ma 1101 fall 2020 unit 1 complex numbers section 12.1
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MA 1101 Fall 2020
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Unit 1 Complex Numbers
Section 12.1 Basic Definitions
Definition 1:
For any real number x (x β₯ 0), βπ₯ = π¦ ππ π¦2 = π₯.
Example:
β9 = 3 πππππ’π π 32 = 9.
Note:
If π₯ < 0, βπ₯ does not exist as a real number.
Example:
ββ9 is undefined (as a real number).
Such numbers, however, do exist and have applications in certain fields (electrical engineering for
example). We can work with such radicals with the following definition.
Definition 2:
The imaginary unit is denoted as j. In particular:
π = ββπ and ππ = βπ
This definition, used in conjunction with the property below, enables us to evaluate and simplify negative
square roots.
Property of Radicals:
βπ Γ π = βπ Γ βπ [provided at least one of a or b is positive]
Example:
ββ9 = β9 Γ β1 =
=
Problem 1:
Express each radical in terms of j.
π) ββ25 π) ββ7 π) 3ββ20
π) β ββ32 π) ββ3
16
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Property of negative radicals:
ββπ = π βπ
Note:
The property βπ Γ βπ = βπ Γ π can be used to combine a product of two radicals in many cases.
As noted on the previous page, however, this property may not be applied if both a and b are negative. In
such cases, first simplify each radical individually.
Problem 2:
Find and simplify each product.
π) β4 Γ β9 π) ββ4 Γ β9
π) ββ4 Γ ββ9 π) ββ3 Γ ββ4
π) β ββ2 Γ ββ10 π) β (β(β2)(β5) ) (ββ10 )
π) ββ5
3 Γ β
4
15 β) β
β9
2 Γ β
β16
25
Do #βs 5,7,9,11,13,17,19,21, p. 360 text
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Simplifying an expression of the form ππ (where n is a whole number)
Example:
Simplify:
π) π8 π) π11
Useful strategy:
If the power n is even, rewrite ππ as (π2)π. Use the definition of π2 = β1 to complete.
If the power n is odd, rewrite ππ as (ππβ1) π. Proceed as before.
Problem 3:
Simplify the following.
π) π26 π) π31 π) π9 β π7
π) β π4 π) (βπ)4
Do #βs 25,27,29,31, p. 360 text
Complex Numbers
Definition 3:
A number of the form π + ππ£, where a and b are real numbers, is called a complex number.
Notes:
If a = 0, the resulting number bj is a pure imaginary number.
If b = 0, the resulting number a is a real number.
Conclusion:
Complex numbers include the set of real numbers as well as the set of pure imaginary
numbers.
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For example:
The pure imaginary number 2π could be rewritten as 0 + 2π.
The real number -5 could be rewritten as β5 + 0π.
Note:
The form π + ππ£ is known as the rectangular form of a complex number.
In this form:
βaβ is the real part.
βbβ is the imaginary part.
Problem 4:
Express each of the following complex numbers in rectangular form.
π) β β9 β ββ9 π)β54 β ββ24
π) π5 β 4 π) ββ25π2 + ββ16
Do #βs 33,35,37,39,41,43, p. 360 text
Note:
Each complex number is unique. Therefore, two complex numbers (π₯ + π¦π) πππ (π + ππ) are equal
only if π₯ = π πππ π¦ = π.
Example:
If π₯ + π¦π = 3 β 2π, then π₯ = 3 πππ π¦ = β2.
Problem 5:
Find the values of x and y which satisfy the following equations.
π) π₯ β π¦π = β5 + π π) 4π₯ + 3ππ¦ = 20 β 6π
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π) 4 β 3π + π₯ = 6π + ππ¦
π) π₯ + 3π₯π + 3π¦ = 5 β π β ππ¦
Do #βs 49,51,53, p. 360 text
One final point:
The conjugate of the complex number (π + ππ£) is (π β ππ£) and vice versa.
Example:
Write the conjugate of the following.
π) 3 β 5π π) 6j π) ββ49 + 4π2
Do #βs 45,47,55,57, pp. 360-361 text
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Section 12.2 Basic Operations with Complex Numbers (See 12.3 β 12.6, p. 361 text)
Case 1: The given numbers are in the form π + ππ£.
Problem 1:
Perform the indicated operation and simplify.
π) (3 + 2π) + (β5 β π) π) (β1 β 3π) β (5 β 2π)
π) 1.5π (2π β 1.4) π) (2 β π)(5 + 3π)
π) (β1 β π)2
Do #βs 5,7,9,13,15,23,25, p. 363 text
Problem 2:
Find the product of 2 + π and its conjugate.
Do # 45, p. 363 text
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Division of Complex Numbers
To divide two complex numbers, use the same process as rationalizing the denominator of a rational expression:
multiply the numerator and the denominator by the conjugate of the denominator.
Example:
7β2π
3+4π
=7β2π
3+4πΓ
3β4π
3β4π
=21β28πβ6π+8π2
9β16π2
=21β34πβ8
9+16
=13β34π
25
=ππ
ππβ
ππ
πππ
Problem 3:
Simplify the following quotients. Express your final answer in the form π + ππ£.
a) 2βj
1+2j π)
4+3π
2π
π) 2π
3βπ+
1
2 π (π + 4)
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Extra Practice:
1. 12+10π
6β8π
2. 1
π+
2
3+π
3. (6π+5)(2β4π)
(5βπ)(4π+1)
Do #βs 27,29,31,35,47, p. 363 text
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Case 2: The given numbers require some simplification before proceeding.
Problem 4:
Simplify.
π) (3ββ16 + β9 ) β (7 β 2ββ4) π) (5 β ββ36)(βββ1)
π) π7βπ
2πβπ2
Do #βs 33,39, p. 363 text
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Miscellaneous Problems
1. Express πβ4 + 2πβ1 in rectangular form.
2. Write the reciprocal of 2 β 3π in rectangular form.
3. Solve for x:
(π₯ + 5π)2 = 11 β 60π
Solutions:
Do #βs 49,51, p. 363 text
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Section 12.3 Graphical representation of a complex number
Recall: Rectangular form of a complex number: π₯ + π¦π
The complex plane:
A complex number π± + π²π£ may be represented by the point (π±, π²) in the complex plane
where:
π₯ = the real part π¦ = the imaginary part
Example:
Represent each number as a point in the complex plane.
π) 2 + π π) β 3 β 2π
Extending the concept: Drawing a line segment from the origin to the point (π±, π²)in the plane gives an
alternate way to represent a complex number β as a vector.
Example: Represent each number as a vector in the complex plane.
π) 3 + π π) β 1 β 4π
Do #βs 3,5,7, p. 363 text
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Adding complex numbers graphically:
Step 1: Find the point in the plane corresponding to the first number. Draw a vector from the origin to this
point.
Step 2: Repeat this process for the second number.
Step 3: Complete a parallelogram with the lines drawn as adjacent sides. The resulting fourth vertex is the
point representing the sum.
Problem:
Perform the indicated operations graphically.
π) (1 + 3π) + (3 + 2π) π)(4 β π) + (β2 + 3π)
π) (3 β π) β (2 + 2π)
Do #βs 9,11,13,15,17,21,23, p. 363 text
β β β
β
β
β
Imaginary
Real
β β β
β
β
β
Imaginary
Real
β β β
β
β
β
Imaginary
Real
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Section 12.4 Polar form of a complex number
Example:
Represent the number 3 + 4π as a vector in the complex plane.
Notes:
As a vector possesses both magnitude and direction, so too does a complex number.
The magnitude is simply the length of the vector.
The direction is the angle π formed between the positive real axis and the vector itself.
The form:
Consider the complex number π + ππ represented as a vector in the complex plane.
Expressing x and y in terms of r using basic trigonometry we get:
cos π = π₯
π β
sin π = π¦
π β
Substituting for x and y, the number π + ππ =
β β β
β
β
β
Imaginary
Real
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To convert the number π + ππ to polar form, we need to find the values of βrβ and βΞΈβ .
How to solve for βrβ and βΞΈβ:
π2 = π₯2 + π¦2 (r > 0)
tan π = π¦
π₯ (0 β€ π < 360Β°)
Problem 1:
Represent each of the following complex numbers graphically and determine its polar form.
π)3 + 4π π) β 2 β πβ5
Do #βs 3,5,7,9,11,15,17, p. 368 text
β β β
β
β
β
Imaginary
Real
β β β
β
β
β
Imaginary
Real
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Alternate notation for polar form:
πβ π = π(cos π + π sin π) Examples:
π) 5(πππ 53.1Β° + ππ ππ 53.1Β°) =
π) 1.82 β 320Β° =
Problem 2:
Express each complex number in rectangular form.
π) 2.57(πππ 112.3Β° + π π ππ 112.3Β°) π) 1.82 β 320Β°
Problem 3:
Represent the number 4(πππ 270Β° + ππ ππ270Β°) graphically and express it in rectangular
form.
Do #βs 19,21,25,27,29,33,35, p. 363 text
Multiplying and Dividing complex numbers in polar form
Basic Results:
π) (π1β π1)(π2β π2) = π1π2 β (π1 + π2)
ππ) (π1β π1)
(π2β π2)=
π1
π2 β (π1 β π2)
β β β
β
β
β
Imaginary
Real
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Problem 4:
Perform the indicated operation.
π) (1.5 β 230Β°)(2.4 β 150Β°) π) 3.2β 67.3Β°
1.3β 154.1Β°
Do #βs 5,7,9,11,17,19,25,27,29, pp. 375-376 text
Consider the problem below:
Find the sum: 2 β 30Β° + β 60Β°
Notes:
β’ Unlike multiplication and division of numbers in polar form, addition cannot be done.
β’ The above sum can only be found by first converting each number to rectangular form,
adding the results and then converting this number to polar form.
Solution:
Do #βs 21,23, p. 375 text
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Applications:
a.) Given that the current in a circuit is 3.90 β 6.04πππ΄ and the impedance is 5.16 + 1.14ππΞ©, find the
magnitude of the voltage.
b.) Given that the voltage in a given circuit is 8.3 β 3.1ππ and the impedance is 2.1 β 1.1πΞ©, find the
magnitude of the current.
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c.) Two resistors have resistances that can be expressed as π 1 = 5β 30π and π 2 = 8β 120π. What is
the total resistance if the resistors are in series? Parallel?
Do # 55, p. 363 text
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Section 12.5 Exponential form of a complex number
Exponential form of a complex number is written as:
ππππ½
As with polar form:
r is the magnitude of the representative vector
π is the angle formed by this vector (in radians)
Recall:
An angle is converted from degrees to radians by multiplying by π
180.
Example:
127.1Β° =127π
180 = 2.22
Probkems:
1. Express the complex number 5.24 β 118.2Β° in:
a) exponential form b) rectangular form
2. Express 4.15 π5.60π in polar form.
Summary of the three forms:
Rectangular: π + ππ
Polar: π(ππ¨π¬ π½ + π π¬π’π§ π½) = πβ π½ (π½ ππ π ππππππ)
Exponential: ππππ½ (π½ ππ πππ ππππ)
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3. Express β3 β 2π in exponential form.
4. Simplify the following product and express the result in rectangular form.
(12.3 π1.54π)(5.9 π3.17π)
Do #βs 3,5,7,9,11,15,17,19,21,23,25,27,29,31,33 pp. 370-371 text
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Section 12.6 Powers and Roots of complex numbers
Topic 1:
Raising a complex number (in polar form) to a power
Demoivreβs Theorem:
(πβ π½)π = ππ β ππ½
Problems:
1. Simplify the following. Leave your answer in polar form.
π) (0.5 β 100Β°) 6 π) [ 2(πππ 15Β° + π π ππ 15Β°)]4
2. Change 2 β π to polar form. Using this result, simplify (2 β π)5.
Do #βs 13,15,33,35, pp. 375-376 text
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Topic 2:
Using DeMoivreβs Theorem, find the βnβ πππ roots of a complex number (in polar form).
Basic Steps:
β’ Express the given number in polar form.
β’ Apply the appropriate fractional exponent (as per the root of interest) and determine
the first rootβ¦using DeMoivreβs Theorem.
β’ Add 360Β° π‘π π and repeat step 2 to find the second root.
β’ Continue adding 360Β° and repeating the process of step 2 until all required roots
have been found.
Example: Find the 3 cube roots of 3 β 4π
Step 1: 3 β 4π = 5β 306.8699Β°
Step 2: cube root = exponent of 1
3
Step 3: (5β 306.8699Β°)1
3 = 51
3β 1
3(306.8699Β°) = 1.71β 102.3Β° (this is the first root)
Step 4: (5β 666.8699Β°)1
3 = 51
3β 1
3(666.8699Β°) = 1.71β 222.3Β° (second root)
(5β 1026.8699Β°)1
3 = 51
3β 1
3(1026.8699Β°) = 1.71β 342.3Β° (third root)
Example:
Find the cube roots of β1:
β1 = β1 + 0π = 1β 180Β°
First root: (1β 180Β°)1
3 = 11
3β 1
3(180Β°) = 1β 60Β° = 0.5 + 0.87π
Second root: (1β 540Β°)1
3 = 11
3β 1
3(540Β°) = 1β 180Β° = β1
Third root: (1β 900Β°)1
3 = 11
3β 1
3(900Β°) = 1β 300Β° = 0.5 β 0.87π
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Example:
Find the square roots of 2π. Give your answers in rectangular form.
Problem:
Find the cube roots of each of the following:
π) 2 β π
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b) 27
Do #βs 37,39,41,43,49, p. 376 text