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When solving quadratic equations, it is common to come across equations without a solution — also said to have ### Catch-Up and Review

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-realnumbers.

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

**Try these practice exercises to warm up for this lesson.**

a What are the solutions to the equation $3x_{2}−27=0?$

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b How many real solutions does the equation $x_{2}=9$ have?

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c How many real solutions does the equation $x_{2}=-25$ have?

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Challenge

Group the following equations depending on whether they have real solutions.
**not real**, but they solve the equations.

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

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class=\"mord\">-<\/span><\/span><span class=\"mord\">1<\/span><span class=\"mord\">2<\/span><\/span><\/span><\/span>"},{"id":1,"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.897438em;vertical-align:-0.08333em;\"><\/span><span class=\"mord\"><span class=\"mord mathdefault\">x<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.8141079999999999em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">\u2212<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" 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class=\"mbin\">+<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">2<\/span><span class=\"mord\">6<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">0<\/span><\/span><\/span><\/span>"}],[{"id":1,"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1.04em;vertical-align:-0.13278em;\"><\/span><span class=\"mord\">\u00b1<\/span><span class=\"mord sqrt\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.90722em;\"><span class=\"svg-align\" 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Discussion

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.

## The Imaginary Unit

The imaginary unit $i$ can also be regarded as a solution to the equation $x_{2}+1=0.$
The above property is true *only* when $a>0.$ Here are some examples of how to use the property to simplify radical expressions.

$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 $i$ is the principal square root of $-1,$ that is, $i=-1 .$ From this definition, it can also be said that $i_{2}=-1.$

**Imaginary Unit**

$i=-1 $ or $i_{2}=-1$

$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. $-a =a ⋅-1 =a ⋅i$

$-5 -4 -20 =5 ⋅i=4 ⋅i=2i=20 ⋅i=4⋅5 ⋅i=25 ⋅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

Rewrite each given radical expression in terms of the imaginary unit $i.$ Write the answer in the form $ab ⋅i,$ where $a$ is a non-zero integer and $b$ is a natural number such that $b $ cannot be simplified further.
### 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

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.

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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 $x?$

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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 $x?$

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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 $-a =a ⋅i$ for $a>0.$

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

c Multiply the area of the triangle by $3.$ To simplify $8 ,$ rewrite $8$ as $4⋅2$ and use the Product Property of Square Roots.

Square | Triangle | |
---|---|---|

Area Formula | $A_{□}=s_{2}$ | $A_{△}=21 bh$ |

Substitution | $A_{□}=x_{2}$ | $A_{△}=21 ⋅3⋅4=6$ |

$A_{□}+A_{△}=2 $

Next, substitute the corresponding expressions and solve the resulting equation for $x.$
$A_{□}+A_{△}=2$

SubstituteII

$A_{□}=x_{2}$, $A_{△}=6$

$x_{2}+6=2$

SubEqn

$LHS−6=RHS−6$

$x_{2}=-4$

SqrtEqn

$LHS =RHS $

$x_{2} =-4 $

$a_{2} =±a$

$x=±-4 $

$-a =a ⋅i,a>0 $

By applying this property, the right-hand side can now be written as imaginary numbers.
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 | $A_{□}=x_{2}$ | $A_{△}=6$ |

$A_{□}+2A_{△}=9 $

Now, substitute the corresponding expressions and solve the equation for $x.$
$A_{□}+2A_{△}=29$

SubstituteII

$A_{□}=x_{2}$, $A_{△}=6$

$x_{2}+2(6)=9$

Multiply

Multiply

$x_{2}+12=9$

SubEqn

$LHS−12=RHS−12$

$x_{2}=-3$

SqrtEqn

$LHS =RHS $

$x_{2} =-3 $

$a_{2} =±a$

$x=±-3 $

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 $x.$
$A_{□}+3A_{△}=10$

SubstituteII

$A_{□}=x_{2}$, $A_{△}=6$

$x_{2}+3(6)=10$

Multiply

Multiply

$x_{2}+18=10$

SubEqn

$LHS−18=RHS−18$

$x_{2}=-8$

SqrtEqn

$LHS =RHS $

$x_{2} =-8 $

$a_{2} =±a$

$x=±-8 $

$x=±-8 $

SqrtNegToISqrt

$-a =ia $

$x=±8 i$

SplitIntoFactors

Split into factors

$x=±4⋅2 i$

SqrtProd

$a⋅b =a ⋅b $

$x=±4 ⋅2 i$

CalcRoot

Calculate root

$x=±22 i$

Discussion

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=-1 ori_{2}=-11 $

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.
$i_{2}=-1⇕(i_{2})_{0}=(-1)_{0}⇕i_{0}=1 Definition ofiRaise both sides to0Zero 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 $i.$
$i_{3}=i_{2}⋅ii_{3}⇕i_{3}=-1⋅ii_{3}⇕i_{3}=-i Product of Powers PropertyDefinition ofiSimplify $

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_{5}i_{6}i_{7} =i_{2}⋅i_{2}=1⋅1=1=i_{4}⋅i_{1}=1⋅i=i=i_{4}⋅i_{2}=1(-1)=-1=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

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 $4.$
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 $4n $ is | Then, $i_{n}$ is equal to |

$0$ | $i_{0}=1$ |

$1$ | $i_{1}=i$ |

$2$ | $i_{2}=-1$ |

$3$ | $i_{3}=-i$ |

$434 =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.
$434 =324⋅8 +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

Rewrite $34$ as $32+2$

$i_{32+2}$

SumInExponent

$a_{m+n}=a_{m}⋅a_{n}$

$i_{32}⋅i_{2}$

Rewrite

Rewrite $32$ as $4⋅8$

$i_{4⋅8}⋅i_{2}$

ProdInExponent

$a_{m⋅n}=(a_{m})_{n}$

$(i_{4})_{8}⋅i_{2}$

SubstituteII

$i_{4}=1$, $i_{2}=-1$

$(1)_{8}(-1)$

BaseOne

$1_{a}=1$

$1(-1)$

IdPropMult

Identity Property of Multiplication

$-1$

Pop Quiz

Compute the required power of $i.$

Discussion

The imaginary unit $i,$ which is equal to $-1 ,$ 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

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.

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

Concept

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

$a+bi=c+di⇔a=candb=d$

Discussion

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

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.

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

$(a+bi)−(c+di)=(a−c)+(b−d)i$

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.

$z_{1}+z_{2}$

SubstituteII

$z_{1}=5+2i$, $z_{2}=3−3i$

$5+2i+3−3i$

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.

3

Combine Imaginary Parts

Example

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?

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

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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 $i.$

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 | $8Ω$ |

Capacitor | $-6iΩ$ |

Inductor | $10iΩ$ |

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

$Impedance=8+4i$

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 $1$ | $6Ω$ | $6Ω$ |

Capacitor | $9Ω$ | $-9iΩ$ |

Inductor | $4Ω$ | $4iΩ$ |

Resistor $2$ | $3Ω$ | $3Ω$ |

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

Simplify right-hand side

AddNeg

$a+(-b)=a−b$

$Impedance=6−9i+4i+3$

CommutativePropAdd

Commutative Property of Addition

$Impedance=6+3+4i−9i$

AssociativePropAdd

Associative Property of Addition

$Impedance=(6+3)+(4i−9i)$

FactorOut

Factor out $i$

$Impedance=(6+3)+(4−9)i$

AddSubTerms

Add and subtract terms

$Impedance=9−5i$

Discussion

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

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.

$(a+bi)(c+di)=(ac−bd)+(ad+$