


1
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AMES: Matrix Multiplication
Teacher Notes:
Exercise 5 Solution:
15 
 
J ×K =
 −1
15 
 
 −20 
I ×(J × K ) = 

 143 
11 53 
I×J = 

26 −65
 −20 
(I × J ) × K = 

 143 
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Exercise 5
2 −9 
 −3 5 2 
3


Given I = 
 , J = 1 4  , and K =   , find I × ( J × K ) and (I × J ) × K .
 5 −8 4 
 −1
6 3 


Exercise 6
 5 −1
 5 −1
Given L = 
 and M = 
 , find 2(L × M ) .
 −2 4 
 −2 4 
Exercise 6 Solution:
 27 −9 
L×M = 

 −18 18 
 54 −18 
2(L × M ) = 

 −36 36 
Exercise 7
 5 7
1 0 
Given N = 
 and O = 
 , find N × O .
0 1 
 −11 2 
Exercise 7 Solution:
 5 7
N ×O =


 −11 2 
Exercise 8 Solution:
 Alyssa  96.25
 =


Jonathan  86.25
Therefore, Alyssa finished
the class with a 96.25, and
Jonathan finished the class
with an 86.25.
Exercise 8
Alyssa and Jonathan have completed a math class consisting of two tests and a final.
Alyssa made a 90 on the first test, a 95 on the second test, and a 100 on the final. Jonathan
made a 75 on the first test, a 100 on the second test, and an 85 on the final. Each test is
worth 25 percent of the grade for the class, and the final is worth 50 percent of the grade.
Using matrix multiplication, determine what grades Alyssa and Jonathan made in the
class. (The matrices have been set up for you below.)
.25 
90 95 100   
 × .25 

75 100 85  .50 
 
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AMES: Applications of Complex Numbers
Notes:
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Lesson Information
Contents
•
•
•
•
•
•
DC Voltage Simulation
AC Voltage Simulation
Complex Numbers and Alternating Current
Basic Circuit Diagrams
Kirchhoff ’s Voltage Law (KVL)
Ohm’s Law
Materials Per Student
In an ideal environment,
each student would have
their own computer.
However, students
could easily share or the
instructor can show the
simulation to the students
via a projector.
•
This material is based upon work
supported by the U.S. Department
of Homeland Security under
Grant Award Number, 2013-PD127-000001, Modification #2.
Copyright © 2015 Cyber Innovation Center
1 computer with access to https://goo.gl/gmEPq5
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AMES: Applications of Complex Numbers
Teacher Notes:
Another option is to use
the physical items to
actually build the circuit.
This would require light
bulbs, a voltmeter (or
multimeter), an AC/
DC power supply, and
wires to connect the
previous component.
For safety reasons, this is
not recommended if the
instructors/facilitators do
not have prior experience
with AC/DC power
supplies.
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INTRODUCTION
Many people may think that the applications of complex numbers, like their name suggests,
are imaginary, but the applications of complex numbers are real and can significantly
simplify many problems. The practical applications of complex numbers exist in many fields,
including physics, biology, economics, chemistry, engineering, statistics, and mathematics.
However, the majority of these applications require the use of calculus. One application
that does not use calculus involves circuits.
To begin learning about complex numbers in circuits, we will use the PhET Interactive
Simulation called “Circuit Construction Kit (AC+DC), Virtual Lab.” To access this
simulation, visit https://goo.gl/gmEPq5 and click “Run Now!” Once the simulation is
running, it will look similar to the image below.
The first thing we will do is pause the simulation using the double bar pause icon at the
bottom of the window. In the image above, the simulation is paused.
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DC VOLTAGE
Solutions:
1. Direct current.
2. Batteries, Solar Cells
DC stands for direct
current but when DC is
used as an adjective, it is
understood to just mean
direct (e.g. DC voltage
means direct voltage).
Before we build a circuit with DC voltage, answer the following questions.
1. What does DC stand for?
2. What are common sources of DC voltage?
Now let’s build a DC circuit with two wires, a light bulb, and a battery. This can be done by
dragging and dropping the components from the right side of the simulation to the blue
area. The completed circuit will resemble the image below.
Once the circuit is built, click the play button. The electrons (blue spheres) will begin to
flow around the circuit and the light bulb will light up. Now we can measure the DC voltage
over the entire circuit by using the voltmeter in the “Tools” section of the simulator. Click
the check box next to “Voltmeter” and it will appear near the circuit.
Solutions:
1. -9V or 9V
2. A 9 Volt battery.
3. 9 Volts
To measure the voltage across this circuit, place the red lead of the voltmeter on one of the
wires and place the black lead on the other wire. Answer the following questions .
1. How much voltage was measured?
2. In terms of voltage, what kind of battery is being used in this DC circuit?
3. What is the magnitude of the measured voltage?
Pause the simulation, remove the voltmeter, and add another battery to the circuit to reflect
the circuit below.
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AMES: Applications of Complex Numbers
Solutions:
1. -18V or 18V
2. 18 Volts
3. 9 Volts
For 4, 5, and 6, answers
may vary. Ultimately, the
students should observe
that DC voltage can be
added (or subtracted)
as one would expect. For
example, three 9 Volt
power sources can be
placed adjacent to each
other and the voltage
behaves the same as if the
circuit had one 27 Volt
power source.
Solutions:
1. Alternating Current
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Resume the simulation with the play button and measure the voltage again by placing
one of the voltmeter’s leads on one wire and the other lead on the other wire. Answer the
following questions.
1. How much voltage was measured?
2. What was the magnitude of the measured voltage?
3. How much larger/smaller was the magnitude of the voltage?
4. Since the voltage has been measured with two batteries, make a hypothesis about what
would happen if a 3rd battery was added or a 4th battery.
5. Test the hypothesis. Is it true?
6. What conclusions can be made about how to add voltage/power sources in a DC circuit?
AC VOLTAGE
Before building an AC circuit, answer the following questions.
1. What does AC stand for (in the context of electricity)?
2. What are common uses for AC?
If the simulation is not paused, be sure to do so before continuing. Remove the voltmeter
and all DC power sources from the circuit. Once the circuit no longer has a power supply,
connect an AC voltage source. The circuit should look similar to the circuit pictured below.
2. Powering homes and
businesses. Wall outlets
supply AC Voltage.
Now press the play button and then use the voltmeter to measure the voltage of the power
source. This can be done by placing the voltmeter leads as shown in the image below.
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AMES: Applications of Complex Numbers
Solutions:
1. They should be
alternating direction.
2. The electrons in direct
current flow in one
direction. The electrons in
alternating current flow in
an alternating pattern.
3. It alternates between -10
and 10 volts.
4. The magnitude is 10
volts.
Solutions:
5. The students may get
different measurements
based on what the
measurement was when
they paused the simulation
in step 1. An explanation
is on the following pages.
6. The magnitude will vary
from student to student as
well.
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Answer the following questions.
1. How are the electrons moving?
2. How does the electron motion differ from direct current?
3. What is the measurement for voltage?
4. What is the magnitude for voltage?
In this case the voltage seems to be alternating between -10 and 10. Thus, we claim that
this is a 10 Volt AC power supply and that the magnitude of the voltage is 10. Pause the
simulation and complete the following.
1. Pause the simulation, and remove the voltmeter.
2. Place another AC power source adjacent to the first one as pictured.
3. Before pressing play, make a hypothesis about the magnitude of the total voltage with
two 10V AC power supplies.
4. Press play and use the voltmeter to measure the voltage.
5. What is the measurement for the voltage?
6. What is the magnitude for the voltage?
7. Repeat steps 1 through 6 to place another AC voltage source. Make a conclusion about
how AC voltage sources are added together.
7. The students should
conclude that AC voltage
does not add in the same
way that DC voltage adds.
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AMES: Applications of Complex Numbers
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AC AND COMPLEX NUMBERS
In the previous activity, we observed that alternating current behaves quite differently from
direct current. The first observation was with the motion of the electrons. Instead of flowing
around the circuit, they stayed relatively close to where they began, moved in one direction,
and then moved back in the other direction.
Another observation was that the voltage had a similar periodicity to it. It would measure
10V, 0V, -10V, 0V; 10V, 0V, -10V, 0V; and repeat this pattern indefinitely. However, in the
DC voltage simulation, it would remain a constant 9V.
Lastly, the way operations (such as addition) were performed on DC voltage and AC
voltage differed. For DC if we began with 9V and added 9V, it resulted in 18V. If we added
9 more volts, we would get 27V. However, for AC voltage we started with a 10V source,
added another 10V source and seemed to get a random resulting power source. The reason
this happens is the periodicity of alternating current. Mathematically, one of the best ways
to describe this phenomena is with complex numbers. Let’s look at what is happening
graphically on the complex plane. When working with circuits, it is customary to use the
letter j in place of i because I is typically used to refer to current. Thus the imaginary axis
below is labeled with j.
j
10

As we have previously seen, complex numbers can be represented with their absolute value,
or magnitude, by plotting the number on the complex plane and connecting it to the origin.
In the simulation with one AC voltage source, the magnitude of the voltage was 10V. Thus
we can represent the voltage as the complex number 10 + 0 j V. This can be graphically
represented as shown above.
Moving forward in time with respect to the AC simulation, after displaying 10V there is a
moment where the voltmeter showed 0 V. This can be represented as the complex number
0 + 10 j V. This is graphically represented as shown below.
j
10j

10
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AMES: Applications of Complex Numbers
Exercise 1 Solution:
−10 + 0 j V
Magnitude is 10
j
10j
-10
10
R
Exercise 2 Solution:
0 − 10 j V
Magnitude is 10. This
representation of 0V points
downward because it falls
at a different part of the
cycle 10, 0, -10, 0. Having
one 0 point upward and
the other point downward
gives a mathematical and
graphical representation
for two different 0s.
j
10j
-10
10
R
Exercise 3 Solution:
10 + 0 j V
Magnitude is 10.
j
10j
10
-10j
Exercise 1
Assume the simulation moves forward in time, and the voltmeter displays -10V. Express
this value as a complex number, find it’s magnitude, and represent this moment on a
complex plane.
Exercise 2
Assume the simulation moves forward in time, and the voltmeter now displays 0V. Express
this value as a complex number, find it’s magnitude, and represent this moment on a
complex plane.
Exercise 3
Assume the simulation moves forward in time, and the voltmeter now displays 10V.
Express this value as a complex number, find it’s magnitude, and represent this moment on
a complex plane.
Based on the two images on the previous page and exercises 1-3, we can see that in the
mathematical representation of AC voltage, the magnitude of the complex number never
actually changes. This can be represented with as circle, as shown below.
j
10j

10
-10
-10j
As previously shown, when the voltmeter reads 0, the voltage can be representing using
only the imaginary component of a complex number. But why would we care to represent
the voltage with a complex number when the voltmeter reads 0 volts? The exercises that
follow will guide us to making a conclusion about the importance of imaginary and
complex numbers.
Before completing the exercises, we will need to be able to understand circuit diagrams and
Kirchhoff ’s Voltage Law.
-10j
-10
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Circuit Diagrams
A circuit diagram is a set of images and lines that represent a circuit like the ones created in
the simulation. The two diagrams below show some an AC circuit diagram and a DC circuit
diagram, each with a power source and a load.
+
-
R
+
The table that follows further describes the symbols used in the examples that follow.
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Symbol
+
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Description
AC voltage source
Battery Cell
Since battery cells produce are DC voltage, we will use this to mean a DC voltage
source.
Resistor (Or a load)
This symbol is usually used to represent
a resistor. The purpose of a resistor is to
reduce the flow of current. Since we will
not be concerned with what exactly is in
the circuit, we will use this more generally
to mean a load. A load can be a light bulb,
a resistor, or any other circuit component
that slows the current.
- +
Kirchhoff’s Voltage Law
Kirchoff ’s Voltage Law can be stated as:
The sum of voltages around a closed loop is zero.
This means that the voltage drop across every component in a loop within a circuit along
with the voltage sources adds to be 0. The image below provides a simple visual of 1 loop.
Load
Exercise 4 solution:
KVL allows us to add the
voltages directly since they
are in the same loop.
Let z1 = 0 + 10 j and
z2 = 0 − 10 j .
10V
+
1 loop
PowerSource + Load = 0
10V + x = 0
x = −10V
In the loop above, the voltage drop across the load is not labeled. However, because of
Kirchhoff ’s Voltage Law we know that it will be 10V since there is a 10V power source.
Although this direct application of KVL is useful for determining the change in voltage for
individual circuit components, we will use a property that KVL implies. That is, KVL shows
that voltage can be added directly under the rules allowed in the set of complex numbers as
long as the voltage sources are in the same loop. For example, the complex voltages in the
circuit below can be added together to give us the resultant voltage source. The resultant
voltage source is the sum of two or more sources.
Then z1 = 10 ,
+
z2 = 10 ,
+
z1 + z2 = 0 + 0 j ,
A Load
-
z1 + z2 = 0 .
5 − 30 j V
The resultant voltage is
( 5 − 30 j ) + ( −3 + 4 j ) = 2 − 26 j V.
−3 + 4 j V
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Max Voltage Output
In the following exercises, we look at moments at which all of the voltage is considered to
be in the imaginary component, thus the voltmeter would read 0.
Exercise 4
Two AC power sources are placed in a circuit as pictured with their voltages. Find the
magnitude of each voltage source and the resulting voltage source.
Exercise 5 solution:
+
KVL allows us to add the
voltages directly since they
are in the same loop.
+
A Load
-
Let z1 = 0 + 10 j and
z2 = 0 + 10 j .
Then z1 = 10 ,
z2 = 10 ,
z1 + z2 = 0 + 20 j ,
z1 + z2 = 20 .
0 + 10 j V
0 − 10 j V
Exercise 5
Two AC power sources are placed in a circuit as pictured with their voltages. Find the
magnitude of each voltage source and the resulting voltage source.
+
A Load
Exercise 6 solutions:
1. 0V
+
-
2. 0V
0 + 10 j V
0 + 10 j V
3. 0V
4. 20V
5. The magnitude of the
complex number that
represents the voltage.
6. During moments where
the voltmeter reads 0, the
imaginary component
maintains the value of
maximum output for the
power source. The complex
number allows us to know
where the AC voltage
source is in its repeating
pattern, or period, (e.g. the
first time the voltage source
outputs 0 in the pattern 10,
0, -10, 0; verses the second
time it outputs 0).
Exercise 6
Answer the following questions to compare the voltage sources in exercises 4 and 5.
1. What value would the voltmeter read for each voltage source in exercise 4?
2. What value would the voltmeter read for each voltage source in exercise 5?
3. Recall that after exercises 1, 2, and 3 we concluded that the magnitude of the voltage
source never changes. As time passes, what is the maximum value the voltmeter would
read for the entire circuit in exercise 4?
4. What is the maximum value the voltmeter would read for the entire circuit in exercise
5?
5. In general, what is the maximum output equivalent to?
6. Make a conclusion about the importance of the complex and imaginary numbers
concerning the magnitude and periodic nature of AC voltage (e.g. in exercises 1, 2, 3
the pattern 10V, 0V, -10V, 0V | 10V, 0V, -10V, 0V| 10V, 0V, -10V, 0V etc.).
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AMES: Applications of Complex Numbers
Exercise 7 solution:
3+ 4 j =
4−3j =
( 4)
2
( 3) + ( 4 ) = 5
2
2
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Exercise 7
Find the maximum voltage output in the circuit pictured below.
+ ( −3) = 5 V
2
Exercise 8 solution:
( 7 − 9 j ) + ( −3 + 6 j ) = 4 − 3 j
+
A Load
Exercise 8
Find the maximum voltage output in the circuit pictured below.
+
4−3j =5 V
A Load
+
-
Exercise 9 solution:
(8 + 3 j ) + ( 2 + 1 j ) = 10 + 4 j
7 −9j V
−3 + 6 j V
Exercise 9
Find the maximum voltage output in the circuit pictured below.
+
10 + 4 j ≈ 10.77 V
+
A Load
-
Exercise 10 solution:
−5 + 12 j = 13 V
3+ 4j V
-
8+3j V
2 +1j V
Exercise 10
Find the maximum voltage output in the circuit pictured below.
+
A Load
+
+
-
−1 + 8 j V
6 + 15 j V
−10 − 11j V
10
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Thus far in this lesson, we have mentioned the periodic nature of AC voltage by recognizing
a pattern (previously it was 10, 0, -10, 0 | 10, 0, -10, 0) with respect to voltage output. Let’s
look at this pattern in a bit more detail to understand how AC voltage can be graphically
represented with respect to time.
Let’s begin by assuming we have an alternating voltage source that at time t=0 seconds
outputs 0 volts, at 2 seconds outputs 5 volts, at 4 seconds outputs 0 volts, at 6 seconds
outputs -5 volts, and then begins to repeat this pattern. This is represented with the table
and graph below.
Exercise 11 Solutions:
1. Sinusoidal, Sine
Function, Cosine Function
2. 8 seconds
3. 5 Volts
4. They are all the same.
Output (Volts)
0
5
0
-5
0
5
0
-5
0
5
Output (V)
Time, t (seconds)
0
2
4
6
8
10
12
14
16
t (sec)
2
4
6
8
10 12 14 16
-5
Exercise 11
Answer the following questions about the table and graph above.
1. What type of function can model this?
2. What is the period of this function (how long does it take to begin repeating)?
3. What is the amplitude of this function (distance from the horizontal to the crest of a
wave, or half the total height)?
4. Describe the relationship between the amplitude, maximum voltage output, and
magnitude of the voltage.
Now let’s consider if we have two voltage sources that output 0 volts at two different times,
one 4 seconds later than the other. Let’s assume both sources have an amplitude of 5 volts.
They can be plotted as follows.
Output (V)
5
2
4
6
8
10
12
14
16
t (sec)
17 18
-5
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AMES: Applications of Complex Numbers
Exercise 12 Solutions:
1. 4 seconds
2. 4 seconds
3. -5 Volts
4. 5 Volts
5. 0 Volts
6. 0 Volts
7. 5 Volts
8. 5 Volts
9. 0 Volts
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Exercise 12
Let source 1 be represented by the red (solid) line and let source 2 be represented by the
blue (dashed) line. Answer the following questions concerning the previous graph.
1. How much later than source 1 does source 2 output 0 volts?
2. How much later than source 1 does source 2 output 5 volts?
3. What is the output voltage of source 1 at six seconds?
4. What is the output voltage of source 2 at six seconds?
5. What is the sum of their output at 6 seconds?
6. Choose a different time between 4 and 16, what is the sum of the voltages?
7. If source 1 was in a loop by itself, what would the maximum voltage output be?
8. If source 2 was in a loop by itself, what would the maximum voltage output be?
9. If these two sources were in the same loop, what would the maximum voltage output
be?
The delay of one source with respect to another can be referred to as phase shift. In exercise
12, the phase shift of source 2 with respect to source 1 causes the total output to always be 0.
Exercise 13 Solution:
All of the graphs. Each
graph represents the two
sources at a different time.
Let’s look at how phase shift is represented with respect to complex numbers.
Exercise 13
Consider the same two voltage sources at any given time. Which of the following graphs
represents them.
j
j
5j
5j

5
-5
5
-5
-5j

-5j
5j
j
5j
5
-5
-5j

j
5
-5

-5j
In the complex plane representation, we can see that the two voltage sources always have an
angle of 180 degrees, or π. Thus, we can establish the phase shift with respect to a source as
the angle between that source and another voltage source.
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Exercise 14 Solutions:
Part 1
Now we have a way to refer to a power source using its maximum voltage output and phase
shift.
Exercise 14
j
3j
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On the complex plane, plot the given voltage source combinations.

3
-3
2. Source 1 has a voltage of 2 2 + 2 2 j . Source 2 has a max voltage output of 4 Volts and
is phase shifted 90 degrees with respect to source 1.
-3j
Part 2
j
4j

4
-4
1. Source 1 has a maximum voltage output of 3 volts. Source 2 has a maximum output of
3 volts and is phase shifted 45 degrees with respect to source 1.
-4j
3. Source 1 has a voltage of −5 + 12 j . Source 2 has a max voltage output of 10 volts and
is phase shifted 135 degrees with respect to source 3. Source 3 has a max voltage of 11
volts and is phase shifted 60 degrees with respect to source 1.
To simplify this concept a bit more, we can set the positive real axis as the basis for the
phase shift and refer to all of the voltages with an amplitude and a phase shift. For example,
the voltage sources below are each labeled with their amplitude and phase shift with
respect to the positive real axis. Note that there are multiple ways one voltage source can
be represented.
Part 3
j
2 45 V or 2 − 315 V
j
5 210 V or 5 − 150 V
j


2

10 11 13
5
When given voltages in the notation introduced above, assume that each phase shift is with
respect to the positive real axis.
Exercise 15 Solution:
1. r is the amplitude and θ
is the phase shift.
(
)
(
2. 2, 45 and 5, 210
)
3.
Exercise 15
Answer the following:
1. How does this new notation (with amplitude and phase shift) relate to the polar form
of complex numbers?
2. From the two images above, represent the sources 2 45 and 5 210 as an ordered
2 cos 45 + 2i sin 45 = 2 + i 2 pair of polar coordinates.
( )
( )
( )
( )


2 cos 45 + 2i sin 45 = 2 + i 2
(
)
(
)
5 cos 210 + 5i sin 210 =
(
)
(
)
cos 210 + 5i sin 210 =
3. Find the corresponding complex coordinates for 2 45 and 5 210 .
5 3 5
− i
2
2
5 3 5
− i
2
2
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Exercise 16 solution on the
following page.
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Exercise 16
Assume that each pair of voltage sources in the following are connected as shown in the
image below. Find the amplitude and phase shift of the sum of both sources. Plot both
sources and their sum on the complex plane.
+
A Load
Exercise 17 Solutions:
Part 1
R=
( 22 + 6 j )
(4 + 2 j)
( 22 + 6 j ) ( 4 − 2 j )
R=
(4 + 2 j) (4 − 2 j)
R = 5− j
Part 2
R=
12315
2 − 135
(
12
R = 315 − −135
2
(
))
R = 6 450 = 690
Part 3
(
)(
)
(
)
V = 890 460
V = 8 ⋅ 4 90 + 60
(
V = 32 150
)
Part 4
V = ( 5 − 8 j ) ( −9 + 2 j )
+
-
Source A
Source B
1. Source A: 830 ; Source B: 890
2. Source A: 460 ; Source B: 6180
3. Source A: 12315 ; Source B: 2 − 135
As the previous exercise shows, it can be rather tedious to add two sources given their polar
form. The benefit of using this form comes with multiplication.
Ohm’s Law
Ohm’s Law relates voltage, current, and resistance through the equation V = IR where V is
voltage, I is current, and R is resistance (impedance for AC cirucits). When we are working
with AC power sources the voltage can be a complex number. Since voltage is a product of
current and impedance (AC resistance), then those can be complex numbers as well.
Exercise 17
Given the following, solve for the unknown component in Ohm’s formula. Provide the
solution in the same form as the givens (i.e. if the givens are polar, then the solution should
be polar).
1. V = 22 + 6 j and I = 4 + 2 j . Find R.
2. V =12315 and I = 2 − 135 . Find R.
3. I = 890 and R = 4 60 . Find V.
4. I = 5 − 8 j and R = −9 + 2 j . Find V.
V = −29 + 82 j
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Teacher Page:
Exercise 16 Part 1 Solution:
( )
( )
8 cos ( 90 ) + 8 j sin ( 90 ) = 0 + 8 j
j
8 cos 30 + 8 j sin 30 = 4 3 + 4 j

(4

8 3  13.9
8

)
3 + 4 j + ( 0 + 8 j ) = 4 3 + 12 j
( 4 3 ) + (12 )
2
4 3 + 12 j =
2
=8 3
For Phase Shift:
Since both the real component and the imaginary component of the sum are
positive, the resulting angle would be in Quadrant I.
 12 

phase shift = tan −1 
 = 60
4 3
Exercise 16 Part 2 Solution:
( )
( )
6 cos (180 ) + 6 j sin (180 ) = −6 + 0 j
j
4 cos 60 + 4 j sin 60 = 2 + 2 3 j

4

(2 + 2 3 j ) + ( −6 + 0 j ) = −4 + 2
2 7  5.29

6
3j
−4 + 2 3 j = 2 7
For Phase Shift:
For the sum, since the real component is negative and the imaginary
component is positive, the resulting angle is in Quadrant II. Since arctangent
is defined from -90 degrees to positive 90 degrees, we must add 180 degrees to
get the solution in quadrant II.
2 3
phase shift = tan −1 
+ 180 = −40 + 180 = 140
 −4 


Exercise 16 Part 3 Solution:
j
(
)
( )
2 cos ( −135 ) + 2 j sin ( −135 ) = −
12 cos 315 + 12 j sin 315 = 6 2 − 6 2 j

2
12

2 37  12.17
(6

) (
)
2 − 2j
2 −6 2 + − 2 − 2j = 5 2 −7 2j
5 2 − 7 2 j = 2 37
For Phase Shift:
For the sum, since the real component is positive and the imaginary
component is negative, the resulting angle is in Quadrant IV.
 −7 2 
phase shift = tan −1 
= −54.46
 5 2 


15
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