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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-realnumbers.
Catch-Up and Review
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
Non-real Solutions to Quadratic Equations
Group the following equations depending on whether they have real solutions.
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.
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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.
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.
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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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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
| Square | Triangle | |
|---|---|---|
| Area Formula | A_(□) = s^2 | A_(△) = 1/2bh |
| Substitution | A_(□) =x^2 | A_(△) =1/2* 3* 4=6 |
A_(□) + A_(△) = 2
SubstituteII
A_(□)= x^2, A_(△)= 6
x^2 + 6 = 2
SubEqn
LHS-6=RHS-6
x^2 = -4
SqrtEqn
sqrt(LHS)=sqrt(RHS)
sqrt(x^2) = sqrt(-4)
sqrt(a^2)=± a
x = ±sqrt(-4)
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_(△) = 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
sqrt(LHS)=sqrt(RHS)
sqrt(x^2) = sqrt(-3)
sqrt(a^2)=± a
x = ±sqrt(-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
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
sqrt(LHS)=sqrt(RHS)
sqrt(x^2) = sqrt(-8)
sqrt(a^2)=± a
x = ±sqrt(-8)
x = ±sqrt(-8)
SqrtNegToISqrt
sqrt(- a)= isqrt(a)
x = ± sqrt(8)i
SplitIntoFactors
Split into factors
x = ± sqrt(4* 2)i
SqrtProd
sqrt(a* b)=sqrt(a)*sqrt(b)
x = ± sqrt(4)*sqrt(2)i
CalcRoot
Calculate root
x = ± 2sqrt(2)i
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.
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).
| 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 |
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
Calculating Powers of i
Compute the required power 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.
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.
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
1
Separate Real Parts From Imaginary Parts
expand_more 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
expand_more Next, combine the real parts.
3
Combine Imaginary Parts
expand_more 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?
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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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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 |
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 | 6 Ω | 6 Ω |
| Capacitor | 9 Ω | -9i Ω |
| Inductor | 4 Ω | 4i Ω |
| Resistor 2 | 3 Ω | 3 Ω |
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
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
1
Distribute Terms
expand_more 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
SubstituteII
z_1= 3+2i, z_2= 4+2i
( 3+2i)( 4+2i)
Distr
Distribute (3+2i)
(3+2i)4 + (3+2i)2i
Distr
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
expand_more 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
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
expand_more By definition, i^2=-1. Thus, substitute -1 for i^2 into the last expression.
12 + 8i + 6i + 4i^2
IDefPow
i^2=- 1
12 + 8i + 6i + 4(-1)
MultPropNegOne
Multiplication Property of -1
12 + 8i + 6i - 4
4
Combine Like Terms
expand_more 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.
Consequently, z_1* z_2 = 8+14i.
12 + 8i + 6i - 4
CommutativePropAdd
Commutative Property of Addition
12 - 4 + 8i + 6i
AssociativePropAdd
Associative Property of Addition
(12 - 4) + (8i + 6i)
FactorOut
Factor out i
(12 - 4) + (8 + 6)i
AddSubTerms
Add and subtract terms
8 + 14i
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.
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?
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b If 5+6i volts are added to the voltage of the second circuit, what is the resulting voltage?
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c What is the difference between the original voltage of the second circuit and the voltage of the first circuit?
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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.
V_1 = C_1 * I_1
SubstituteII
C_1= 4+3i, I_1= 8-3i
V_1 = ( 4+3i)( 8-3i)
▼
Multiply 4+3i by 8-3i
Multiply
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)
MultPow
a^m*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
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)
FactorOut
Factor out i
V_1 = (32 + 9) + (24 - 12)i
AddSubTerms
Add and subtract terms
V_1 = 41 + 12i
b First, start by finding V_2 using a similar procedure as the one made in the previous part.
V_2 = C_2 * I_2
SubstituteII
C_1= 5+2i, I_1= 7+5i
V_2 = ( 5+2i)( 7+5i)
▼
Multiply 5+2i by 7+5i
Multiply
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)
MultPow
a^m*a^n=a^(m+n)
V_2 = 5(7) + 5(5i) + 2(7)i + 2(5)i^2
IDefPow
i^2=- 1
V_2 = 5(7) + 5(5i) + 2(7)i + 2(5)(-1)
Multiply
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)
FactorOut
Factor out i
V_2 = (35 - 10) + (25 + 14)i
AddSubTerms
Add and subtract terms
V_2 = 25 + 39i
5+6i + V_2
▼
Substitute 25+39i for V_2 and simplify
Substitute
V_2= 25 + 39i
5+6i + 25 + 39i
CommutativePropAdd
Commutative Property of Addition
5 + 25 + 6i + 39i
FactorOut
Factor out i
5 + 25 + (6 + 39)i
AddTerms
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_2 - V_1
▼
Substitute values and simplify
SubstituteII
V_2= 25+39i, V_1= 41+12i
25+39i - ( 41+12i)
Distr
Distribute -1
25 + 39i - 41 - 12i
CommutativePropAdd
Commutative Property of Addition
25 - 41 + 39i - 12i
FactorOut
Factor out i
25 - 41 + (39 - 12)i
SubTerms
Subtract terms
-16 + 27i
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.
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
z* z
▼
Substitute a+bi for z and simplify
Substitute
z= a+bi
( a+bi)(a+bi)
Substitute
a+bi= a-bi
(a+bi)( a-bi)
Multiply
Multiply
a^2-abi+abi-b^2i^2
AddTerms
Add terms
a^2-b^2i^2
IDefPow
i^2=- 1
a^2-b^2(-1)
NegNeg
- (- a)=a
a^2+b^2
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.
Rationalizing a denominator using complex conjugates leads to the following formula to simplify quotients of complex numbers.
1
Multiply by the Conjugate of the Denominator
expand_more 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
expand_more The obtained expression can now be simplified.
The denominator has now been rationalized and the quotient rewritten as a complex number in standard form.
(5+2i)(3+4i)/(3-4i)(3+4i)
(5+2i)(3+4i)/25
▼
Simplify numerator
Multiply
Multiply
15 + 20i + 6i + 8i^2/25
IDefPow
i^2=- 1
15 + 20i + 6i + 8(-1)/25
MultPropNegOne
Multiplication Property of -1
15 + 20i + 6i - 8/25
AddSubTerms
Add and subtract terms
7 + 26i/25
WriteSumFrac
Write as a sum of fractions
7/25 + 26/25i
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?
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b Write 203+i as a complex number in standard form.
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c Find the real and imaginary parts of 10+5i2+4i.
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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* z
SubstituteII
z= -8+4i, z= -8-4i
( -8+4i)( -8-4i)
MultPar
Multiply parentheses
-8(-8) - 8(-4i) + 4i(-8) +4i(-4i)
Multiply
Multiply
64 + 32i - 32i - 16i^2
SubTerms
Subtract terms
64 - 16i^2
IDefPow
i^2=- 1
64 - 16(-1)
MultNegNegOnePar
- a(- b)=a* b
64 + 16
AddTerms
Add terms
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)/(3+i)(3-i)
20(3-i)/10
▼
Simplify
6-2i
c To find the real and imaginary parts, the given quotient has to be simplified first.
(10+5i)(2-4i)/(2+4i)(2-4i)
(10+5i)(2-4i)/20
▼
Simplify numerator
Multiply
Multiply
20-40i+10i-20i^2/20
IDefPow
i^2=- 1
20-40i+10i-20(-1)/20
MultPropNegOne
Multiplication Property of -1
20-40i+10i+20/20
AddTerms
Add terms
40-30i/20
▼
Rewrite
WriteSumFrac
Write as a sum of fractions
40/20-30/20i
CalcQuot
Calculate quotient
2-30/20i
ReduceFrac
a/b=.a /10./.b /10.
2-3/2i
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+ 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.
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