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Algebra 2 View details

1. The Basics of Complex Numbers

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Chapter 1

1.

The Basics of Complex Numbers

Complex numbers are a type of number that have both a real part and an imaginary part. They are particularly useful in various fields such as engineering, physics, and computer science. The real part is a regular number, while the imaginary part is a number multiplied by the imaginary unit, often denoted as i. This imaginary unit is unique because it is defined as the square root of - 1, a concept not possible with just real numbers. Understanding complex numbers allows for solving equations that don't have real number solutions and modeling phenomena in natural sciences. For example, electrical engineers use them to analyze circuits, and they are also used in quantum mechanics.

Student Learning Objectives:
Understand imaginary and complex numbers Determine the value of i^n Operate with complex numbers Rationalize a denominator using complex conjugates Graph complex numbers in the complex plane

	18 Theory slides
	14 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

The Basics of Complex Numbers

Slide of 18

When solving quadratic equations, it is common to come across equations without a solution — also said to have no solution. For example, x^2=-4 has no solution because no real number exists such that squaring it results in a negative number. Wait, what about numbers that are not real? Are there numbers other than real ones? This lesson will teach and explore such non-real numbers.

Catch-Up and Review

Here are a few recommended readings to do before beginning this lesson.

Quadratic Equations
Solution to an Equation
Number of Solutions of a Quadratic Equation

Try these practice exercises to warm up for this lesson.

a

What are the solutions to the equation 3x^2-27=0?

b

How many real solutions does the equation x^2=9 have?

c

How many real solutions does the equation x^2=-25 have?

Challenge

Non-real Solutions to Quadratic Equations

Group the following equations depending on whether they have real solutions.

Quadratic Equations: x^2+x-2=0; x^2 = 15; x^2-4x+16=0; x^2=-16; 2x^2-7=1; x^2-2x+5=0

Although the equations grouped on the right-hand side do not have real solutions, using the Quadratic Formula, their solutions can be written in terms of square roots of negative numbers. With this in mind, pair each of the following equations with its solutions. Be aware that the numeric expressions on the right-hand side column are not real, but they solve the equations.

Discussion

Expanding the Real Numbers Set

When solving equations, it is possible to encounter equations that do not have a solution in a certain number system. This does not mean that these equations will remain unsolved forever, but that a solution cannot be reached with numbers in the known number system. x + 2 = 1 ⇓ No solution in natural numbers These situations have led mathematicians to define more general number systems where such equations can be solved. The table shows some examples.

Equation	Unsolvable in	Solvable in
x+2=1	Natural numbers N	Integer numbers Z
2x-3=0	Integer numbers Z	Rational numbers Q
x^2=5	Rational numbers Q	Irrational numbers
x^2=-2	Real numbers R	?

Equations like x^2=-2 do not have real solutions. At this point, the big question is: Does a number system more general than the real number system in which such equations can be solved exist?

Is it possible to expand the real number system so that x^2 = - 2 has solutions?

Mathematicians' minds were occupied with such questions for years. Finally, they figured out that calling i the solution of x^2=-1 allowed them to solve any equation — the solutions could be real numbers or combinations of real numbers and i. This led them to create the imaginary unit. The term imaginary was coined by René Descartes in 1637.

Concept

The Imaginary Unit

The imaginary unit i is the principal square root of -1, that is, i=sqrt(-1). From this definition, it can also be said that i^2=-1.

Imaginary Unit
i=sqrt(-1) or i^2=-1

The imaginary unit i can also be regarded as a solution to the equation x^2+1=0. x^2+1 = 0 ⇒ i^2+1 = 0 The imaginary unit allows to rewrite the square root of any negative number. Once i replaces the square root of - 1, the square root of the remaining positive number can be evaluated as usual.

sqrt(- a) = sqrt(a) * sqrt(- 1) = sqrt(a) * i

The above property is true only when a>0. Here are some examples of how to use the property to simplify radical expressions. sqrt(-5) &= sqrt(5)* i [0.25em] sqrt(-4) &= sqrt(4)* i = 2i [0.25em] sqrt(-20) &= sqrt(20)* i = sqrt(4* 5)* i = 2sqrt(5)* i

The combination of real numbers and any expression of the form bi, with b≠ 0, creates a new set of numbers called imaginary numbers.

Pop Quiz

Rewriting Square Roots of Negative Numbers

Rewrite each given radical expression in terms of the imaginary unit i. Write the answer in the form asqrt(b)* i, where a is a non-zero integer and b is a natural number such that sqrt(b) cannot be simplified further.

Random radical expressions

Extra

How to Input the Answer

If a=1, the left box must be left empty.
If a=-1, the left box must contain only the minus sign (-).
If b=1, the radical symbol must not be included.

Example

Real World Situations With Imaginary Solutions

Tadeo just learned that imaginary numbers are given that name because they do not exist in the real world — they are imaginary. Therefore, if an equation that models a real-life situation has imaginary solutions, then it cannot be solved in the real world. To illustrate this concept, Tadeo's math teacher drew the following polygons and asked three questions.

a

If the sum of the areas of both polygons is equal to 2 square centimeters, what are all the possible values for x?

b

If the sum of the area of the square and twice the are of the triangle is equal to 9 square centimeters, what are all the possible values for x?

c

If the sum of the are of the square and three times the are of the triangle is equal to 10 square centimeters, what are all the possible values for x?

Hint

a

The area of a square equals the side length squared and the area of a right triangle equals half the product of the lengths of the legs. When solving the equation, remember that sqrt(- a) = sqrt(a)* i for a> 0.

b

This time the area of the triangle must be multiplied by 2. Remember, if a> 0 then sqrt(- a) = sqrt(a)* i. Note that the imaginary unit is outside of the radical symbol.

c

Multiply the area of the triangle by 3. To simplify sqrt(8), rewrite 8 as 4* 2 and use the Product Property of Square Roots.

Solution

a

The area of a square is equal to the side length squared. Whereas the area of a right triangle is half the product of the lengths of its legs. With this in mind, the areas of the polygons drawn by Tadeo's teacher can be written algebraically.

	Square	Triangle
Area Formula	A_(□) = s^2	A_(△) = 1/2bh
Substitution	A_(□) =x^2	A_(△) =1/2* 3* 4=6

The sum of these areas has to be set equal to 2. A_(□) + A_(△) = 2 Next, substitute the corresponding expressions and solve the resulting equation for x.

A_(□) + A_(△) = 2

A_(□)= x^2, A_(△)= 6

x^2 + 6 = 2

LHS-6=RHS-6

x^2 = -4

sqrt(LHS)=sqrt(RHS)

sqrt(x^2) = sqrt(-4)

sqrt(a^2)=± a

x = ±sqrt(-4)

Note that the radicand of the final expression is negative, implying that sqrt(- 4) is an imaginary number. Recall how the square root of a negative number is rewritten. sqrt(- a) = sqrt(a)* i, a> 0 By applying this property, the right-hand side can now be written as imaginary numbers.

x = ±sqrt(-4)

sqrt(- a)= isqrt(a)

x = ± sqrt(4)i

Calculate root

x = ±2i

For the given situation, there are two imaginary solutions — namely, 2i and -2i.

b

As shown in Part A, the areas of the given polygons are given by the following expressions.

	Square	Triangle
Area	A_(□) = x^2	A_(△) =6

This time, however, the area of the square has to be added to twice the area of the triangle, and their sum has to be equal to 9 square centimeters. A_(□) + 2A_(△) = 9 Now, substitute the corresponding expressions and solve the equation for x.

A_(□) + 2A_(△) = 29

A_(□)= x^2, A_(△)= 6

x^2 + 2( 6) = 9

Multiply

x^2 + 12 = 9

LHS-12=RHS-12

x^2 = -3

sqrt(LHS)=sqrt(RHS)

sqrt(x^2) = sqrt(-3)

sqrt(a^2)=± a

x = ±sqrt(-3)

Since the radicand is negative, it can be rewritten in terms of the imaginary unit.

x = ±sqrt(-3)

sqrt(- a)= isqrt(a)

x = ± sqrt(3)i

Once more, there are two imaginary solutions to the given situation — namely, sqrt(3)i and -sqrt(3)i.

c

For this third situation, the area of the square must be added to three times the area of the triangle, and their sum must be equal to 10 square centimeters.

A_(□) + 3A_(△) = 10

Substitute the expressions for the areas that were shown in Part A and solve the equation for x.

A_(□) + 3A_(△) = 10

A_(□)= x^2, A_(△)= 6

x^2 + 3( 6) = 10

Multiply

x^2 + 18 = 10

LHS-18=RHS-18

x^2 = -8

sqrt(LHS)=sqrt(RHS)

sqrt(x^2) = sqrt(-8)

sqrt(a^2)=± a

x = ±sqrt(-8)

Again, the radicand is negative but can be rewritten using the imaginary unit.

x = ±sqrt(-8)

sqrt(- a)= isqrt(a)

x = ± sqrt(8)i

SplitIntoFactors

Split into factors

x = ± sqrt(4* 2)i

sqrt(a* b)=sqrt(a)*sqrt(b)

x = ± sqrt(4)*sqrt(2)i

Calculate root

x = ± 2sqrt(2)i

For the third time, the two solutions found are imaginary numbers — namely, 2sqrt(2)i and -2sqrt(2)i.

Note that all the solutions to the situations presented by Tadeo's teacher are imaginary. Therefore, even though the equations representing the situations have solutions, those solutions do not make sense in the real world — Tadeo cannot draw a square using the dimensions found.

Discussion

Powers of i

Continuing with Tadeo's journey into this new universe of imaginary numbers, he wonders if it is possible to use them in a similar way as real numbers. Too excited to wait until the next class, he writes the definition of the imaginary unit. i=sqrt(-1) or i^2=-1 Tadeo notices that the mere definition gives him two different powers of i — namely, i^1 and i^2. This motivates him to compute other powers of i. Since he knows that any non-zero number raised to the zero power equals 1, he decides to check whether this is true for the imaginary unit too. ccl i^2 =-1 & &Definition ofi ⇕ & & (i^2)^0 = (-1)^0 & &Raise both sides to0 ⇕ & & [0.25em] i^0 = 1 & &Zero Exponent Property It seems like the Zero Exponent Property is also true for the imaginary unit. Next, Tadeo continues with i^3. By using the Product of Powers Property, he rewrites i^3 as the product of i^2 and i. lcl i^3 = i^2* i & &Product of Powers Property ⇕ & & i^3 = -1* i & &Definition ofi ⇕ & & [0.25em] i^3 = - i & &Simplify Tadeo then writes down the powers in a more organized way so can better analyze them.

Tadeo notices that the results alternate between a real number and an imaginary number. He wonders if this might be a pattern. To verify his suspicions, he goes on to find the next four powers of i. Once again, he will apply the Product of Powers Property. i^4 &= i^2* i^2 = 1* 1 = 1 [0.25em] i^5 &= i^4* i^1 = 1* i = i [0.25em] i^6 &= i^4* i^2 = 1(-1) = -1 [0.25em] i^7 &= i^4* i^3 = 1(- i) = - i Tadeo's intuition was correct! The powers of i alternate between a real and an imaginary number. When the exponent is even, the result is a real number, while odd exponents produce imaginary results. And there is more! After comparing the first group of powers i^0 through i^3 with the second group of powers i^4 through i^7, Tadeo is on to discovering something spectacular.

The results of the second group are the same as the first. This amazed Tadeo so much that he emailed his teacher right away. Excited by Tadeo's discovery, the teacher responded that this pattern repeats over and over in cycles of 4 and allows finding any power of i. Shocking, right?

Discussion

Looking for a Pattern

With the purpose of mastering the calculus of powers of i, Tadeo asked his teacher to give him a homework problem. As such, the teacher asked him to find i^(34).

Initially, Tadeo plans to find i^8, i^9, i^(10), all the way to i^(34), but he realizes that would take forever! For that reason, he decides to analyze the values of the first 8 powers of i — values he already knows. Because the cycle repeats every four iterations, he thinks there is a relation between the exponents to the number 4.

0 and 4 are both multiples of 4; 1 and 5 equal to a multiple of 4 plus 1; 2 and 6 are equal to a multiple of 4 plus 2; 3 and 7 are equal to a multiple of 4 plus 3.

Tadeo concludes that because of the recognized pattern, it is convenient to divide the given exponent from the homework problem by 4, and depending on the remainder, he will find the result.

Finding i^n
If the remainder of n4 is	Then, i^n is equal to
0	i^0 = 1
1	i^1 = i
2	i^2 = -1
3	i^3 = -i

Therefore, to find the value of i^(34), his first move is to divide 34 by 4. 34/4 = 8.5 The result is not an integer. This implies that 34 is not a multiple of 4. However, the integer part of the result, which is 8, gives him a clue as to how 34 can be rewritten. 34/4 = 4* 8_(32) + 2 From this, the remainder is 2. Consequently, the value of i^(34) is the same as the value of i^2. i^(34)=i^2=-1 Right after finding the value of i^(34), the teacher asked Tadeo how his homework is going. Tadeo told him how he solved the exercise and the teacher was so happy about it. His teacher also told him that he could solve the problem using the Product of Powers Property. Oh, teachers and their math tricks!

i^(34)

▼

Simplify

Rewrite 34 as 32+2

i^(32+2)

a^(m+n)=a^m*a^n

i^(32)* i^2

Rewrite 32 as 4* 8

i^(4* 8)* i^2

a^(m* n)=(a^m)^n

(i^4)^8* i^2

i^4= 1, i^2= -1

( 1)^8( -1)

1^a=1

1( -1)

Identity Property of Multiplication

-1

So far, Tadeo has discovered how to calculate different powers of i, which encourages him to continue exploring how to operate with imaginary numbers.

Pop Quiz

Calculating Powers of i

Compute the required power of i.

Random powers of i

Discussion

Imaginary and Complex Numbers

The imaginary unit i, which is equal to sqrt(-1), not only allows square roots to be used in calculations of negative numbers, it also allows for the construction of the set of imaginary numbers.

Concept

Imaginary Numbers

The set of imaginary numbers, represented by the symbol I, is formed by all the numbers that can be written as a+bi, where a is any real number, b is a non-zero real number, and i is the imaginary unit.

All sets of numbers known so far can be organized as follows.

Diagram with Real Numbers and Imaginary Numbers at the top. From the Real Numbers box, two boxes are derived, the Rational and Irrational Numbers. Below the Rational Numbers box, the Integer box is drawn, below it the Whole Numbers box is drawn, and below it, the Natural Numbers box is drawn.

Additionally, imaginary numbers and real numbers can be grouped into a single set called the set of complex numbers.

Concept

Complex Numbers

The set of complex numbers, represented by the symbol C, is formed by all numbers that can be written in the form z=a+bi, where a and b are real numbers and i is the imaginary unit. Here, a is called the real part and b is called the imaginary part of the complex number.

If b≠ 0, the number is an imaginary number. Conversely, if b=0, the number is real. Additionally, if a=0 and b≠ 0, the number is a pure imaginary number. Both real and imaginary numbers are subsets of the complex number set.

A big set divided into two parts. The left-hand part is the Real Numbers set; the right-hand part is the Imaginary Numbers set; inside the Imaginary Numbers set, there is a small set labeled as the Pure Imaginary Numbers set.

Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.

a+bi = c+di ⇔ a=c and b=d

Discussion

Combining Complex Numbers

Now that Tadeo figured out the pattern for the powers of i, he feels confident in learning the other mathematical operations for complex numbers. He heads to the library, asks for a math textbook, explores the text and charts for a few minutes, and focuses on the following.

Method

Adding and Subtracting Complex Numbers

Two complex numbers a+bi and c+di can be added or subtracted by using the commutative and associative properties of real numbers. To add or subtract two complex numbers, combine their real parts and their imaginary parts separately.

( a+ bi)+( c+ di) = ( a+ c) + ( b+ d)i
( a+ bi)-( c+ di) = ( a- c) + ( b- d)i

Consider for example the complex numbers z_1=5+2i and z_2=3-3i. To add z_1 and z_2, the above formula can be used or, equivalently, the next three steps can be followed. In a similar way the numbers can be subtracted.

1

Separate Real Parts From Imaginary Parts

Applying the commutative and associative properties of real numbers, group the real parts on the left and the imaginary parts on the right.

z_1+z_2

z_1= 5+2i, z_2= 3-3i

5+2i + 3-3i

CommutativePropAdd

Commutative Property of Addition

5+3 + 2i-3i

AssociativePropAdd

Associative Property of Addition

(5+3) + (2i-3i)

2

Combine Real Parts

Next, combine the real parts.

(5+3) + (2i-3i)

Add terms

8 + (2i-3i)

3

Combine Imaginary Parts

Finally, combine the imaginary parts. It could be convenient to factor out i first, and then combine the remaining numbers.

8 + (2i-3i)

Factor out i

8 + (2-3)i

Subtract term

8 + (-1)i

a+(- b)=a-b

8-i

Consequently, z_1+z_2 = 8-i.

Note that the complex numbers are closed under addition and subtraction — that is, the sum or difference of two complex numbers is always a complex number. If z_1+z_2=0=0+0i, then z_2 is said to be the opposite, or additive inverse, of z_1.

Example

Impedance for a Series Circuit

Excited to continue learning about complex numbers, Tadeo ran to his brother's room and asked if he knew of any real-life applications. His brother, an electrical engineer, reached for his favorite book with a diagram of a series circuit. In the case of resistors, the number next to each component indicates its resistance. In the case of capacitors and inductors, it indicates its reactance.

Tadeo's brother went on telling him that the impedance, or opposition to the current flow, of the circuit shown is equal to the sum of the impedances of each component.

a

What is the impedance of the series circuit?

b

In addition to the first diagram, Tadeo's brother drew another series circuit, but this time one that has two resistors.

Again, he asked Tadeo to find the impedance.

Hint

a

Add the numbers in the table. Remember to combine the real parts and the imaginary parts separately.

b

The impedance of a resistor equals its resistance. The impedance of a capacitor equals its reactance multiplied by - i. The impedance of an inductor equals its reactance multiplied by i.

Solution

a

According to Tadeo's brother, the impedance of the series circuit equals the sum of the impedance of the three components of the circuit — a resistor, a capacitor, and an inductor.

Component	Impedance
Resistor	8 Ω
Capacitor	-6i Ω
Inductor	10i Ω

Here, the symbol Ω represents ohms, which is a unit for measuring the electrical resistance between two points. To find the impedance of the circuit, the three impedances need to be added.

Impedance = 8 + ( -6i) + 10i

▼

Simplify right-hand side

a+(- b)=a-b

Impedance = 8 - 6i + 10i

Factor out i

Impedance = 8 + (-6 + 10)i

Add terms

Impedance = 8 + 4i

Consequently, the impedance of the series circuit is 8+4i ohms.

b

This time, the impedance of each component is not specified in the diagram. However, it can be derived from the numbers written next to each component.

The impedance of a resistor equals its resistance, the impedance of a capacitor equals its reactance multiplied by - i, and the impedance of an inductor equals its reactance multiplied by i. All of these quantities are measured in ohms.

Component	Resistance or Reactance	Impedance
Resistor 1	6 Ω	6 Ω
Capacitor	9 Ω	-9i Ω
Inductor	4 Ω	4i Ω
Resistor 2	3 Ω	3 Ω

Finally, these four impedances will be added to find the impedance of the series circuit.

Impedance = 6 + ( -9i) + 4i + 3

▼

Simplify right-hand side

a+(- b)=a-b

Impedance = 6 - 9i + 4i + 3

CommutativePropAdd

Commutative Property of Addition

Impedance = 6 + 3 + 4i - 9i

AssociativePropAdd

Associative Property of Addition

Impedance = (6 + 3) + (4i - 9i)

Factor out i

Impedance = (6 + 3) + (4 - 9)i

Add and subtract terms

Impedance = 9 - 5i

Therefore, the impedance of the series circuit is 9-5i ohms.

Discussion

Multiplication of Complex Numbers

Tadeo is feeling great about complex numbers so far but wants to learn even more. He suspects that complex numbers can also be multiplied, which causes him to wonder if there is a method to do that. Thirsty for knowledge, he looked in his e-book and found the answer.

Method

Multiplying Complex Numbers

Two complex numbers a+bi and c+di can be multiplied by using the Distributive Property of real numbers. When two complex numbers are multiplied, the resulting expression could contain i^2. Using the definition of the imaginary unit, it is replaced with - 1 so that the resulting number is in standard form.

( a + bi)( c + di) = ( a c - b d) + ( a d + b c)i

Consider for example the complex numbers z_1=3+2i and z_2=4+2i. To multiply z_1 by z_2, the above formula can be used or, equivalently, the next four steps can be followed.

1

Distribute Terms

First, distribute the terms of one number to the terms of the other. This can be done either by using the Distributive Property or the FOIL method.

z_1* z_2

z_1= 3+2i, z_2= 4+2i

( 3+2i)( 4+2i)

Distribute (3+2i)

(3+2i)4 + (3+2i)2i

Distribute 4 & 2i

(3)4 + (2i)4 + (3)2i + (2i)2i

CommutativePropMult

Commutative Property of Multiplication

4(3) + 4(2i) + 3(2i) + (2* 2)(i* i)

2

Perform the Multiplications

Perform any multiplication. Use the Product of Powers Property to rewrite i* i as one single power.

4(3) + 4(2i) + 3(2i) + (2* 2)(i* i)

Multiply

12 + 8i + 6i + 4(i* i)

ProdToPowTwoFac

a* a=a^2

12 + 8i + 6i + 4i^2

3

Use the Definition of the Imaginary Unit

By definition, i^2=-1. Thus, substitute -1 for i^2 into the last expression.

12 + 8i + 6i + 4i^2

i^2=- 1

12 + 8i + 6i + 4(-1)

Multiplication Property of -1

12 + 8i + 6i - 4

4

Combine Like Terms

Finally, combine like terms. Here, it is useful to separate the real parts from the imaginary parts and to factor out i to write the final answer.

12 + 8i + 6i - 4

CommutativePropAdd

Commutative Property of Addition

12 - 4 + 8i + 6i

AssociativePropAdd

Associative Property of Addition

(12 - 4) + (8i + 6i)

Factor out i

(12 - 4) + (8 + 6)i

Add and subtract terms

8 + 14i

Consequently, z_1* z_2 = 8+14i.

Note that the complex numbers are closed under multiplication — that is, the product of two complex numbers is always a complex number.

Example

Voltage of Series Circuit

Tadeo's brother, excited about Tadeo's interest in complex numbers, wants to teach him Ohm's law — a formula for calculating the voltage of a circuit. According to this law, the voltage of a circuit, in volts, is equal to the electric current multiplied by the impedance, that is, V=C* I. He goes on to draw two circuits.

Circuit 1: Current1=4+3i, Impedance1=8-3i; Circuit 2: Current2=5+2i, Impedance2=7+5i

a

The current of the first circuit is 4+3i amps and the impedance is 8-3i ohms. What is the voltage of the circuit?

b

If 5+6i volts are added to the voltage of the second circuit, what is the resulting voltage?

c

What is the difference between the original voltage of the second circuit and the voltage of the first circuit?

Hint

a

Multiply the current by the impedance. Follow the steps to multiply complex numbers.

b

Start by finding V_2. Then, add 5+6i to obtain the voltage.

c

Subtract V_1 from V_2.

Solution

a

According to Ohm's law, the voltage is found by multiplying the current by the impedance. The current and the impedance of the first circuit are given.

C_1 &= (4+3i) A I_1 &= (8-3i) Ω Now, multiply these two quantities by following the steps to multiply complex numbers. Here, the units of measure will be omitted during the computations.

V_1 = C_1 * I_1

C_1= 4+3i, I_1= 8-3i

V_1 = ( 4+3i)( 8-3i)

▼

Multiply 4+3i by 8-3i

Multiply

V_1 = 4(8) + 4(-3i) + 3i(8) + 3i(-3i)

CommutativePropMult

Commutative Property of Multiplication

V_1 = 4(8) + 4(-3)i + 3(8)i + 3(-3)(i* i)

a^m*a^n=a^(m+n)

V_1 = 4(8) + 4(-3)i + 3(8)i + 3(-3)i^2

i^2=- 1

V_1 = 4(8) + 4(-3)i + 3(8)i + 3(-3)(-1)

Multiply

V_1 = 32 - 12i + 24i + 9

CommutativePropAdd

Commutative Property of Addition

V_1 = 32 + 9 + 24i - 12i

AssociativePropAdd

Associative Property of Addition

V_1 = (32 + 9) + (24i - 12i)

Factor out i

V_1 = (32 + 9) + (24 - 12)i

Add and subtract terms

V_1 = 41 + 12i

Therefore, the voltage of the first circuit is 41+12i volts.

b

First, start by finding V_2 using a similar procedure as the one made in the previous part.

C_2 &= (5+2i) A I_2 &= (7+5i) Ω The voltage of the circuit equals the product of these two quantities.

V_2 = C_2 * I_2

C_1= 5+2i, I_1= 7+5i

V_2 = ( 5+2i)( 7+5i)

▼

Multiply 5+2i by 7+5i

Multiply

V_2 = 5(7) + 5(5i) + 2i(7) + 2i(5i)

CommutativePropMult

Commutative Property of Multiplication

V_2 = 5(7) + 5(5i) + 2(7)i + 2(5)(i* i)

a^m*a^n=a^(m+n)

V_2 = 5(7) + 5(5i) + 2(7)i + 2(5)i^2

i^2=- 1

V_2 = 5(7) + 5(5i) + 2(7)i + 2(5)(-1)

Multiply

V_2 = 35 + 25i + 14i - 10

CommutativePropAdd

Commutative Property of Addition

V_2 = 35 - 10 + 25i + 14i

AssociativePropAdd

Associative Property of Addition

V_2 = (35 - 10) + (25i + 14i)

Factor out i

V_2 = (35 - 10) + (25 + 14)i

Add and subtract terms

V_2 = 25 + 39i

Therefore, the voltage of the second circuit is 25+39i volts. Finally, add 5+6i volts to V_2.

5+6i + V_2

▼

Substitute 25+39i for V_2 and simplify

V_2= 25 + 39i

5+6i + 25 + 39i

CommutativePropAdd

Commutative Property of Addition

5 + 25 + 6i + 39i

Factor out i

5 + 25 + (6 + 39)i

Add terms

30 + 45i

c

In the previous parts, the voltage of the first circuit V_1 and the original voltage of the second circuit V_2 were found.

V_1 &= 41+12i [0.25em] V_2 &= 25+39i To find the difference between these complex numbers, the steps used to subtract complex numbers will be followed.

V_2 - V_1

▼

Substitute values and simplify

V_2= 25+39i, V_1= 41+12i

25+39i - ( 41+12i)

Distribute -1

25 + 39i - 41 - 12i

CommutativePropAdd

Commutative Property of Addition

25 - 41 + 39i - 12i

Factor out i

25 - 41 + (39 - 12)i

Subtract terms

-16 + 27i

The difference between the voltage of the second circuit and the voltage of the first circuit is -16+27i volts.

Discussion

Dividing Complex Numbers

Up to this point, Tadeo learned how to add, subtract, and multiply complex numbers. In this process of learning the operations of complex numbers, two things stood out to him.

When performing any of the operations, the imaginary unit i is treated as a variable.
The final result must be in the form of a + bi.

There is just one more operation to cover. It is time to investigate the division of complex numbers.

Tadeo searched for an answer on the Internet. Most of the results contained the following explanation.

While he was glad to find this explanation, Tadeo could not understand it because he does not know what the complex conjugate of a number is. Being his eager self, he looks up the definition.

Discussion

Complex Conjugate

The complex conjugate of a complex number has the same real part, but the imaginary part is the opposite of its original sign. Therefore, changing the sign of the imaginary part of a complex number creates its complex conjugate. It is denoted by a line drawn above the complex number.

a+bi = a-bi or a-bi = a+bi

For example, the complex conjugate of z = 3 - 5i is z= 3 + 5i. It is worth noting that the product of a complex number and its conjugate is a real number.

z* z

▼

Substitute a+bi for z and simplify

z= a+bi

( a+bi)(a+bi)

a+bi= a-bi

(a+bi)( a-bi)

Multiply

a^2-abi+abi-b^2i^2

Add terms

a^2-b^2i^2

i^2=- 1

a^2-b^2(-1)

- (- a)=a

a^2+b^2

This fact is very useful when simplifying quotients whose denominators are complex numbers.

Now that Tadeo knows about complex conjugates, there is nothing that can stop him from learning how to divide complex numbers.

Method

Rationalizing a Denominator Using Complex Conjugates

When a rational expression has a denominator that is a complex number, instead of performing a division, the fraction is rewritten so that the denominator is a real number. That is, the denominator has to be rationalized. For example, consider the following quotient. 5+2i/3-4i To simplify the quotient, multiply the numerator and the denominator by the complex conjugate of the denominator.

1

Multiply by the Conjugate of the Denominator

To get rid of the complex number in the denominator, use the fact that the product of a complex number and its conjugate is a real number. (a+bi)(a-bi) = a^2+b^2 Therefore, multiply the numerator and the denominator by the complex conjugate of the denominator. In this case, multiply by 3+4i. 5+2i/3-4i = (5+2i)( 3+4i)/(3-4i)( 3+4i) Notice that 3+4i3+4i equals 1. Subsequently, the value of the original fraction does not change by the Identity Property of Multiplication.

2

Simplify the Expression

The obtained expression can now be simplified.

(5+2i)(3+4i)/(3-4i)(3+4i)

▼

Simplify denominator

(a+bi)(a-bi)=a^2+b^2

(5+2i)(3+4i)/3^2+4^2

Calculate power

(5+2i)(3+4i)/9+16

Add terms

(5+2i)(3+4i)/25

▼

Simplify numerator

Multiply

15 + 20i + 6i + 8i^2/25

i^2=- 1

15 + 20i + 6i + 8(-1)/25

Multiplication Property of -1

15 + 20i + 6i - 8/25

Add and subtract terms

7 + 26i/25

Write as a sum of fractions

7/25 + 26/25i

The denominator has now been rationalized and the quotient rewritten as a complex number in standard form.

Rationalizing a denominator using complex conjugates leads to the following formula to simplify quotients of complex numbers.

a+ bi/c+ di = a c + b d/c^2+ d^2 + (b c - a d/c^2+ d^2)i

Example

Conjugates and Divisions

The weekend is here and Tadeo still wants to continue practicing operations with complex numbers. Unfortunately, his brother is not at home to keep giving him cool examples. However, this does not stop Tadeo from picking up a book and looking for exercises.

From the book, he chose three exercises that he found interesting.

a

If z=-8+4i, what is z* z equal to?

b

Write 203+i as a complex number in standard form.

c

Find the real and imaginary parts of 10+5i2+4i.

Hint

a

The conjugate of a complex number is obtained by changing the sign of the imaginary part.

b

Rationalize the denominator multiplying each part of the fraction by the conjugate of the denominator.

c

First, rationalize the denominator. Then, separate the real and the imaginary parts. The imaginary part is the number next to the imaginary unit.

Solution

a

The conjugate of a complex number z, denoted by z, is obtained by changing the sign of the imaginary part of z. The imaginary part is the term containing the imaginary unit.

z = -8 +4i The imaginary part of z is 4. Therefore, the conjugate of z will have -4 as its imaginary part. z = -8 - 4i Now that the conjugate of z is identified, the product of the conjugates can be calculated.

z* z

z= -8+4i, z= -8-4i

( -8+4i)( -8-4i)

Multiply parentheses

-8(-8) - 8(-4i) + 4i(-8) +4i(-4i)

Multiply

64 + 32i - 32i - 16i^2

Subtract terms

64 - 16i^2

i^2=- 1

64 - 16(-1)

MultNegNegOnePar

- a(- b)=a* b

64 + 16

Add terms

80

Consequently, z* z=80.

b

To rewrite the given quotient, the denominator has to be rationalized. In other words, the quotient has to be rewritten so that its denominator is a real number.

20/3+i To rationalize the denominator, each part of the fraction has to be multiplied by the conjugate of the denominator. Denominator: & 3+i Conjugate: & 3-i Now, multiply the numerator and denominator of the given fraction by 3-i. 20/3+i=20( 3-i)/(3+i)( 3-i) Using the fact that (a+bi)(a-bi)=a^2+b^2, the product in the denominator can be rewritten as 3^2+1^2, which gives a real number. Then, the quotient can be simplified and written in standard form.

20(3-i)/(3+i)(3-i)

▼

Simplify denominator

(a+bi)(a-bi)=a^2+b^2

20(3-i)/3^2+1^2

Calculate power

20(3-i)/9+1

Add terms

20(3-i)/10

▼

Simplify

a/b=.a /10./.b /10.

2(3-i)/1

a/1=a

2(3-i)

Distribute 2

6-2i

c

To find the real and imaginary parts, the given quotient has to be simplified first.

10+5i/2+4i As in Part B, the denominator has to be rationalized. To do so, the numerator and the denominator of the fraction should be multiplied by 2-4i, which is the conjugate of 2+4i. 10+5i/2+4i = (10+5i)( 2-4i)/(2+4i)( 2-4i) Next, continue simplifying the quotient.

(10+5i)(2-4i)/(2+4i)(2-4i)

▼

Simplify denominator

(a+bi)(a-bi)=a^2+b^2

(10+5i)(2-4i)/2^2+4^2

Calculate power

(10+5i)(2-4i)/4+16

Add terms

(10+5i)(2-4i)/20

▼

Simplify numerator

Multiply

20-40i+10i-20i^2/20

i^2=- 1

20-40i+10i-20(-1)/20

Multiplication Property of -1

20-40i+10i+20/20

Add terms

40-30i/20

▼

Rewrite

Write as a sum of fractions

40/20-30/20i

Calculate quotient

2-30/20i

a/b=.a /10./.b /10.

2-3/2i

Consequently, the real part of the given complex number is 2 and the imaginary part is - 32. Re(10+5i/2+4i) &= 2 [1em] Im(10+5i/2+4i) &= -3/2

Pop Quiz

Operating With Complex Numbers

Perform the required operation and write the result in standard form. Round each part to two decimal places, if needed.

Random Operations with Complex Numbers

Extra

How to Input the Answer

Write the real part in the first box.
Write the imaginary part in the second box.
To change the sign of the imaginary part, click on the square at the top-right corner.

Closure

Graphing Complex Numbers in the Complex Plane

Just as Tadeo thought he knew all about complex numbers, his teacher told him that unlike real numbers, complex numbers cannot be represented on a number line. However, they can be represented on the complex plane — similar to the coordinate plane but the horizontal axis represents the real part and the vertical axis the imaginary part of a complex number.

Note that the number -3+ 2i is represented by the point ( -3, 2). Complex Number: & a + b i & ↓ ↓ Point on Complex Plane: & ( a , b ) Also, the absolute value of a complex number is its distance to the origin in the complex plane and is given by |a+bi|=sqrt(a^2+b^2). This formula is derived by using the Distance Formula. Additionally, the conjugation of a number z can be perceived geometrically as the reflection of z about the real axis.

Although there are still many things to learn about complex numbers, Tadeo's teacher is quite happy with all the knowledge Tadeo has learned and encourages him to continue his eagerness to discover new things.

The Basics of Complex Numbers

Exercise 2.1

Pair the given powers of i with their simplified forms.

The Product of Powers Property holds true for the imaginary unit. We can use it to find the powers of i. Let's consider the first four powers of i. Recall that any number raised to the power of one equals itself, so i^1= i. By definition, i is equal to sqrt(- 1), so we know that i^2= - 1. l i^1= i i^2= - 1 i^3= i^2* i^1= - 1* i = - i i^4= i^2* i^2= - 1*( - 1)= 1 Note that the fourth power of i is equal to 1. Therefore, we can simplify the given powers of i by rewriting them so that i^4 is a factor. This will make the process in simplifying the given expressions easier! Using this trick together with the Product of Powers Property, let's calculate i^6.

We will compute i^(35) by applying the same procedure.

Let's now compute i^(1872).

Finally, we will simplify i^(1905).

Let's take a final look at all the pairings! i^6 &→ - 1 i^(35) &→ - i i^(1872) &→ 1 i^(1905) &→ i

Electrical circuit components such as resistors, inductors, and capacitors oppose the current flow. For resistors, this opposition is called resistance. For capacitors and inductors, it is called reactance. The total electrical impedance of a series circuit is equal to the sum of the opposition of each component.

Component	Resistance or Reactance (Ω)	Impedance (Ω)
Resistor	R	R
Inductor	L	Li
Capacitor	C	- Ci

With this in mind, consider the following circuits.

What is the impedance of the circuit?

What is the impedance of the series circuit?

The impedance for a circuit is the sum of the impedances for the individual components. Impedance of Circuit: R+Li+(- Ci) In this case, the contribution of each component to the impedance of the circuit can be listed as below.

Component	Resistance or Reactance (Ω)	Impedance (Ω)
Resistor	12	12
Inductor	6	6i
Capacitor	7	- 7i

Now that we know the impedance of each component, we can find the total impedance of the circuit.

The impedance of the circuit is 12-i ohms.

In this case, we will again start by listing the contribution of each component to the impedance of the circuit.

Component	Resistance or Reactance (Ω)	Impedance (Ω)
Resistor I	14	14
Resistor II	4	4
Inductor	9	9i
Capacitor	2	- 2i

We will calculate the impedance of the circuit by adding up the impedances of the individual components.

The impedance of the circuit is 18+7i ohms.

Consider the following complex numbers. z_1 = 9+5i z_2 = 3-2i Calculate each operation and write the result in standard form.

z_1 * z_2

z_2 * z_2

z_1/z_2

We will first substitute z_1=9+5i and z_2=3-2i into the given operation. Then we can multiply the expressions.

We will now simplify the expression by combining the real parts and the imaginary parts separately.

We will start by finding the conjugate of z_2. Recall that the conjugate of a complex number z, denoted by z, is found by changing the sign of the imaginary part of z. z_2=3-2i = 3+( -2)i The imaginary part of z_2 is - 2. Therefore, its conjugate will have 2 as its imaginary part. z_2=3+2i We can now substitute z_2= 3-2i and z_2= 3+2i into the given operation. z_2* z_2=( 3-2i)( 3+2i) Recall the fact that the product of a complex number and its conjugate is a real number. (a+bi)(a-bi) = a^2+b^2 We can use this fact to simplify the given product.

We will start by substituting z_1= 9+5i and z_2= 3-2i into z_1z_2. z_1/z_2=9+5i/3-2i We need to rationalize the denominator. To do so, we will multiply both numerator and denominator of the fraction by the conjugate of the denominator. We already found the conjugate of z_2 in Part B. z_2=3+2i We will now multiply the numerator and denominator of the given fraction by 3+2i. 9+5i/3-2i=(9+5i)( 3+2i)/(3-2i)( 3+2i) In Part B we also found that the product in the denominator is 13. This will help us in simplifying the quotient.

Now we need to write this expression in the standard form of a complex number. We can do this by rewriting it as a sum of fractions. 17+33i/13 ⇒ 17/13 + 33/13i

The voltage V of a circuit is related to the electric current C and impedance I by the formula V=C* I. Use this formula to solve the following problems. Write the answers in standard form.

The current in the circuit is 2+4i amps and the impedance is 1-5i ohms. What is the voltage of the circuit?

The voltage in the circuit is 30+12i volts and the current is 5-i amps. What is the impedance of the circuit?

To find the voltage of the circuit, let's substitute the given current C=2+4i amps and the given impedance I=1-5i ohms into the formula V=C* I. We will then simplify the expression.

The voltage of the circuit is 22-6i volts.

To find the impedance of the circuit, let's substitute the given voltage V=30+12i volts and the given current C=5-i amps into the formula V=C* I. We will then solve for I.

To calculate the quotient, we need to rationalize the denominator. To do so, we will multiply both the numerator and the denominator of the fraction by the conjugate of the denominator. Denominator: & 5-i Conjugate: & 5+i We will now multiply the numerator and denominator of the fraction by 5+i. 30+12i/5-i=(30+12i)( 5+i)/(5-i)( 5+i) Using the fact that (a+bi)(a-bi)=a^2+b^2, we can rewrite the product in the denominator as 5^2+1^2, which results in a real number. Then we can simplify the quotient and write the result in standard form.

The impedance of the circuit is 6913 + 4513i ohms.

Emily plots the numbers -5, 1, -2-2i, and -2+5i on a complex plane and then connects them so that she gets a quadrilateral. What type of quadrilateral does Emily get?

We are given four complex numbers. - 5, 1, -2-2i, -2+5i Recall that in the complex plane, the horizontal axis represents the real part and the vertical axis the imaginary part of a complex number. This means that a complex number a+bi corresponds to the point (a,b). ccc Complex Number& &Point on Complex Plane - 5 & & (- 5, 0) 1 & & (1,0) -2-2i & & (-2,-2) -2+5i & & (-2,5) Let's plot these points on the complex plane.

Now let's connect the points so that we get a quadrilateral. To do so, we start at any of the points and connect them counterclockwise.

As we can see in the graph, the quadrilateral formed by these points is a kite.

The Basics of Complex Numbers

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