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Polar Equations and Complex
Numbers
Raja Almukahhal
Larame Spence
Mara Landers
Nick Fiori
Art Fortgang
Melissa Vigil
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Printed: October 2, 2013
AUTHORS
Raja Almukahhal
Larame Spence
Mara Landers
Nick Fiori
Art Fortgang
Melissa Vigil
www.ck12.org
Chapter 1. Polar Equations and Complex Numbers
C HAPTER
1
Polar Equations and
Complex Numbers
C HAPTER O UTLINE
1.1
Polar Coordinates
1.2
Polar and Cartesian Transformation
1.3
Systems of Polar Equations
1.4
Imaginary Numbers
1.5
Complex Numbers
1.6
Quadratic Formula and Complex Sums
1.7
Products and Quotients of Complex Numbers
1.8
Polar Form of Complex Numbers
1.9
Product and Quotient Theorems
1.10
Powers and Roots of Complex Numbers
Introduction
When you first began graphing mathematical equations and values, you probably used a Cartesian graph, also known
as an x, y or rectangular graph. In this chapter you will be graphing quite a bit on a polar graph, or circular graph.
Polar equations and polar graphs can be a bit intimidating, particularly at first. With practice, however, you will
likely come to appreciate a number of situations where a polar graph is easier or just makes more sense than a
rectangular graph.
Another major topic in this chapter is imaginary numbers. You may be thinking, "If the numbers are imaginary
anyway, why should I need to learn about them?" One reason is that imaginary numbers can become real numbers
when multiplied together. It seems strange, but imaginary numbers actually ’convert’ to real numbers all the time in
mathematics, and every number you have ever seen can be written as a "complex number," which is the combination
of a real and an imaginary number!
Another interesting section in this chapter is the lesson on using the quadratic formula. You probably remember
using the formula in the past, and perhaps even remember the "discriminant," which can be used to identify the
number of real solutions to a quadratic equation. Now you will get to learn why all of your prior lessons always
specified "real" answers, when asking you to find the roots!
1
1.1. Polar Coordinates
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1.1 Polar Coordinates
Here you will learn about the polar coordinate system, which is similar in some ways to the (x, y) graphs you have
worked with in the past, but is specialized for visually exploring angles.
Everyone has dreamed of flying at one time or another. Not only would there be much less traffic to worry about,
but directions would be so much simpler!
Walking or driving: "Go East 2 blocks, turn left, then North 6 blocks. Wait for the train. Turn right, East 3 more
blocks, careful of the cow! Turn left, go North 4 more blocks and park."
Flying: "Fly 30deg East of North for a little less than 11 and 1/4 blocks. Land."
Nice daydream, what does it have to do with polar coordinates?
Watch This
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- Khan Academy: Polar Coordinates 1
Guidance
The Polar Coordinate System is alternative to the Cartesian Coordinate system you have used in the past to graph
functions. The polar coordinate system is specialized for visualizing and manipulating angles.
Angles are identified by travelling counter-clockwise around the circular graph from the 0deg line, or r-axis (where
the + x axis would be) to a specified angle.
2
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Chapter 1. Polar Equations and Complex Numbers
To plot a specific point, first go along the r-axis by r units. Then, rotate counterclockwise by the given angle,
commonly represented "θ". Be careful to use the correct units for the angle measure (either radians or degrees).
Radians
Usually polar plots are done with radians (especially if they include trigonometric functions), but sometimes degrees
are used.
A radian is the angle formed between the r axis and a polar axis drawn to meet a section of the circumference that
is the same length as the radius of a circle.
Given that the circumference of a circle is 2π · r, and since r is the radius, that means there are 2π radians in a
complete circle, and 1π radians in 1/2 of a circle.
If 1/2 of a circle is π radians, and is 180deg, that means that there are
180
π
degrees in each radian.
That translates to approximately 57.3 degrees = 1 radian.
Graphing using technology
Polar Equations can be graphed using a graphing calculator: With the graphing calculator- go to MODE. There
select RADIAN for the angle measure and POL (for Polar) on the FUNC (function)line. When Y = is pressed, note
that the equation has changed from y = to r = . There input the polar equation. After pressing graph, if you can’t see
the full graph, adjust x- and y- max/min, etc in WINDOW.
Example A
Plot the points on a polar coordinate graph:
Point A 2, π3
Point B (4, 135o )
Point C −2, π6
Solution
Below is the pole, polar axis and the points A, B and C.
3
1.1. Polar Coordinates
Example B
Plot the following points:
a. (4, 30o )
b. (2.5, π)
c. −1, π3
d. 3, 5π
6
e. (−2, 300o )
Solution
4
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Chapter 1. Polar Equations and Complex Numbers
Example C
Use a graphing calculator or plotting program to plot the following equations:
a. r = 1 + 3sinθ
b. r = 1 + 2cosθ
Solution
a.
b.
Review the steps above under Graphing using technology if you are having trouble.
5
1.1. Polar Coordinates
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Vocabulary
The polar coordinate system is a specialized graph used for angles and angle manipulations.
The pole is the center point on a polar graph.
One radian is the angle formed by moving counter-clockwise around the circumference of a circle by the length of
the radius. It is equal to apx 57.3 degrees.
The polar axis is a ray drawn from the pole at the 0o angle on a polar graph.
Guided Practice
1) Plot the points on a polar graph:
a) 2, π3
b) (3, 90o )
c) (1.5, π)
2) Convert from radians to degrees:
a) π2
b) 5.17
c) 3π
2
3) Convert from degrees to radians:
a) 251o
b) 360o
c) 327o
4) Convert from degrees to radians, answer in terms of π:
a) 90o
b) 270o
c) 45o
Answers
1) The points are plotted on the graph below:
6
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Chapter 1. Polar Equations and Complex Numbers
2) Recall that πrad = 180o and 1rad =
180
π
≈ 57.3o
a) If πrad = 180o then π2 rad = 90o
b) If 1rad ≈ 57.3o then 5.17rad ≈ 296o
o
c) If πrad = 180o then 3π
2 rad = 270
3) Recall that
180o
π
= 57.3o ≈ 1rad
a) If 57.3o ≈ 1rad then 251o ≈ 4.38rad ≈ 1.4πrad
b) If 57.3o ≈ 1rad then 360o ≈ 6.28rad
327o
c) If 57.3o ≈ 1rad then 57.3
o ≈ 5.71rad
4) Recall that 2πrad = 360o and therefore πrad = 180o
a) If πrad = 180o then π2 rad = 90o
b) If πrad = 180o and π2 rad = 90o then 1 21 πrad → 32 π →
c) If π2 rad = 90o then π4 rad = 45o
3π
2 rad
= 270o
Practice
1. Why can a point on the plane not be labeled using a unique ordered pair (r, θ)
2. Explain how to graph (r, θ) if r < 0 and/or θ > 360
Graph Each Point in the Polar Plane
3. A (6, 145o )
4. B −2, 13π
6
7
1.1. Polar Coordinates
o
5. C 47 , −210
6. D 5, π2 7. E 3.5, −π
8
Name Two Other Pairs of Polar Coordinates for Each Point
8. (1.5, 170o )
π
9. −5, −3
10. (3, 305o )
Graph Each Polar Equation
11.
12.
13.
14.
15.
r=3
θ = π5
r = 15.5
r = 1.5
θ = −175o
Find the Distance Between Points
16. P1 5, π2 and P2 7, 3π
9
17. P1 (1.3, −52o ) and P2 (−13.6, −162o )
18. P1 (3, 250o )P2 (7, 90o )
8
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Chapter 1. Polar Equations and Complex Numbers
1.2 Polar and Cartesian Transformation
Here you will learn how to convert points and expressions between polar form and rectangular (Cartesian coordinate)
form.
You will see during this lesson that points can be converted from rectangular form to polar form with a little Algebra
and Trigonometry.
Can the equation of a shape be converted also? How about a circle, for instance?
Watch This
Embedded Video:
MEDIA
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- Khan Academy: Polar Coordinates 2
Guidance
Polar Form to Rectangular Form
Sometimes a problem will be given as a coordinate in polar form but rectangular form may be needed.
To transform the polar point 4, 3π
4 into rectangular coordinates: first identify (r, θ)
r = 4 and θ =
3π
4 .
9
1.2. Polar and Cartesian Transformation
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Second, draw a vertical line from the point to the polar axis (the horizontal axis). The distance from the pole to
where the line you just drew intersects the polar axis is the x value, and the length of the line segment from the point
to the polar axis is the y value.
These distances can be calculated using Trigonometry:
x = r cos θ and y = r sin θ
√
√
3π
and
y
=
4
sin
or
x
=
−2
x = 4 cos 3π
2
y
=
2
2
4
4
√
√
4, 3π
4 in polar coordinates is equivalent to (−2 2, 2 2) in rectangular coordinates.
Rectangular Form to Polar Form
Going from rectangular coordinates to polar coordinates is also possible, but it takes a bit more work. Suppose we
want to find the polar coordinates of the rectangular point (2, 2). To begin doing this operation, the distance that the
point (2, 2) is from the origin (the radius, r) can be found by
p
r = x 2 + y2
p
r = 22 + 22
√
√
r= 8=2 2
The angle that the line segment between the point and the origin can be found by
tan θ =
tan θ =
y
x
2
2
tan θ = 1
θ = tan−1 1
θ=
π
4
Since this point is in the first quadrant (both the x and y coordinate are positive) the angle must be 45o or π4 radians.
It is also possible that when tan θ = 1 the angle can be in the third quadrant, or 5π
4 radians. But this angle will not
satisfy the conditions of the problem, since a third quadrant angle must have both x and y negative.
Note: when you use using tan θ = xy to find the measure of θ you should consider, at first, the quotient tan θ = xy and find the first quadrant angle that satisfies this condition. This angle will be called the reference angle, denoted
10
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Chapter 1. Polar Equations and Complex Numbers
θre f . Find the actual angle by analyzing which quadrant the angle must be given the signs of x and y.
Example A
Transform the polar coordinates 2, 11π
to rectangular form
6
Solution
r = 2 and θ =
11π
6
x = r cos θ and y = r sin θ
√
11π
x = 2 cos 11π
and
y
=
2
sin
or
x
=
3
2 y = −1
6
6
√
2, 11π
is equivalent to (3 2, −1) or in decimal form, approximately (4.342, −1)
6
Example B
√
Find the Polar coordinates for (3, −3 3)
√
x = 3 and y = −3 3
Solution
Draw a right triangle in standard form. Find the distance the point is from the origin and the angle the line segment
that represents this distance makes with the +x axis:
11
1.2. Polar and Cartesian Transformation
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q
√
r = 32 + (−3 3)2
√
= 9 + 27
√
= 36
=6
And for the angle,
√ tan θre f = (−3 3 3) √
tan θre f = 3
√
θre f = tan−1 3
θre f =
π
3
So, θre f =
π
3
and we can look at the signs of x and y – (+, -) – to see that θ =
√
The rectangular point (3, −3 3) is equivalent to the polar point 6, 5π
3 .
5π
3
since it is a 4th quadrant angle.
Recall that when solving for θ, we used
√ √
tan θ = (−3 3 3) or tan θ = 3
We found
2π
θ = 5π
3 . BUT, θ could also be θ = 3 . You must examine the signs of each coordinate to see that the angle must be in
the fourth quadrant in rectangular units or between 3π
2 and 2π in polar units. Of the two possible angles for θ, only
5π
−1 on a calculator you will always get an answer in the range − π ≤ θ ≤ π .
is
valid.
Note
that
when
you
use
tan
3
2
2
Example C
Convert the following rectangular coordinates to polar coordinates
√
a. (3, 3 3)
b. (−2, 2)
Convert the following polar coordinates to rectangular coordinates:
c. 4, 2π
3
d. −1, 5π
6
Solutions
a. (6, 66◦ )
√
b. (2 2, 225◦ )
√
c. (−2, 2 3)
√
3
1
d.
2 ,−2
TABLE 1.1:
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Chapter 1. Polar Equations and Complex Numbers
TABLE 1.1: (continued)
Concept question wrap up
Equation of a circle
x2 + y2 = k2 is the equation of a circle with a radius of k in rectangular coordinates.
The equation of a circle is extremely simple in polar form. In fact, a circle on a polar graph is analogous to a
horizontal line on a rectangular graph!
You can transform this equation to polar form by substituting the polar values for x, y. Recall x = r cos θ and y =
r sin θ
(r cos θ)2 + (r sin θ)2 = k2 ,
square the terms: r2 cos2 θ + r2 sin2 θ = k2 ,
factor the r2 from both terms on the left: r2 (cos2 θ + sin2 θ) = k2
recall the identity: cos2 θ + sin2 θ = 1
r 2 = k2
Therefore: r = ±k is an equation for a circle in polar units.
When r is equal to a constant, the polar graph is a circle.
Vocabulary
Quadrants are each of the four "corners" of a graph.
The sine of an angle may be found by dividing the length of the side opposite the angle in question by the length of
the hypotenuse.
The cosine of an angle may be found by dividing the length of the side adjacent (next to) the angle in question by
the length of the hypotenuse.
The tangent of an angle may be found by dividing the length of the side opposite the angle in question by the length
of the side adjacent to it.
Guided Practice
1) Change (3, −45o ) to rectangular coordinates.
2) Change −3, π4 to rectangular coordinates.
3) Express the equation in rectangular form: r = 6cosθ
4) Express the equation in rectangular form: r = 6
Answers
1) To change the description of the point to rectangular, first find the x-value, then the y-value, as follows:
x-coordinate:
x = r · cosθ : the x coordinate is found by multiplying r by the cosine of θ
x = 3 · cos(−45o ) : substitute the given information for r and θ
x = 2.121
13
1.2. Polar and Cartesian Transformation
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y-coordinate:
y = r · sinθ : the y coordinate is found by multiplying r by the sine of θ
y = 3 · sin(−45o ) : substitute the given information for r and θ
y = −2.121
∴ (2.121, −2.121) is the location of (3, −45o ) in rectangular form.
2) To change −3, π4 to rectangular coordinates, use the same process as Q #1:
x-coordinate:
−3cos π4 : substituting the values from the problem into x = r · cosθ
−3cos45 : π4 = 45o
x = −2.121
y-coordinate
−3sin π4 : substituting the values from the problem into y = r · sinθ
−3sin45 : π4 = 45o
y = −2.121
∴ (−2.121, −2.121) is the location of −3, π4 in rectangular form.
3) To express r = 6cosθ in rectangular form:
r2 = 6rcosθ : multiply both sides by r
x2 + y2 = 6x : Using x2 + y2 = r2 and x = rcosθ
∴ x2 + y2 = 6x is the equation in rectangular form.
4) This one is easy:
r = 6 is the polar form of the equation for a circle
r2 = 62 : square both sides
x2 + y2 = 36 : Using x2 + y2 = r2 and simplifying
∴ x2 + y2 = 36 is the equation in rectangular form.
Practice
1. How is the point with polar coordinates (5, π) represented in rectangular coordinates?
Plot each point below in polar coordinates (r, θ). Then write the rectangular coordinates (x, y) for the point.
2. (3, 60o ) 3. −10, π3
4. (15, π)
The rectangular coordinates (x, y) are given. For each question: a) Find two pairs of polar coordinates (r, θ), one
with r >0 and the other with r <0. b) Express θ in radians, and round to the nearest hundredth.
5. (5, −5)
14
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Chapter 1. Polar Equations and Complex Numbers
6. (0, 10)
7. (−8, 6)
Transform each polar equation to an equation using rectangular coordinates. Identify the graph, and give a rough
sketch or description of the sketch.
8.
9.
10.
11.
π
θ = 10
r=8
rsinθ = 7
rcosθ = −3
Transform each rectangular equation to an equation using polar coordinates. Identify the graph, and give a rough
sketch or description of the sketch.
12.
13.
14.
15.
x2 + √
y2 − 2x = 0
y = 3x
y = −5
xy = 15
15
1.3. Systems of Polar Equations
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1.3 Systems of Polar Equations
Here you will learn about finding points of intersection of shapes on polar graphs.
You likely recall that when you graph multiple equations on the same line, you usually end up with locations on the
graph where they intersect (unless you are graphing parallel lines!).
The same is true when graphing equations in polar form, and/or on a polar graph. When you graph the intersection
of multiple polar equations, you treat them just as you would rectangular equations, graph both and find the areas
that are true for both equations.
Watch This
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- anortonMovies: Example: Finding Points of Intersection on a polar graph
Guidance
Polar equations can be graphed using polar coordinates. Graphing two polar equations on the same set of axes may
result in having point(s) of intersection.
All points on a polar graph are coordinates that make the equation valid. The coordinates of point(s) of intersection
when substituted into each equation will make both of the equations valid.
One method to find point(s) of intersection for two polar graphs is by setting the equations equal to each other.
Call the first equation r1 and the second equation r2 .
Points of intersection are when r1 = r2 , so set the equations equal and then solve the resulting trigonometric equation.
Example A
Find the intersection of r1 = 3 sinθ
r2 =
√
3 cos θ
Solution
Set the equations equal to each other: 3 sin θ =
√
3 cos θ
divide both sides of the equation by cos θ and 3:
√
sin
θ
Simplify: cos θ = 3 3
sin θ = tan θ
Use the identity: cos
θ
16
3 sin θ
3 cos θ
=
√
3 cos θ
3 cos θ
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tan θ =
Chapter 1. Polar Equations and Complex Numbers
√
3
3
7π
θ = or 6
substitute π6 in either equation to obtain r = 1.5
substitute 7π
6 in either equation to obtain -1.5
Note: the coordinates 1.5, π6 and −1.5, 7π
6 represent
π
6
the SAME polar point so there is only one solution to this
equation.
Are we done? If we look at the graphs of r1 and r2 , we can see that there is another point of intersection:
when θ = 0, r1 = 3 sin θ = 3 sin(0) = 0
That means r1 = 3 sin θ goes through the pole (0, 0).
√
For r2 : when θ = π2 , r2 = 0 that is r2 = 3 cos θ goes through the point 0, π2 .
Therefore, both graphs go through the pole and the pole is a point of intersection.
The pole was NOT revealed as a point of intersection using the first step! (Why? Hint: How many ways are there to
represent the pole in polar coordinates?) This shows us that after you use algebraic methods to find intersections at
points other than the pole, you should also check for intersections at the pole.
Example B
Find the point(s) of intersection for the two graphs:
r1 = 1
r2 = 2 sin 2θ
Solution
Set r1 = r2 and solve:
1 = 2 sin 2θ
1
2
= sin 2θ
Use a substitution α = 2θ to solve
1
2
= sin α
α = π6 , 5π
6
Since α = 2θ , solving for θ gives us
θ=
π 5π
12 , 12
But, recall that θ has the range 0 ≤ θ ≤ 2π. Since we solved with 0 ≤ α ≤ 2π we actually need to consider values of
θ with 0 ≤ θ ≤ 4π. Why? Recall that sin(2θ) has two cycles between 0 and 2π, and so we add two more solutions,
17
1.3. Systems of Polar Equations
α=
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13π 17π
6 , 6
and since α = 2θ,
θ=
13π 17π
12 , 12
Finally, we need to consider solutions when r = -1 because r = 1 and r = -1 are the same polar equation. So, solving
− 12 = sin α
α=
7π 11π
6 , 6
Again, using α = 2θ and adding solutions for the repetition gives us four more solutions,
θ=
7π 11π 19π 21π
12 , 12 , 12 , 12
So in total, there are eight solutions to this set of equations.
Note: Recall-when solving trigonometric equations where the angle is θ requires looking at all potential values
between 0 and 2π. When the angle is 2θ as it is in this case, be sure to look for all potential values between 0 and
4π. When the angle is 3θ as it is in this case, be sure to look for all potential values between 0 and 6π., and so on.
2. Since r1 cannot equal 0, the pole is not on its graph and not a point of intersection.
3. The graph reveals eight points of intersection which were found in step (1).
Example C
Find point(s) of intersection, if any exist, for the following pair of equations:
r1 = 2
r2 = secθ
Solution
Here we will use a table of values for each function, solving by quadrant. Recall that the period of sec θ is 2π.
For the first quadrant
TABLE 1.2:
θ (angle)
r1 (distance)
r2
18
0
2
1
π/6
2
1.15
π/4
2
1.4
π/3
2
2
π/2
2
und
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Chapter 1. Polar Equations and Complex Numbers
For the second quadrant
TABLE 1.3:
θ (angle)
r1 (distance)
r2
2π/3
2
-2
3π/4
2
-1.4
5π/6
2
-1.15
π
2
-1
For the third quadrant:
TABLE 1.4:
θ (angle)
r1 (distance)
r2
7π/6
2
-1.15
5π/4
2
-1.4
4π/3
2
-2
3π/2
2
und
For the fourth quadrant
TABLE 1.5:
θ (angle)
r1 (distance)
r2
5π/3
2
2
7π/4
2
1.4
11π/6
2
1.15
2π
2
1
Note that the 3rd and 4th quadrant repeat 1st and 2nd quadrant values.
Observe in the table of values that (2, π/3) and (2, 5π/3) are the points of intersection. Look at the curves- the first
equation yields a circle while the second yields a line. The maximum number of intersecting points for a line and a
circle is 2. The two points have been found.
Vocabulary
Points of intersection are locations where two different equations have the same solutions.
Conic sections are members of a family of curved figures including circles, ellipses, parabolas, and hyperbolas
Guided Practice
Questions
1) Find the point(s) of intersection for this pair of Polar equations:
19
1.3. Systems of Polar Equations
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r = 2 + 4 sinθ
θ = 60°
2) Find the point(s) of intersection for this pair of Polar equations:
r1 = 2 cosθ
r2 = 1
Answers
1) The equation θ = 60o is a line making a 60° angle with the r axis.
Make a table of values for r = 2 + 4 sin θ
For the first quadrant:
TABLE 1.6:
θ (angle)
R (distance)
0
2
30
4
45
4.83
60
5.46
90
6
For the second quadrant:
TABLE 1.7:
θ (angle)
R (distance)
120
5.46
135
4.83
150
4
180
2
For the third quadrant:
TABLE 1.8:
θ (angle)
R (distance)
210
0
225
-.83
240
-1.46
270
-2
For the fourth quadrant
TABLE 1.9:
θ (angle)
R (distance)
300
-1.46
315
-.83
330
0
360
2
Notice there are two solutions in the table., (60, 5.46) and (240, -1.46) = (60, 1.46). Recall that when r <0, you plot
a point (r, θ), by rotating 180o (or π).
Finally, we need to check the pole: r = 2 + 4 sin θ passes through the pole for θ = 330o , and θ = 60o also passes
through the pole. Thus, the third point of intersection is (0, 0).
20
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Chapter 1. Polar Equations and Complex Numbers
2) Make a table:
For the first quadrant:
TABLE 1.10:
θ (angle)
r1 (distance)
r2
0
2
1
π/6
√
3
1
π/4
√
2
1
π/3
1
1
π/2
0
1
For the second quadrant:
TABLE 1.11:
θ (angle)
r1 (distance)
r2
2π/3
-1
1
3π/4
√
− 2
1
5π/6
√
− 3
1
π
-2
1
For the third quadrant:
TABLE 1.12:
θ (angle)
r1 (distance)
r2
7π/6
√
− 3
1
5π/4
√
− 2
1
4π/3
-1
1
3π/2
0
1
For the fourth quadrant:
TABLE 1.13:
θ (angle)
r1 (distance)
r2
5π/3
1
1
7π/4
√
2
1
11π/6
√
3
1
2π
2
1
So the unique solutions are at θ = π3 , 4π
3 . There are also two repeated solutions in this set (can you find them?).
Here is a graph showing the two solutions:
21
1.3. Systems of Polar Equations
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Practice
The graphs of r = 1 and r = 2cosθ are shown below.
1. How many times do they intersect?
2. In which quadrants do they intersect?
3. At what points do the intersections occur?
Based on the image below and the following information: the intersection of the graphs of r = cosθ and r = 1 − cosθ
22
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Chapter 1. Polar Equations and Complex Numbers
4. Identify how many times they intersect?
5. At what points do the intersections occur?
Find the points of intersection of the following pairs of curves.
6.
7.
8.
9.
r = 2r = 2cosθ
r = sin2θr = 2sinθ
r = 2 + 2sinθr = 2 − 2cosθ
r = 3cosθr = 2 − cosθ
Find the point(s) of intersection for each system of equations. Graph to verify your solution.
10.
11.
12.
13.
14.
15.
r1 = cscθr2 = 2sinθ
r1 = cosθr2 = 1 + sinθ
r1 = sinθr2 = sin2θ
r1 = −4sinθr2 = −4cosθ
p
r1 = 1 − 2sinθr2 = 9 cos(θ)
r1 = 1 − cosθr2 = 4cos(3θ)
23
1.4. Imaginary Numbers
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1.4 Imaginary Numbers
Here you will learn how to recognize and evaluate imaginary numbers, and you will explore the definition of the
term "complex number".
Have you ever had an imaginary pet? Many people have, particularly as young children.
Wouldn’t you be surprised if you and your real friend left your imaginary pet dogs alone together, and you came
back to find a real puppy?
Silly thought, what does it have to do with imaginary numbers?
Watch This
’Note:’ For a very detailed explanation of i and the complex numbers, watch the video below, then visit: http://bette
rexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/
Embedded Video:
MEDIA
Click image to the left for more content.
- Khan Academy: Introduction to i and imaginary numbers
Guidance
What is the square root of -1?
You may recall running into roots of negatives in Algebra, when attempting to solve equations like: x2 + 4 = 0 Since
there are no real numbers that can be squared to equal -4, this equation has no real solution. Enter the imaginary
constant: "i".
√
The definition of "i" : i = −1
The use of the word imaginary does not mean these numbers are useless. For a long period in the history of
mathematics, it was thought that the square root of a negative number was in fact only within the mathematical
imagination, without real-world significance hence, imaginary. That has changed. Mathematicians now consider the
imaginary number as another set of numbers that have real significance, but do not fit on what is called the number
line, and engineers, scientists, and others solve real world problems using combinations of real and imaginary
numbers (called complex numbers) every day.
√
√
√
√
Imaginary values such as −16 can be simplified by simplifying the radical into 16 · −1, yielding: 4 −1 or
4i.
The uses of i become more apparent when you begin working with increased powers of i, as you will see in the
examples below.
Complex numbers
24
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Chapter 1. Polar Equations and Complex Numbers
When you combine i’s with real numbers, you get complex numbers:
The definition of complex numbers: Complex
numbers are of the form a + bi, where a is a real number, b is a
√
constant, and i is the imaginary constant −1.
Example A
Simplify
√
−5
Solutions
p
√
−5 = (−1) · (5)
√ √
= −1 5
√
=i 5
Example B
Simplify
√
−72
Solution
p
√
−72 = (−1) · (72)
√ √
= −1 72
√
= i 72
But, we’re not done yet! Since 72 = 36 · 2
√ √
√
i 72 = i 36 2
√
= i(6) 2
√
= 6i 2
Example C
Strange things happen when the imaginary constant i is multiplied by itself different numbers of times.
a) What is i2 ?
b) What is i3 ?
c) What is i4 ?
Solutions
√
i2 is the same as ( −1)2 . When you square a square root, they cancel and you are left with the number originally
inside the radical, in this case −1
a) ∴ i2 = −1
i3 is the same thing as i2 · i, which is −1 · i or −i
b) ∴ i3 = −i
i4 = i2 · i2 which is −1 · −1
c) ∴ i4 = 1
25
1.4. Imaginary Numbers
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TABLE 1.14:
Concept question wrap up
Do you see the application of the crazy analogy from the introduction?
Two imaginary pets creating a real puppy is an oddly effective metaphor for the behavior of the powers of i.
One i is imaginary, but two i’s multiply to be a real number. In fact, every even power of i results in a real number!
Vocabulary
The imaginary constant i is the square root of -1.
A complex number is the sum of a real number and an imaginary number, written in the form: a + bi.
Simplifying the radical involves factoring the term(s) under the root symbol, so that perfect squares may be "pulled
out", and simplified.
Guided Practice
Simplify the following radicals
√
1) −9
√
2) −12
√
3) −17
√
4) 108 − 140
Multiply the imaginary numbers
5) 4i · 3i
√
6 16i · 3
√
√
7) 4i2 · 12i
Solutions
1) To simplify the radical
√
−9
√
√9 · −1
√ : Rewrite −9 as −1 · 9
√9 · −1 : Rewrite
√ as a product of radicals
9 · i : Substitute
−1 → i
√
3i : Simplify 9
2) To simplify the radical
√
−12
√
√12 · −1
√ : Rewrite −12 as −1 · 12
12
·
−1 : Rewrite
√
√ as a product of radicals
√12 · i : Substitute −1 → i
3 · 4 · i : Factor 12√
√
2 3 · i : Simplify 4
√
3) To simplify the radical −17
√
√17 · −1
√ : Rewrite −17 as −1 · 17
17 · −1 : Rewrite as a product of radicals
26
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Chapter 1. Polar Equations and Complex Numbers
√
√
√17 · i : Substitute −1 → i
17i : Simplify
4) To simplify the radical
√
108 − 140
√
√−32 : Subtract within the parenthesis
√32 · −1
√ : Rewrite −32 as −1 · 32
√32 · −1 : Rewrite
√ as a product of radicals
32
·
i
:
Substitute
−1 → i
√
16
·
2
·
i
:
Factor
32
√
√
4i 2 : Simplify 16
5) To multiply 4i · 3i
4 · 3 · i · i : Using the commutative law for multiplication
12 · i2 : Simplify
12 · −1 : Recall i2 = −1
−12
6) To multiply
√
16i · 3
4i · 3 : Simplify
12i
7) To multiply
√
16
√
√
4i2 · 12i
√ √ √ √
2·
4 · i√
4 · 3 · i : Factor
2 ·√i · 2 · 3 · i : Simplify the roots
4 √3 · i2 : Collect terms and simplify
4 √
3 · −1 : Recall i2 = −1
−4 3
Practice
Simplify:
1.
2.
3.
4.
5.
6.
7.
√
√−49
√−81
√−324
−121
√
− √−16
−
√ −1
−1.21
Simplify:
8.
9.
10.
11.
i8
i12
i3
24i20
27
1.4. Imaginary Numbers
12. i225
13. i1024
Multiply:
14.
15.
16.
17.
18.
28
i4 · i11
6 · 5i8
5i√
√
3 √−75 · 5 √−3
2 √
−12 · 6 √
−27 √
−4 −10 · 5 −3 · 6 −18
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Chapter 1. Polar Equations and Complex Numbers
1.5 Complex Numbers
Here you will explore complex numbers: the sums of real and imaginary numbers. You will also learn about the
complex plane, and how to use it to graph complex numbers.
Plotting points was something you did back in early Algebra, and it likely seems pretty simple now. For instance,
plotting the point (4, 5) meant starting at the origin and moving 4 units to the right - the x direction, and 5 units up the y direction.
In this lesson, one of the things we will consider is the graphing of complex numbers such as 4 + 3i.
In essence it doesn’t sound that hard, but which direction do you only imagine moving 3 units?
Watch This
Embedded Video:
MEDIA
Click image to the left for more content.
- Khan Academy: Complex Numbers (part 1)
Guidance
Sometimes when you solve a quadratic equation, the solution has both a real part and an imaginary part. For example,
if you want to solve
(x − 1)2 + 4 = 0
then
(x − 1)2 = −4
√
x − 1 = ± −4
√ √
x − 1 = ± −1 4
x − 1 = ±2i
x = 1 ± 2i
x = 1 + 2i or 1 − 2i
Notice that these two solutions involve a real part, 1, and an imaginary part, ±2i
a + bi is the standard or rectangular form of a complex number.
A complex number is a number that has a real part (in this case a) and an imaginary part, that is, the imaginary
number i with a coefficient b.
29
1.5. Complex Numbers
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The complex numbers are a superset of the real numbers, meaning that all of the real numbers are part of the set of
complex numbers:
Given ’a + bi, if b = 0 (meaning there is no imaginary part to the complex number), then all you have remaining is a
real number. Viewed this way, every real number can be written as a complex number (just let it equal a), but there
are many more complex numbers than real numbers. Hence the complex numbers are a superset of the real numbers.
Graphing complex numbers
In standard form, a + bi, a complex number can be graphed using rectangular coordinates (a, b). The x - coordinate
represents “real number” values, while the y - coordinate represents the “imaginary” values.
Example A
Graph the complex number: z = 2 + 2i, in rectangular form.
Solution
The coordinate (2, 2) is graphed as shown below:
30
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Chapter 1. Polar Equations and Complex Numbers
Example B
Solve each equation and express it as a complex number. (Note: If the imaginary part is 0, express the solution as a
+ 0i)
a) x2 + 24 = 0
b) 2x2 - 4x + 7 = 0
Solutions
√
a) x = ±2 6i
√
b) x = 1 ± 2 3i
Example C
Plot each of the following complex numbers in rectangular form:
a. (4 + 2i)
b. (-3 + i)
c. (3 - 4i)
d. 3i
Solution
Your graph should look like:
31
1.5. Complex Numbers
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TABLE 1.15:
Concept question wrap up
Of course you don’t actually just imagine moving in one direction, but it is kind of a funny thought!
When graphing complex numbers in standard form: a + bi, the real component a is plotted on the horizontal or x
axis. The imaginary component bi is plotted on the y axis in just the same way a real number would be.
∴ 4 + 3i would be located 4 units to the right and 3 units up from the origin.
Vocabulary
Complex numbers are the combination (technically the ’sum’) of a real number and an imaginary number. They are
written in standard form as a + bi, where a is the real number and b is the coefficient of the imaginary number.
The complex plane is the graphical representation of the set of all complex numbers.
A superset is a set that includes other sets within it. The complex number set includes all classic complex numbers
a + bi, and also, because bi may equal 0, it also includes all of the real numbers as well. This makes it a superset of
the set of real numbers.
Guided Practice
√
√
1) Simplify and express as a complex number − 60 + −121
2) Solve the equation and express the answer as a simplified complex number x(4x) + 4 = 0
3) Graph the complex numbers:
a) 3 + 2i
b) 2 - 3i
c) -2 + 2i
Solutions
√
√
1) To simplify − 60 + −121
√
√
− 4 · 15 + 121 · −1 Factor the numbers under the roots
32
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Chapter 1. Polar Equations and Complex Numbers
√
√
−2 √15 + 11 −1 Simplify the perfect square roots
−2 15 + 11i
2) To solve x(4x) + 4 = 0
4x2 + 4 = 0 Distribute the x
4x2 = −4 Subtract 4 from both sides
x2 = −1 Divide both sides by 4
x = i Take the square root of both sides
3) To graph the complex numbers, graph the real value on the x axis, and the imaginary value on the y axis. The
image below shows the points correctly graphed.
Practice
1. What are complex numbers technically the sum of?
2. What does the complex plane represent?
Express in simplest form in terms of i
√
3. 13 − −49
r
√
−16
4. 75 +
r 256
−25
5. −3 −
169
√
√
6. − 36
+
−64
r
−4
7. 10 −
36
33
1.5. Complex Numbers
8.
9.
10.
11.
12.
13.
14.
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√
4 +√ −250
2 −0.0009
3√
√
−0.16
√ − (− 27)
10 + √−0.49
−3
p+ −0.0196
√
9p −8 j9 + 3 25
√
−676b3 a1 c8 + 112
Graph Complex Numbers
15. Two complex numbers are graphed below. What are the numbers expressed in standard complex number
form?
16.
17.
18.
19.
34
Graph the complex number 3 − 4i
Graph the complex number −2 + 3i
Graph the complex point (3 + i)
Graph the complex point (−1 − 2i)
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Chapter 1. Polar Equations and Complex Numbers
1.6 Quadratic Formula and Complex Sums
Here you will explore complex number solutions of the quadratic formula, and will learn how to add and subtract
complex numbers.
You probably remember when you first learned to use the Quadratic formula in Algebra and you ended up with a
negative number under the root symbol. Chances are good that your instructor simply said something like "If you
get a negative number under the root, there are no real answers, since there is no such thing as the root of a negative!"
Now that you are familiar with imaginary numbers, you can probably see that although it would have been an easy
assumption at the time that "no real answers" just meant "no answers at all," that isn’t true. "No real answers" may
well mean that there ARE some "unreal" or imaginary answers.
Watch This
Embedded Video:
MEDIA
Click image to the left for more content.
- James Sousa: Ex. 1: Adding and Subtracting Complex Numbers
Guidance
Review: The Quadratic Formula and the Discriminant
If ax2 + bx + c = 0
√
−b± b2 − 4ac
then x =
2a
Recall that b2 - 4ac is called the discriminant.
TABLE 1.16:
If b2 - 4ac >0 then
there are two unequal real solutions.
If b2 - 4ac = 0 then
there are two equal
real solutions.
If b2 - 4ac <0 then
there are two unequal complex solutions.
35
1.6. Quadratic Formula and Complex Sums
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Sums and Differences of Complex Numbers
When adding (or subtracting) two or more complex numbers the fastest method is to add (or subtract) the real
components to obtain the sum of the real numbers, and then separately add (or subtract) the imaginary coefficients
to obtain the sum of the imaginary numbers, e.g.:
(a + bi) + (c + di) = [a + c] + [b + d]i
Example A
Combine the complex numbers using addition or subtraction
a) (5 + 3i) + (6 − 8i)
b) (3 − 2i) − (2 − 4i)
c) (6) + (4 − 3i)
Solutions
a) Applying the commutative property: (5 + 6) + (3i + −8i) = 11 − 5i
MEDIA
Click image to the left for more content.
b) Distribute the negative: (3 − 2i) + (−2 + 4i), then apply commutation: (3 − 2) + (−2i + 4i) = 1 + 2i
c) The imaginary coefficient in the first term is 0, so applying commutative property gives: (6 + 4) + (0i − 3i) =
10 − 3i
Example B
Given: x2 + 4x + 6 = 0
a) Use the discriminant to predict the nature of the roots
b) Use the Quadratic Formula to solve and identify the roots
c) Express the roots as complex numbers in standard form
Solutions
x2 + 4x + 6 = 0
a = 1, b = 4, c = 6
p
(4)2 − (4)(1)(6)
−4±
x=
2(1)
a) Since b2 − 4ac = −8, there will be 2 complex solutions (no real solutions)
√
x = −4± 2 −8
√
2
b) x =
√
c) x = −2 ± i 2
−4±2i
2
36
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Chapter 1. Polar Equations and Complex Numbers
Example C
Graphing calculator exercise
Add or subtract the complex numbers using a graphing calculator:
a) (4 − 5i) − (3 + 2i)
b) (3 − 7i) + (2 + i)
c) (−2i) + (2 + 6i)
Solution
A graphing calculator can perform operations with complex numbers. Press mode. Scroll down and select a + bi
Press Quit. Now the calculator is able to perform operations with complex numbers in a + bi form. When the
calculator is in complex number mode, be sure to use parenthesis to group the parts of the complex numbers.
Your answers should look like:
a) 1 − 7i
b) 5 − 6i
c) 2 + 4i
Vocabulary
The discriminant is the part of the Quadratic Formula under the radical: b2 − 4ac. A positive discriminant suggests
2 real roots to the quadratic equation, a zero suggests 1 real root and 1 complex root, and a negative indicates 2
complex roots.
A complex root is a complex number that, when used as an input (x) value, results in an output (y) value of zero.
Guided Practice
1) Add the complex numbers
a) (3 + 4i)√+ (9 + 13i) √
b) (16 − −64) + (7 − −81)
2) Subtract the complex numbers
a) (9 + 7i)√− (6 + 3i)
√
b) (18 + −81) − (18 − −64)
3) Solve the equations and express them as complex numbers
a)18x2 − 2x + 24 = 0
b)12 45 x2 = 14 25 x − 11 15
Solutions
1 a) To add the complex numbers (3 + 4i) + (9 + 13i)
(3 + 9)(4i + 13i) Group the real parts and the imaginary parts
12 + 17i Combine like terms and simplify
37
1.6. Quadratic Formula and Complex Sums
√
√
b) To add (16 − −64) + (7 − −81)
(16 − 8i) + (7 − 9i) Simplify the roots in terms of i
16 − 8i + 7 − 9i Distribute the negative
23 − 17i Collect like terms and simplify
2 a) To subtract(9 + 7i) − (6 + 3i)
(9 + 7i) + (−6 − 3i) Distribute the negative
(9 + −6) + (7i + −3i) Group the real part and the imaginary part of each
3 + 4i Combine like terms
√
√
b) To subtract (18 + −81) − (18 − −64)
(18 + 9i) − (18 − 8i) Simplify the roots
18 + 9i − 18 + 8i Distribute the negative
18 − 18 + 9i + 8i Group real and imaginary parts
17i Simplify
3 a) To solve 18x2 − 2x + 24 = 0
9x2 − x + 12 = 0 Divide both sides by 2
A =p9|B = −1|C = 12 Identify A, B, and C using standard form: Ax2 + Bx +C = 0
√
1±
1 − 4(9)(12)
−B± B2 − 4AC
Substitute
the
terms
into
the
Quadratic
Formula
2A
2(9)
x=
1±i
√
431 By the Quadratic Formula
18
b) To solve 12 45 x2 = 14 52 x − 11 51
64 2
72
56
5 x − 5 x + 5 = 0 Convert to improper fractions
2
64x − 72x + 56 = 0 Multiply both sides by 5
8x2 − 9x + 7 = 0 Divide both sides by 8
A = 8|B√= −9|C = 7 Extract values for the Quadratic Formula
x = 9±i 143 By the Quadratic Formula
16
Practice
Add the complex numbers
1.
2.
3.
4.
5.
(4 + 4i) + (3 + 6i)
(4 +
√8i) + (8 +√8i)
20√ −100 + √
−9
3 −121
√ + 15 √−81
−25 −9 + 23 −196
Subtract Complex Numbers
6.
7.
8.
9.
38
(9 + 8i) − (8 + 5i)
(6 −√8i) − (12 +√
9i)
−5√ −49 − 16√ −1
22 −144 − 9 −4
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Chapter 1. Polar Equations and Complex Numbers
√
√
10. 27 −196 − −1
Solve each equation and express the result as a complex number.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
12x 45 x = 3 15 x2 + 16
3x2 − 6x + 15 = 0
8x2 − 5x + 11 = 0
34 21 x2 − 23x + 19 16 = 0
−36x2 − 18x + 6 = 27
When the sum of -4 + 8i and 2 - 9i is graphed, in which quadrant does it lie?
If z1 = −3 + 2i and z2 = 4 − 3i, in which quadrant does the graph of (z2 − z1 ) lie?
On a graph, if point A represents 2 − 3i and point B represents −2 − 5i, which quadrant contains A − B ?
Find the sum of (−3 + 4i) and (−4 − 7i) and graph the result
Graph the difference of (4 + 7i) and (2 − 3i)
39
1.7. Products and Quotients of Complex Numbers
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1.7 Products and Quotients of Complex Numbers
Here you will learn how to multiply and divide complex numbers and explore the concept of conjugates.
Working with complex numbers can be decidedly strange. It is natural to wonder why we put so much time and
effort into learning how to properly manipulate numbers that don’t even exist! There are, however, many real and
rather fun and interesting applications of complex numbers in the "real world."
This website: http://www.picomonster.com/ has a fascinating interactive application using complex numbers. Check
it out.
Watch This
Embedded Video:
MEDIA
Click image to the left for more content.
- James Sousa: Complex Number Operations
Guidance
In this lesson, we will explore the multiplication and division of complex numbers. In general, operations on complex
numbers behave just like operations on regular polynomials: combine like terms, FOIL binomials, distribute, and so
on. The differences are generally apparent when the products and quotients of complex numbers are simplified.
Recall the unusual case of i2 :
i=
√
√
−1 ∴ i2 = ( −1)2 = −1
Since i2 = −1, multiplying complex numbers often results in imaginary terms becoming real terms.
Multiplying Complex Numbers
When multiplying complex numbers in rectangular form, recall the method for multiplying two binomials (sometimes called FOIL): (m + n)(x + y) = mx + my + nx + ny. We use the same procedure for multiplying complex
numbers:
(a + bi) + (c + di) = ac + adi + bci + bdi2
But, unlike the algebraic expression, the above expression contains the number, i.
Recall that i2 = -1, so bdi2 = bd(-1) = -bd and
adi + bci can be combined and then factored as (ad + bc)i. Thus we have the general result,
(a + bi) + (c + di) = (ac - bd) + (ad + bc)i
Dividing Complex Numbers
40
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Chapter 1. Polar Equations and Complex Numbers
To divide two complex numbers is similar to dividing two irrational numbers. Recall that in that problem, the
procedure was to find the irrational conjugate of the denominator and then multiply both the numerator and the
denominator by that conjugate, for example:
3√
Divide:
1+
2
First find the irrational conjugate of the denominator: 1 −
by this value:
√
√
1− √2
3−3 2
3√
×
= 1−2
1+ 2
1− 2
this reduces to
√
3−3 2
= −1
√
2, then multiply both the numerator and the denominator
or
√
= −3 + 3 2
The process is very much the same with complex numbers. With complex numbers, since you are interested in
eliminating the complex numbers from the denominator, you find the complex conjugate of the denominator and
multiply BOTH the numerator AND the denominator by it.
You find the complex conjugate in the same way you found the conjugate of irrational numbers, change the sign of
the imaginary part. For instance, the complex conjugate of 4 + 3i is 4 – 3i
A complex number multiplied by its complex conjugate will yield a real number. By recalling (a + b)(a - b) = a2 b2 simplifying a complex number and its conjugate can be easy, for example:
The conjugate of 4 + 3i is found by retaining the real part (4) and reversing only the sign of the imaginary part (that
is, 3i becomes -3i)
(4 + 3i)(4 - 3i) = 16 - 12i + 12i - 9i2 . Notice that -12i and 12i cancel. Also recall that i2 = -1
That gives us:
16 + 9 = 25
Therefore: (4 + 3i)(4 – 3i) = 25
The product of this complex number and its conjugate is 25.
When multiplying complex numbers sometimes intuition about the nature of the product can mislead.
For example in (a + b)(a + b), where all of the terms are real numbers, no terms of each of the four products will
cancel. Some of the terms may be combined. However in (1 + i)(1 - i), where some terms are real numbers and some
terms are imaginary numbers, this is no longer true. Two of these terms cancel: the first product yields 1 while the
last product yields i2 or -1, and those terms cancel!
Example A
Multiply: (6 + 3i)(2 − 3i)
Solution
(6 + 3i)(2 − 3i) = 12 − 18i + 6i − 9i2
= 21 − 12i
(Note when combining like terms: -9i2 reduces to -9(-1) or 9)
41
1.7. Products and Quotients of Complex Numbers
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Example B
Multiply: (5 − 7i)(5 + 7i)
Solution
(5 − 7i)(5 + 7i) = 25 + 35i − 35i − 49i2
or
= 25 + 0 − 49(−1)
= 74
Example C
Find the quotient:
6−3i
4+3i
Solution
First, observe that the complex conjugate of the denominator is 4 – 3i
Multiply both the numerator and the denominator by 4 – 3i:
=
6−3i
4+3i
× 4−3i
4−3i =
24−18i+12i+9i2
16−12i+12i−9i2
15−6i
25
6−3i
4+3i
=
15−6i
25
Vocabulary
Conjugates are binomial terms which are equal aside from inverse operations between them, e.g. (3 + 2x) and (3 2x).
Complex conjugates such as: (3 + 2i) and (3 - 2i) result in real numbers when multiplied.
Guided Practice
1) Multiply (6 + 7i)(2 + 3i)
2) Multiply (3 + 6i)(2 − 8i)
3) Divide
5+2i
7+4i
4) Divide
4+i
5−i
Solutions
1) To multiply (6 + 7i) · (2 + 3i) treat each complex number as a binomial and apply the FOIL method to find the
product.
TABLE 1.17:
6 · 2 = 12
6 · 3i = 18i
7i · 2 = 14i
7i · 3i = −21
multiply the First terms
multiply the Outer terms
multiply the Inner terms
multiply the Last terms and simplify the i2
12 + 18i + 14i − 21 → 12 + 32i − 21 → −9 + 32i Combine like terms and simplify
∴ (6 + 7i) · (2 + 3i) = −9 + 32i
42
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Chapter 1. Polar Equations and Complex Numbers
2) To multiply (3 + 6i) · (2 − 8i) treat each complex number as a binomial and apply the FOIL method to find the
product.
TABLE 1.18:
3·2 = 6
3 · −8i = −24i
6i · 2 = 12i
6i · −8i = +48
multiply the First terms
multiply the Outer terms
multiply the Inner terms
multiply the Last terms and simplify the i2
6 − 24i + 12i + 48 → 6 − 12i + 48 → 54 − 12i Combine like terms and simplify
∴ (3 + 6i) · (2 − 8i) = 54 − 12i
3) To divide divide 5+2i
7+4i first determine the conjugate of the denominator (7 + 4i) the conjugate is the same two terms
with the opposite sign between them, in this case (7 − 4i)
TABLE 1.19:
5+2i
7+4i
•
7−4i
Multiply the numerator and denominator by the conjugate
FOIL
Simplify
Simplify
7−4i
35+14i−20i−8i2
49−28i+28i−16i2
35−6i+8
49+16
43−6i
65
∴
5+2i
7+4i
=
43−6i
65
4) To divide
4+i
5−i
first identify the conjugate of the denominator: (5 + i)
TABLE 1.20:
4+i
5−i
5+i
5+i
•
20+4i+5i+i2
25+5i−5i−i2
35+9i−1
25+1
34+9i
26
∴
4+i
5−i
=
Multiply the numerator and denominator by the conjugate
FOIL
Simplify
Simplify
34+9i
26
Practice
1. (−6i) + (−4i)
2. (12i + 1) + (6i + 11)
Multiply Complex Numbers
3.
4.
5.
6.
7.
8.
−7i(7i + 5)
(2i + 3)2
(−2i)(−11i)(−12i)
(8i√
− 5)(11i3 −√9)
(9 √
−9) · (14 √
−4)
(10 −1) · (24 −144)
43
1.7. Products and Quotients of Complex Numbers
√
√
9. (18 −49) · (5 −144)
Divide Complex Numbers
16.
11i−2
i+1
10i+10
−8i−10
7i+12
−12i−8
10i+5
−i−7
4i2 +4
−6i2 −1
5i2 −5
2 +3
−8i√
−4 √−60
17.
−100√ −154
18.
48 √−170
10.
11.
12.
13.
14.
15.
44
−2
10
√
8
−20
√
−22
−17
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Chapter 1. Polar Equations and Complex Numbers
1.8 Polar Form of Complex Numbers
Here you will learn how to convert complex numbers from rectangular form to polar form. You will also explore the
graphs of complex numbers on a polar graph.
Complex numbers can be graphed on a polar graph just like real numbers can. You will discover during this lesson
that there are actually a few different ways of doing this.
Watch This
Embedded Video:
MEDIA
Click image to the left for more content.
- James Sousa: Complex Numbers in Trigonometric Form
Guidance
You have learned that rectangular graphs can be put into polar form, and that points in rectangular coordinates can
be plotted in the polar coordinate system. In this section you will learn how to do the same process with complex
numbers.
There are three common forms of complex numbers that you will see when graphing:
a) In the standard form of: z = a + bi, a complex number z can be graphed using rectangular coordinates (a, b). ’a’
represents the x - coordinate, while ’b’ represents the y - coordinate.
b) The polar form: (r, θ) which we explored in a previous lesson, can also be used to graph a complex
p number.
2
2
Recall that you
can use x and y to convert between rectangular and polar forms with: r = x + y and
y
tan θre f = x . Unfortunately, there is a problem with using a conversion from rectangular form to polar form
like:
a + bi → (r, θ)
or
√
−1 − i 3 → 2, 4π
3
The problem is that we have lost the i. So, in order to “keep track” of the imaginary part, we can use another form.
c) The third form, often abbreviated as rcisθ, short for: z = r(cosθ + isinθ), will be used quite often as you progress.
This form comes from the substitutions: x = r cos θ and y = r sin θ.
Using this fact, and sample values of 2 for r and π3 for θ, we can write
45
1.8. Polar Form of Complex Numbers
√
4π
z = −1 − i 3 = 2 cos 4π
3 + 2 i sin 3
Finally, factoring the 2, we get: z = 2 cos
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4π
3
+ i sin
4π
3
Summary of forms
√
√
The complex number: z = −1− 3i, the rectangular point (−1, − 3), the polar point: 2, 4π
3 , and 2 cos
or 2 cis 4π
3 all represent the same number.
4π
3
+ i sin
Steps for Conversion
To convert from polar to rectangular form, the distance that the point (2, 2) is from the origin can be found by
p
p
√
√
d = x2 + y2 or 22 + 22 d = 8 or 2 2
The reference angle (i.e. the corresponding angle in the first quadrant) that the line segment between the point and
the origin can be found by
tan θre f = y x
for z = 2 + 2i,
tan θre f =
2
2
tan θre f = 1.
Since this point is in the first quadrant (both the x and y coordinate are positive) the angle must be 45o or
5π
4
π
4
radians.
It is also possible that when tan θ = 1 the angle can be in the third quadrant or
radians. But this angle will not
satisfy the conditions of the problem, since a third quadrant angle must have both x and y as negatives.
Note: When using tan θ = xy , you should first consider, the quotient xy and find the first quadrant angle that satisfies
this condition. This angle will be called the reference angle, denoted θre f . Find the actual angle by analyzing which
quadrant the angle must be given the x and y signs.
√
The complex number 2 + 2i or (2, 2) in rectangular form has polar coordinates 2 2, π4
Example A
√
Graph in polar form: z = −1 − i 3.
Here is what it looks like in the rectangular coordinate system:
46
4π
3
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Chapter 1. Polar Equations and Complex Numbers
Solution
In Polar form, we find r with
p
r = a2 + b2
q
√
= (−1)2 + (− 3)2
√
= 1+3
√
= 4
=2
and to find θ,
√ tan θre f = −−1 3 √
tan θre f = 3
√
θre f = tan−1 3
θre f =
π
3
Since this angle is in the 4th quadrant, θ =
4π
3 .
Example B
√
Find the Polar coordinates that represent the complex number z = 3 − 3 3i
Solution
√
√ a = 3 and b = −3 3: the rectangular coordinates of the point are 3, −3 3 .
Now, draw a right triangle in standard form. Find the distance the point is from the origin and the angle the line
segment that represents this distance makes with the +x axis:
47
1.8. Polar Form of Complex Numbers
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√
We know a = 3, b = −3 3
q
√
r = 32 + (−3 3)2
√
= 9 + 27
√
= 36
=6
And for the angle,
√ tan θre f = (−3 3 3) √
tan θre f = 3
θre f =
π
3
But, since it is a 4th quadrant angle
θ=
5π
3
√
The rectangular point (3, −3 3i) is equivalent to the polar point 6, 5π
3 .
√
5π
In rcisθ form, (3, −3 3i) is 6 cos 5π
3 + i sin 3 .
Example C
Convert the following complex numbers into polar form, use a TI-84 equivalent graphing calculator:
√
a) √
3−i
b) 9 3 + 9i
Solution
On the TI-84: go to [ANGLE] (or [2nd] function) [APPS]. Scroll down to 5 or “R-Pr(“ and press [Enter] . Next,
enter the rectangular coordinates and close the parenthesis. Press [Enter], the “r” value appears. Scroll down to
6R-Pθ, and the polar angle appears in decimal radian form.
48
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Chapter 1. Polar Equations and Complex Numbers
Note: Also under the [ANGLE] menu, commands 7 and 8 allow transformation from polar form to rectangular
form.
Vocabulary
A reference angle is used to identify the location of a calculated angle once the proper quadrant for the calculated
angle is identified.
The shorthand rcisθ is an abbreviation for: z = r(cosθ + isinθ), a common notation for the polar form of a complex
number.
Guided Practice
1) Plot the complex number z = 12 + 9i
a) What is needed in order to plot this point on the polar plane?
b) How could the r-value be determined?
c) What is the r for this point?
d) How could θ be determined?
e) What is θ for this point?
f) What would z = 12 + 9i look like on the polar plane?
2) What quadrant does z = −3 + 2i occur in when graphed?
3) What are the coordinates of z = -3 + 2i in polar form and trigonometric form?
4) What would be the polar coordinates of the point graphed below?
49
1.8. Polar Form of Complex Numbers
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Solutions
1) To plot z = 12 + 9i on a polar graph
a) First we will need to know r and θ
b) The r value is the hypotenuse of a triangle with other two sides A = 12 and B = 9 it can be determined with the
2
2
Pythagorean Theorem A2 +
√
√B = C
c) The r value for this point is 144 + 81 → 225 = 15
9
d) θ can be calculated using either sinθ = 15
or cosθ = 12
15
e) For this point, sinθ = 53 → 37o or cosθ = 45 → 37o
f) z = 12 + 9i looks like the image below when plotted on a polar plane.
50
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Chapter 1. Polar Equations and Complex Numbers
2) The point z = −3 + 2i occurs 3 units to the left and 2 units up, placing it in Quadrant II
3) To identify the coordinates of z = −3 + 2i in polar form and trigonometric form
q
√
r = (−32 ) + (22 ) → 13 First find r
sinθ = √2 → 33.7o Second, find θ
13
√
∴ [ 13, 33.7o ] are the coordinates in polar form.
√
∴ rcis 13
π
5
are the coordinates in rcis form
4) The rectangular coordinates are (4.5, 3i) therefore the complex number would be z = 4.5 + 3i
r = 5.4 Using the Pythagorean Theorem as in Q #3
θ = 33.75o Using sin = opp
hyp as in Q #3
∴ [5.4, 33.65o ] is the point in polar form ∴ rcis5.4
π
5
are the coordinates in rcis form
Practice
Plot each complex number in the complex plane. Find its polar form, [r, θ] and give the argument θ in degrees.
1.
2.
3.
4.
5.
a) 1 + i b) i c) (1 + i)i
a) −2 b) 3i c)(−2)(3i)
a) 1 + i √
b) 1 − i√c)(1 + i)(1 − i)√ √
a) 1 + i 3 b) 3 − i c)(1 + i 3)( 3 − i)
What are the rectangular coordinates for the point graphed below?
51
1.8. Polar Form of Complex Numbers
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Compute and convert to rcis form.
6. −2−2i
1−i
7. 1√+ i6
3 + 1 i10
8.
2
2
Change to polar form
9. −3
√− 2i
10. 2 3 − 2i
Change to rectangular form
o)
11. 15(cos120o + isin120
12. 12 cos π3 + isin π3
13. For the complex number in standard form x + iy find: a) Polar Form b) Trigonometric Form (Hint: Recall that
x = rcosθ and y = rsinθ)
52
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Chapter 1. Polar Equations and Complex Numbers
1.9 Product and Quotient Theorems
Here you will learn how to multiply and divide complex numbers in polar form. You will also explore the process
of checking answers using rectangular form and using a graphing calculator.
Complex numbers are found in real world calculations involving: quantum mechanics, signal analysis, fluid dynamics, control theory, and many other fields.
In electrical engineering, complex numbers are used for calculations involving impedance (the resistance to electric
flow in a circuit).
Electrical engineers are familiar with the formula:
V = V0 e jωt = V0 (cos ωt + j sin ωt)
by comparing it to the similar expression below, which is explored in this lesson, can you identify the variable j?
r2 (cosθ2 + isinθ2 )
Watch This
Embedded Video:
MEDIA
Click image to the left for more content.
- James Sousa: The Product and Quotient of Complex Numbers in Trigonometric Form
Guidance
The Product Theorem
Since complex numbers can be transformed in polar form, the multiplication of complex numbers can also be done
in polar form. Suppose we know z1 = r1 (cos θ1 + i sin θ1 ) z2 = r2 (cos θ2 + i sin θ2 )
To multiply the two complex numbers in polar form:
z1 · z2 = r1 (cos θ1 + i sin θ1 ) · r2 (cos θ2 + i sin θ2 )
= r1 r2 (cos θ1 + i sin θ1 )(cos θ2 + i sin θ2 )
= r1 r2 · (cos θ1 cos θ2 + i cos θ1 sin θ2 + i sin θ1 cos θ2 + i2 sin θ1 sin θ2 )
= r1 r2 (cos θ1 cos θ2 + i cos θ1 sin θ2 + i sin θ1 cos θ2 − sin θ1 sin θ2 )
= r1 r2 (cos θ1 cos θ2 − sin θ1 sin θ2 + i cos θ1 sin θ2 + i sin θ1 cos θ2 )
= r1 r2 ([cos θ1 cos θ2 − sin θ1 sin θ2 ] + i[cos θ1 sin θ2 + sin θ1 cos θ2 ])
(Use i2 = -1 Gather like terms, factor out i, Substitute the angle sum formulas for both sine and cosine)
z1 · z2 = r1 r2 cis [(θ1 + θ2 )]
53
1.9. Product and Quotient Theorems
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This last equation states that the product of two complex numbers in polar form can be obtained by multiplying
the polar r values of each of the complex numbers and then multiplying that value by cis of the sum of each of the
two angles of the individual complex numbers. This is more concise than the rectangular form for multiplication of
complex numbers.
Quotient Theorem
Dividing complex numbers in polar form can be shown using a similar proof that was used to show multiplication
of complex numbers. Here we omit the proof and give the result. For z1 = r1 (cos θ1 + i sin θ1 ) and z2 = r2 (cos θ2 +
i sin θ2 ), then zz21 = rr12 × cis [θ1 − θ2 ]
Example A
Multiply z1 · z2 where z1 = 2 + 2i and z2 = 1 −
√
3i
Solution
For z1 ,
p
r1 = 22 + 22
√
= 8
√
=2 2
and
tan θ1 =
2
2
tan θ1 = 1
θ1 =
π
4
Note that θ1 is in the first quadrant since a, and b >0.
For z2 ,
q
√
r2 = 12 + (− 3)2
√
= 1+3
√
= 4
=2
and
tan θ2 =
θ2 =
−
√
3,
1
5π
3
Now we can use the formula z1 · z2 = r1 · r2 cis (θ1 + θ2 )
Substituting gives:
√
z1 · z2 = 2 2 × 2 cis π4 + 5π
3
√
= 4 2 cis 23π
12
So we have
√
z1 · z2 = 4 2 cos
23π
12
+ i sin
23π
12
Re-writing in approximate decimal form:
5.656 (0.966 – 0.259i)
54
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Chapter 1. Polar Equations and Complex Numbers
5.46 - 1.46i
If the problem was done using only rectangular units then
√
z1 × z2 = (2 + 2i)(1 − 3i) or
√
√
= 2 − 2 3i + 2i − 2 3i2
Gathering like terms and using i2 = -1
√
√
= (2 + 2 3) − (2 3 + 2)i
or
5.46 − 1.46i
Example B
√
√
Using polar multiplication, find the product (6 − 2 3i)(4 + 4 3i)
Solution
√
√
Let z1 = 6 − 2 3i and z2 = 4 + 4 3i
q
q
√
√
r1 = (6)2 − (2 3)2 and r2 = (4)2 + (4 3)2
√
√
√
√
√
r1 = 36 + 12 = 48 = 4 3 and r2 = 16 + 48 = 64 = 8
For θ1 , first find tan θre f = xy √
(2 3)
tan θre f = 6
√
tan θre f = 3 3
θre f = π6 .
Since x >0 and y <0 we know that θ1 is in the in the 4th quadrant:
θ1 =
11π
6
For θ2 ,
tan θre f =
tan θre f
θre f =
(4
√
3)
4
√
= 3,
π
3
Since θ2 is in the first quadrant,
θ2 =
π
3
Using polar multiplication,
√
π
z1 × z2 = 4 3 × 8 cis 11π
6 +3
√
z1 × z2 = 32 3 cis 13π
6
subtracting 2π from the augment:
√
z1 × z2 = 32 3 cis π6
√
or in expanded form: 32 3 cos π6 + i sin π6
In decimal form this becomes: 55.426(0.866 + 0.500i) or 48 + 27.713i
Check:
55
1.9. Product and Quotient Theorems
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√
√
√
√
(6 − 2 3i)(4 + 4 3i) = 24 + 24 3i − 8 3i − 24i2
√
= 24 + 16 3i + 24
= 48 + 27.713i
Example C
Using polar division, find the quotient of
z1
z2
Solution
Given that
z1 = 5 − 5i
√
z2 = −2 3 − 2i
q
√
for z1 : r1 = 52 + (−5)2 or 5 2 and tan θ1 = −5
5 , so θ1 =
q
√
√
for z2 : r2 = (−2 3)2 + (−2)2 or 16 = 4 and tan θ2 =
7π
4
(4th quadrant)
−2
√
(−2
3)
, so θ2 =
7π
6
(3rd quadrant)
Using the formula, zz12 = rr12 × cis [θ1 − θ2 ] or
√
7π
= 5 4 2 × cis 7π
4 − 6
√
5 2
= 4 × cis 7π
12
√ 7π
= 5 4 2 cos 7π
12 + i sin 12
= 1.768[−0.259 + (0.966)i]
= −0.458 + 1.708i
Check by using the complex conjugate to do the division in rectangular form:
√
√
√
3
+2i
3
+10i+10
3i−10i2
−2
−10
5−5i
√
√
· √
=
2
−2 3−2i −2 3+2i
(−2 3) −(2i)2
√
√
3i+10
= −10 3+10i+10
12+4
√
√
3)i
= (−10 3+10)+(10+10
16
=
(−17.3+10)+(10+17.3)i
16
=
(−7.3)+(27.3)i
16
or
−0.456 + 1.706i
The two radically different approaches yield the same answer. The small difference between the two answers is a
result of decimal rounding.
TABLE 1.21:
Concept question wrap-up
Electrical engineers are familiar with the formula:
V = V0 e jωt = V0 (cos ωt + j sin ωt)
by comparing it to the similar expression below, which is explored in this lesson, can you identify the variable j?
In electrical calculations, the letter I is commonly used to denote current, therefore imaginary numbers are
identified with a j.
Note the similar usage of i in r(cosθ + isinθ)
56
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Chapter 1. Polar Equations and Complex Numbers
TABLE 1.21: (continued)
Vocabulary
The product theorem is used for the multiplication of complex numbers.
The quotient theorem is used for the division of complex numbers.
Complex conjugates are pairs of complex binomials which result in a single real number product.
Guided Practice
π
6
1+2i
2−i
1) Find the product: 7
2) Find the quotient:
• 5
−π
4
Solutions
1) This one is easier than it looks: Recall z1 · z2 = r1 · r2 cis(θ1 + θ2 )
a) r1 · r2 → 7 · 5 =
35 .......
By substitution and multplication
..... Substitute
b) θ1 + θ2 → π6 + −π
4
2π
−3π
..... Find common denominators
12 + 12
−π
.....
Simplify
12
∴ 35cis
−π
12
is the product
2) First finding the quotient by polar multiplication:
q
q
√
√
r1 = (1)2 + (2)2 = 5
r2 = (2)2 + (−1)2 = 5
tan θ1 =
2
1
tan θ1 = 2
θre f = 1.107 radians
since the angle is in the 1st quadrant
θ1 = 1.107 radians
for θ2 ,
−1
2
θre f = 21
tan θ2 =
tan
θre f = 0.464 radians
since θ2 is in the 4th quadrant, between 4.712 or
3π
2
and 6.282 radians (or 2 π)
θ2 = 5.820 radians
Finally, using the division formula,
√
z1
√5 [cis (1.107 − 5.820)]
z2 =
5
z1
z2 = [cis (−4.713)]
z1
z2
z1
z2
= [cos (−4.713) + i sin (−4.713)]
= [cos (1.570) + i sin (1.570)]
57
1.9. Product and Quotient Theorems
If we assume that π2 = 1.570, then
≈ zz12 = cos π2 + i sin π2
z1
z2
= 0 + 1i = i
Practice
1.
2.
3.
4.
5.
√
Find the product using polar form: (2 + 2i)( 3 − i)
2cis(40)•4cis(20)
π
π
Multiply: 2 cos π8 + i sin π8 • 2 cos 10
+ i sin 10
2cis(80)
6cis(200)
Divide: 3cis(130o ) ÷ 4cis(270o )
If z1 = 7
π
6
and z2 = 5
−π
4
and z2 = 5
π
6
find:
6. z1 · z2
7. zz21
8. zz21
If z1 = 8
π
3
find:
9. z1 z2
10. zz21
11. zz21
12. (z1 )2
13. (z2 )3
Find the products
√
14. Find the product using polar form:(2 + 2i)( 3 − i)
o
o + isin20o )
15. 2(cos40o + isin40
) • 4(cos20
π
π
π
π
16. 2 cos 8 + isin 8 • 2 cos 10 + isin 10
Find the quotients
17. 2(cos80o + isin80o ) ÷ 6(cos200o + isin200o )
18. 3cis(130o ) ÷ 4cis(270o )
58
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Chapter 1. Polar Equations and Complex Numbers
1.10 Powers and Roots of Complex Numbers
Here you will explore "De Moivre’s Theorem," which will allow you to calculate powers and roots of complex
numbers.
p
Manually calculating (simplifying) a statement such as: (14 − 17i)5 or 4 (3 − 2i) in present (rectangular) form
would be a very intensive process at best.
Fortunately you will learn in this lesson that there is an alternative: De Moivre’s Theorem. De Moivre’s Theorem is
really the only practical method for finding the powers or roots of a complex number, but there is a catch...
What must be done to a complex number before De Moivre’s Theorem can be utilized?
Watch This
This lesson reviews the two opposite operations involving De Moivre’s Theorem: finding powers, and finding
roots. The video lessons review each operation separately.
Embedded Video:
TABLE 1.22:
a. Raising Complex Numbers to a
Power:
b. Finding Roots of Complex Numbers:
- James Sousa: De Moivre’s Theor
em
59
1.10. Powers and Roots of Complex Numbers
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TABLE 1.22: (continued)
- James Sousa: Determining the N
th Roots of a Complex Number
Guidance
Powers of Complex Numbers
How do we raise a complex number to a power? Let’s start with an example
(-4 - 4i)3 = (-4 - 4i) x (-4 - 4i) x (-4 - 4i) =
In rectangular form, this can get very complex. What about in r cis θ form?
√
(−4 − 4i) = 4 2 cis 5π
4
So the problem becomes
√
√
4 2 cis 5π
4 · 4 2 cis
5π
4
√
· 4 2 cis
5π
4
and using our multiplication rule from the previous section,
√
(−4 − 4i)3 = (4 2)3 cis 15π
4
Notice, (a + bi)3 = r3 cis 3 θ
In words: Raise the r-value to the same degree as the complex number is raised and then multiply that by cis of the
angle multiplied by the number of the degree.
Reflecting on the example above, we can identify De Moivre’s Theorem:
TABLE 1.23:
Let z = r(cos θ + i sin θ) be a complex number in rcisθ form. If n is a positive integer, zn is
zn = rn (cos(nθ) + i sin(nθ))
It should be clear that the polar form provides a much faster result for raising a complex number to a power than
doing the problem in rectangular form.
Roots of Complex Numbers
I imagine you noticed long ago that when an new operation is presented in mathematics, the inverse operation often
follows. That is generally because the inverse operation is often procedurally similar, and it makes good sense to
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Chapter 1. Polar Equations and Complex Numbers
learn both at the same time.
This is no exception:
The inverse operation of finding a power for a number is to find a root of the same number.
a) Recall from Algebra that any root can be written as x1/n
b) Given that the formula for De Moivre’s theorem also works for fractional powers, the same formula can be
used for finding roots:
z1/n = (a + bi)1/n = r1/n cis θn
Example A
√
Find the value of (1 + 3i)4
q
√
r = (1)2 + ( 3)2 = 2
√
tan θre f = 1 3 ,
and θ is in the 1st quadrant, so
θ=
π
3
Using our equation from above:
z4 = r4 cis 4θ
z4 = (2)4 cis 4 π3
Expanding cis form:
z4 = 16 cos 4π
3 + i sin
4π
3
= 16 ((−0.5) − 0.866i)
Finally we have
z4 = -8 - 13.856i
Example B
Find
√
1+i
Solution
1
First, rewriting in exponential form: (1 + i) 2
And now in polar form:
√
√
1/2
1+i =
2 cis π4
Expanding cis form,
√
=
2 cos π4 + i sin
π
4
1/2
Using the formula:
= (21/2 )1/2 cos 12 · π4 + i sin
= 21/4 cos π8 + i sin π8
1
2
· π4
In decimal form, we get
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1.10. Powers and Roots of Complex Numbers
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=1.189( 0.924 + 0.383i)
=1.099 + 0.455i
To check, we will multiply the result by itself in rectangular form:
(1.099 + 0.455i) · (1.099 + 0.455i) = 1.0992 + 1.099(0.455i) + 1.099(0.455i) +(0.455i)2
= 1.208 + 0.500i + 0.500i + 0.208i2
= 1.208 + i − 0.208 or
= 1+i
Example C
Find the value of x :
x3
√ = 1 − 3i
Solution
√
3i in polar form.
√
Use x = 1, y = − 3 to obtain r = 2, θ =
√ let z = 1 − 3i in rectangular form
z = 2 cis 5π
3 in polar form
√ 1/3
x = 1 − 3i
First we put 1 −
x = 2cis
5π
3
5π
3
1/3
Use De Moivre’s Equation to find the first solution:
x1 = 21/3 cis 5π/3
or 21/3 cis 5π
3
9
Leave answer in cis form to find the remaining solutions:
n = 3 which means that the 3 solutions are 2π
3 radians apart or
2π
1/3 cis 5π + 2π + 2π
x2 = 21/3 cis 5π
9 + 3 and x3 = 2
9
3
3
2π
Note: It is not necessary to add 2π
3 again. Adding 3 three times equals 2 π. That would result in rotating around a
full circle and to start where it all began- that is the first solution.
The three solutions are:
x1 = 21/3 cis 5π
9
x2 = 21/3 cis 11π
9
x3 = 21/3 cis 17π
9
Each of these solutions, when graphed will be
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2π
3
apart.
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Chapter 1. Polar Equations and Complex Numbers
Check any one of these solutions to see if the results are confirmed.
Checking the second solution:
x2 = 21/3 cis 11π
9
11π
= 1.260 cos 11π
9 + i sin 9
= 1.260[−0.766 − 0.643i]
= −0.965 − 0.810i
Does (-0.965 – 0.810i)3 or (-0.965 – 0.810i) (-0.965 – 0.810i) (-0.965 – 0.810i)
√ = 1 − 3i ?
TABLE 1.24:
Concept question follow-up
A complex number operation written in rectangular form, such as: (13 − 4i)3 must be converted to polar form to
utilize De Moivre’s Theorem.
Vocabulary
De Moivre’s Theorem is the only practical manual method of identifying the powers or roots of complex numbers.
Guided Practice
1) What are the two square roots of i?
p
2) Calculate 4 (1 + 0i) What are the four fourth roots of 1?
√
3) Calculate ( 3 + i)7
Solutions
1) Let z =
√
0+i
1/2
r = 1, θ = π/2 or z = 1 × cis π2
Utilizing De Moivre’s Theorem:
z1 = 1 × cis π4 or z2 = 1 × cis 5π
4
5π
z1 = 1 cos π4 + i sin π4 or z2 = 1 cos 5π
4 + i sin 4
z1 = 0.707 + 0.707i or z2 = −0.707 − 0.707i
Check for z1 solution: (0.707 + 0.707i)2 = i?
0.500 + 0.500i + 0.500i + 0.500i2 = 0.500 + i + 0.500(-1) or i
2) Let z = 1 or z = 1 + 0i Then the problem becomes find z1/4 = (1 + 0i)1/4
Since r = 1 θ = 0, z1/4 = [1 × cis 0]1/4 with z1 = 11/4 cos 40 + i sin
0
4
or 1(1 + 0) or 1
That root is not a surprise. Now use De Moivre’s to find the other roots:
z2 = 11/4 cos 0 + π2 + i sin 0 + π2 Since there are 4 roots, dividing 2π by 4 yields 0.5π
or 0 + i or just i z3 = 11/4 cos 0 + 2π
+ i sin 0 + 2π
which yields z3 = -1
2
2
3π
Finally z4 = 11/4 cos 0 + 3π
+
i
sin
0
+
or
z
=
−i
4
2
2
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1.10. Powers and Roots of Complex Numbers
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The four fourth roots of 1 are 1, i, -1 and -i
q
√
√ 2
3) To calculate ( 3 + i)7 start by converting to rcis form First find r Recall r =
3 + 12
√
r = 3+1
r=2
√
If cosθ = 2 3 and sinθ = 12 then θ = 30o and is in quadrant I. Now that we have trigonometric form, the rest is easy:
√
( 3 + i)7 = [2(cos30o + isin30o )]7 ..... Write the original problem in rcis form
o
o
27 [(cos(7
√ · 30 ) + isin(7 · 30 )] ..... De Moivre’s Theorem
128[− 2 3 + −1
..... Simplify
2 i] √
√
( 3 + i)7 = −64 3 − 64i ..... Simplify again
√
√
∴ ( 3 + i)7 = −64 3 − 64i
Practice
Perform indicated operation on these complex numbers:
1. Divide: 2+3i
1−i
2. Multiply: (−6 − i)(−6 + i)
√
2
3
1
3. Multiply:
2 − 2i
4.
5.
6.
7.
8.
√
Find the product using polar form: (2 + 2i)( 3 − i)
◦
Multiply: 2(cos 40◦ + i sin 40
20◦ + i sin 20◦ )
) • 4(cos
π
π
π
π
Multiply: 2 cos 8 + i sin 8 • 2 cos 10 + i sin 10
Divide: 2(cos 80◦ + i sin 80◦ ) ÷ 6(cos 200◦ + i sin 200◦ )
Divide: 3 cis(130◦ ) ÷ 4 cis(270◦ )
Use De Moivre’s Theorem:
9. [3(cos 80◦ + i sin 80◦ )]3
h√
i4
5π
2 cos 5π
10.
+
i
sin
16
16
6
√
11.
3−i
12. Identify the 3 complex cube roots of 1 + i
13. Identify the 4 complex fourth roots of −16i
14. Identify the five complex fifth roots of i
Summary
This chapter focuses on the various uses of cartesian (x, y) graph and polar graph methods.
Through the material in this chapter, students should become proficient at:
Converting values and graphs between polar and cartesian methods.
Identifying and graphing imaginary and complex numbers.
Using the Quadratic Formula to find the imaginary roots of quadratic equations.
Multiplying and dividing complex numbers and graphing the solutions on polar and cartesian graphs.
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