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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

3 x^{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 $Z$
$2 x - 3 = 0$	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, $i = - 1 .$ From this definition, it can also be said that $i^{2} = - 1 .$

Imaginary Unit
$i = - 1$ 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 \Rightarrow 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.

$- a = a \cdot - 1 = a \cdot i$

The above property is true only when

a > 0 .

Here are some examples of how to use the property to simplify radical expressions.

- 5 - 4 - 20 = 5 \cdot i = 4 \cdot i = 2 i = 20 \cdot i = 4 \cdot 5 \cdot i = 25 \cdot i

The combination of real numbers and any expression of the form

b i,

with

b \neq = 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

a b \cdot 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

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 square with side x and a right triangle with legs 3 and 4.

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

- a = a \cdot i

for

a > 0 .

b This time the area of the triangle must be multiplied by

2 .

Remember, if

a > 0

then

- a = a \cdot 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

4 \cdot 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	$A_{□} = s^{2}$	$A_{△} = \frac{1}{2} b h$
Substitution	$A_{□} = x^{2}$	$A_{△} = \frac{1}{2} \cdot 3 \cdot 4 = 6$

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

SubstituteII

$A_{□} = x^{2}$ , $A_{△} = 6$

x^{2} + 6 = 2

SubEqn

$LHS - 6 = RHS - 6$

x^{2} = - 4

SqrtEqn

$LHS = RHS$

x^{2} = - 4

$a^{2} = \pm a$

x = \pm - 4

Note that the radicand of the final expression is negative, implying that

- 4

is an imaginary number. Recall how the square root of a negative number is rewritten.

- a = a \cdot i, a > 0

By applying this property, the right-hand side can now be written as imaginary numbers.

x = \pm - 4

SqrtNegToISqrt

$- a = i a$

x = \pm 4 i

CalcRoot

Calculate root

x = \pm 2 i

For the given situation, there are two imaginary solutions — namely,

2 i

and

- 2 i .

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_{□} + 2 A_{△} = 9

Now, substitute the corresponding expressions and solve the equation for

x .

A_{□} + 2 A_{△} = 29

SubstituteII

$A_{□} = x^{2}$ , $A_{△} = 6$

x^{2} + 2 (6) = 9

Multiply

x^{2} + 12 = 9

SubEqn

$LHS - 12 = RHS - 12$

x^{2} = - 3

SqrtEqn

$LHS = RHS$

x^{2} = - 3

$a^{2} = \pm a$

x = \pm - 3

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

x = \pm - 3

SqrtNegToISqrt

$- a = i a$

x = \pm 3 i

Once more, there are two imaginary solutions to the given situation — namely,

3 i

and

- 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_{□} + 3 A_{△} = 10

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

x .

A_{□} + 3 A_{△} = 10

SubstituteII

$A_{□} = x^{2}$ , $A_{△} = 6$

x^{2} + 3 (6) = 10

Multiply

x^{2} + 18 = 10

SubEqn

$LHS - 18 = RHS - 18$

x^{2} = - 8

SqrtEqn

$LHS = RHS$

x^{2} = - 8

$a^{2} = \pm a$

x = \pm - 8

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

x = \pm - 8

SqrtNegToISqrt

$- a = i a$

x = \pm 8 i

SplitIntoFactors

Split into factors

x = \pm 4 \cdot 2 i

SqrtProd

$a \cdot b = a \cdot b$

x = \pm 4 \cdot 2 i

CalcRoot

Calculate root

x = \pm 22 i

For the third time, the two solutions found are imaginary numbers — namely,

22 i

and

- 22 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 = - 1 or i^{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 of i Raise both sides to 0 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 .

i^{3} = i^{2} \cdot i i^{3} ⇕ i^{3} = - 1 \cdot i i^{3} ⇕ i^{3} = - i Product of Powers Property Definition of 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^{5} i^{6} i^{7} = i^{2} \cdot i^{2} = 1 \cdot 1 = 1 = i^{4} \cdot i^{1} = 1 \cdot i = i = i^{4} \cdot i^{2} = 1 (- 1) = - 1 = i^{4} \cdot 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.

i^0=1, i^1 = i, i^2 = -1, i^3=-i; i^4=1, i^5 = i, i^6 = -1, and i^7=-i

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 $\frac{n}{4}$ 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

4 .

\frac{3 4}{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.

\frac{3 4}{4} = 32 4 \cdot 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

$a^{m + n} = a^{m} \cdot a^{n}$

i^{32} \cdot i^{2}

Rewrite

Rewrite $32$ as $4 \cdot 8$

i^{4 \cdot 8} \cdot i^{2}

ProdInExponent

$a^{m \cdot n} = (a^{m})^{n}$

(i^{4})^{8} \cdot i^{2}

SubstituteII

$i^{4} = 1$ , $i^{2} = - 1$

(1)^{8} (- 1)

BaseOne

$1^{a} = 1$

1 (- 1)

IdPropMult

\IdPropMult

- 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 $i .$

Discussion

Imaginary and Complex Numbers

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.

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.

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.

A series circuit with a resistor of 8 ohms, a capacitor of 6 ohms, and an inductor of 10 ohms. The impedances are 8, -6i, and 10i ohms respectively.

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.

A series circuit with a two resistors, one of 6 ohms and the other of 3 ohms, a capacitor of 9 ohms, and an inductor of 4 ohms.

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	$- 6 i Ω$
Inductor	$10 i Ω$

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 + (- 6 i) + 10 i

Simplify right-hand side

AddNeg

$a + (- b) = a - b$

Impedance = 8 - 6 i + 10 i

FactorOut

Factor out $i$

Impedance = 8 + (- 6 + 10) i

AddTerms

Add terms

Impedance = 8 + 4 i

Consequently, the impedance of the series circuit is

8 + 4 i

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	$9 Ω$	$- 9 i Ω$
Inductor	$4 Ω$	$4 i Ω$
Resistor $2$	$3 Ω$	$3 Ω$

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

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

Simplify right-hand side

AddNeg

$a + (- b) = a - b$

Impedance = 6 - 9 i + 4 i + 3

CommutativePropAdd

\CommutativePropAdd

Impedance = 6 + 3 + 4 i - 9 i

AssociativePropAdd

\AssociativePropAdd

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

FactorOut

Factor out $i$

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

AddSubTerms

Add and subtract terms

Impedance = 9 - 5 i

Therefore, the impedance of the series circuit is

9 - 5 i

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.

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 \cdot 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 + 3 i

amps and the impedance is

8 - 3 i

ohms. What is the voltage of the circuit?

b If

5 + 6 i

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 + 6 i

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} I_{1} = (4 + 3 i) A = (8 - 3 i) 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} \cdot I_{1}

SubstituteII

$C_{1} = 4 + 3 i$ , $I_{1} = 8 - 3 i$

V_{1} = (4 + 3 i) (8 - 3 i)

Multiply

4 + 3 i

8 - 3 i

Multiply

V_{1} = 4 (8) + 4 (- 3 i) + 3 i (8) + 3 i (- 3 i)

CommutativePropMult

\CommutativePropMult

V_{1} = 4 (8) + 4 (- 3) i + 3 (8) i + 3 (- 3) (i \cdot i)

MultPow

$a^{m} \cdot a^{n} = a^{m + n}$

V_{1} = 4 (8) + 4 (- 3) i + 3 (8) i + 3 (- 3) i^{2}

IDefPow

$i^{2} = - 1$

V_{1} = 4 (8) + 4 (- 3) i + 3 (8) i + 3 (- 3) (- 1)

Multiply

V_{1} = 32 - 12 i + 24 i + 9

CommutativePropAdd

\CommutativePropAdd

V_{1} = 32 + 9 + 24 i - 12 i

AssociativePropAdd

\AssociativePropAdd

V_{1} = (32 + 9) + (24 i - 12 i)

FactorOut

Factor out $i$

V_{1} = (32 + 9) + (24 - 12) i

AddSubTerms

Add and subtract terms

V_{1} = 41 + 12 i

Therefore, the voltage of the first circuit is

41 + 12 i

volts.

b First, start by finding

V_{2}

using a similar procedure as the one made in the previous part.

C_{2} I_{2} = (5 + 2 i) A = (7 + 5 i) Ω

The voltage of the circuit equals the product of these two quantities.

V_{2} = C_{2} \cdot I_{2}

SubstituteII

$C_{1} = 5 + 2 i$ , $I_{1} = 7 + 5 i$

V_{2} = (5 + 2 i) (7 + 5 i)

Multiply

5 + 2 i

7 + 5 i

Multiply

V_{2} = 5 (7) + 5 (5 i) + 2 i (7) + 2 i (5 i)

CommutativePropMult

\CommutativePropMult

V_{2} = 5 (7) + 5 (5 i) + 2 (7) i + 2 (5) (i \cdot i)

MultPow

$a^{m} \cdot a^{n} = a^{m + n}$

V_{2} = 5 (7) + 5 (5 i) + 2 (7) i + 2 (5) i^{2}

IDefPow

$i^{2} = - 1$

V_{2} = 5 (7) + 5 (5 i) + 2 (7) i + 2 (5) (- 1)

Multiply

V_{2} = 35 + 25 i + 14 i - 10

CommutativePropAdd

\CommutativePropAdd

V_{2} = 35 - 10 + 25 i + 14 i

AssociativePropAdd

\AssociativePropAdd

V_{2} = (35 - 10) + (25 i + 14 i)

FactorOut

Factor out $i$

V_{2} = (35 - 10) + (25 + 14) i

AddSubTerms

Add and subtract terms

V_{2} = 25 + 39 i

Therefore, the voltage of the second circuit is

25 + 39 i

volts. Finally, add

5 + 6 i

volts to

V_{2} .

5 + 6 i + V_{2}

Substitute

25 + 39 i

for

V_{2}

and simplify

Substitute

$V_{2} = 25 + 39 i$

5 + 6 i + 25 + 39 i

CommutativePropAdd

\CommutativePropAdd

5 + 25 + 6 i + 39 i

FactorOut

Factor out $i$

5 + 25 + (6 + 39) i

AddTerms

Add terms

30 + 45 i

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} V_{2} = 41 + 12 i = 25 + 39 i

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

SubstituteII

$V_{2} = 25 + 39 i$ , $V_{1} = 41 + 12 i$

25 + 39 i - (41 + 12 i)

Distr

Distribute $- 1$

25 + 39 i - 41 - 12 i

CommutativePropAdd

\CommutativePropAdd

25 - 41 + 39 i - 12 i

FactorOut

Factor out $i$

25 - 41 + (39 - 12) i

SubTerms

Subtract terms

- 16 + 27 i

The difference between the voltage of the second circuit and the voltage of the first circuit is

- 16 + 27 i

volts.

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.

\overline{a + b i} = a - b i or \overline{a - b i} = a + b i

For example, the complex conjugate of

z = 3 - 5 i

\overset{z}{ˉ} = 3 + 5 i .

It is worth noting that the product of a complex number and its conjugate is a real number.

z \cdot \overset{z}{ˉ}

Substitute

a + b i

for

z

and simplify

Substitute

$z = a + b i$

(a + b i) (\overline{a + b i})

Substitute

$\overline{a + b i} = a - b i$

(a + b i) (a - b i)

Multiply

a^{2} - a b i + a b i - b^{2} i^{2}

AddTerms

Add terms

a^{2} - b^{2} i^{2}

IDefPow

$i^{2} = - 1$

a^{2} - b^{2} (- 1)

NegNeg

$- (- a) = a$

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.

\frac{5 + 2 i}{3 - 4 i}

To simplify the quotient, multiply the numerator and the denominator by the complex conjugate of the denominator.

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 + b i) (a - b i) = a^{2} + b^{2}

Therefore, multiply the numerator and the denominator by the complex conjugate of the denominator. In this case, multiply by

3 + 4 i .

\frac{5 + 2 i}{3 - 4 i} = \frac{( 5 + 2 i ) ( 3 + 4 i )}{( 3 - 4 i ) ( 3 + 4 i )}

Notice that

\frac{3 + 4 i}{3 + 4 i}

equals

1 .

Subsequently, the value of the original fraction does not change by the Identity Property of Multiplication.

Simplify the Expression

The obtained expression can now be simplified.

\frac{( 5 + 2 i ) ( 3 + 4 i )}{( 3 - 4 i ) ( 3 + 4 i )}

Simplify denominator

$(a + b i) (a - b i) = a^{2} + b^{2}$

\frac{( 5 + 2 i ) ( 3 + 4 i )}{3 ^{2} + 4 ^{2}}

CalcPow

Calculate power

\frac{( 5 + 2 i ) ( 3 + 4 i )}{9 + 1 6}

AddTerms

Add terms

\frac{( 5 + 2 i ) ( 3 + 4 i )}{2 5}

Simplify numerator

Multiply

\frac{1 5 + 2 0 i + 6 i + 8 i ^{2}}{2 5}

IDefPow

$i^{2} = - 1$

\frac{1 5 + 2 0 i + 6 i + 8 ( - 1 )}{2 5}

MultPropNegOne

Multiplication Property of $- 1$

\frac{1 5 + 2 0 i + 6 i - 8}{2 5}

AddSubTerms

Add and subtract terms

\frac{7 + 2 6 i}{2 5}

WriteSumFrac

Write as a sum of fractions

\frac{7}{2 5} + \frac{2 6}{2 5} i

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.

$\frac{a + b i}{c + d i} = \frac{a c + b d}{c ^{2} + d ^{2}} + (\frac{b c - a d}{c ^{2} + d ^{2}}) i$

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 + 4 i,

what is

z \cdot \overset{z}{ˉ}

equal to?

b Write

\frac{2 0}{3 + i}

as a complex number in standard form.

c Find the real and imaginary parts of

\frac{1 0 + 5 i}{2 + 4 i} .

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

\overset{z}{ˉ},

is obtained by changing the sign of the imaginary part of

z .

The imaginary part is the term containing the imaginary unit.

z = - 8 + \underline{\underline{4}} i

The imaginary part of

z

4 .

Therefore, the conjugate of

z

will have

- 4

as its imaginary part.

\overset{z}{ˉ} = - 8 - 4 i

Now that the conjugate of

z

is identified, the product of the conjugates can be calculated.

z \cdot \overset{z}{ˉ}

SubstituteII

$z = - 8 + 4 i$ , $\overset{z}{ˉ} = - 8 - 4 i$

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

MultPar

Multiply parentheses

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

Multiply

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

SubTerms

Subtract terms

64 - 16 i^{2}

IDefPow

$i^{2} = - 1$

64 - 16 (- 1)

MultNegNegOnePar

$- a (- b) = a \cdot b$

64 + 16

AddTerms

Add terms

80

Consequently,

z \cdot \overset{z}{ˉ} = 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.

\frac{2 0}{3 + i}

To rationalize the denominator, each part of the fraction has to be multiplied by the conjugate of the denominator.

Denominator : Conjugate : 3 + i 3 - i

Now, multiply the numerator and denominator of the given fraction by

3 - i .

\frac{2 0}{3 + i} = \frac{2 0 ( 3 - i )}{( 3 + i ) ( 3 - i )}

Using the fact that

(a + b i) (a - b i) = 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.

\frac{2 0 ( 3 - i )}{( 3 + i ) ( 3 - i )}

Simplify denominator

$(a + b i) (a - b i) = a^{2} + b^{2}$

\frac{2 0 ( 3 - i )}{3 ^{2} + 1 ^{2}}

CalcPow

Calculate power

\frac{2 0 ( 3 - i )}{9 + 1}

AddTerms

Add terms

\frac{2 0 ( 3 - i )}{1 0}

Simplify

ReduceFrac

$\frac{a}{b} = \frac{a / 1 0}{b / 1 0}$

\frac{2 ( 3 - i )}{1}

DivByOne

$\frac{a}{1} = a$

2 (3 - i)

Distr

Distribute $2$

6 - 2 i

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

\frac{1 0 + 5 i}{2 + 4 i}

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 - 4 i,

which is the conjugate of

2 + 4 i .

\frac{1 0 + 5 i}{2 + 4 i} = \frac{( 1 0 + 5 i ) ( 2 - 4 i )}{( 2 + 4 i ) ( 2 - 4 i )}

Next, continue simplifying the quotient.

\frac{( 1 0 + 5 i ) ( 2 - 4 i )}{( 2 + 4 i ) ( 2 - 4 i )}

Simplify denominator

$(a + b i) (a - b i) = a^{2} + b^{2}$

\frac{( 1 0 + 5 i ) ( 2 - 4 i )}{2 ^{2} + 4 ^{2}}

CalcPow

Calculate power

\frac{( 1 0 + 5 i ) ( 2 - 4 i )}{4 + 1 6}

AddTerms

Add terms

\frac{( 1 0 + 5 i ) ( 2 - 4 i )}{2 0}

Simplify numerator

Multiply

\frac{2 0 - 4 0 i + 1 0 i - 2 0 i ^{2}}{2 0}

IDefPow

$i^{2} = - 1$

\frac{2 0 - 4 0 i + 1 0 i - 2 0 ( - 1 )}{2 0}

MultPropNegOne

Multiplication Property of $- 1$

\frac{2 0 - 4 0 i + 1 0 i + 2 0}{2 0}

AddTerms

Add terms

\frac{4 0 - 3 0 i}{2 0}

Rewrite

WriteSumFrac

Write as a sum of fractions

\frac{4 0}{2 0} - \frac{3 0}{2 0} i

CalcQuot

Calculate quotient

2 - \frac{3 0}{2 0} i

ReduceFrac

$\frac{a}{b} = \frac{a / 1 0}{b / 1 0}$

2 - \frac{3}{2} i

Consequently, the real part of the given complex number is

2

and the imaginary part is

- \frac{3}{2} .

Re (\frac{1 0 + 5 i}{2 + 4 i}) Im (\frac{1 0 + 5 i}{2 + 4 i}) = 2 = - \frac{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.

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 + 2 i

is represented by the point

(- 3, 2) .

Complex Number : Point on Complex Plane : a + b i ↓ ↓ (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 + b i ∣ = 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.

Number 4+3i plotted as point (3,4). Its absolute value is 5.

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.

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