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C H A P T E R
16
E
Polar coordinates and
complex numbers
Objectives
PL
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To describe points on the plane using polar coordinates
To describe graphs with polar coordinates
To transform polar coordinates to cartesian coordinates
To transform cartesian coordinates to polar coordinates
To understand the imaginary number i
SA
M
To understand the set of complex numbers C
To understand the real-valued functions of the complex numbers, Re(z) and Im(z)
To represent complex numbers graphically on an Argand diagram
To understand the rules which define equality, addition, subtraction and
multiplication of complex numbers
To understand the concept of the complex conjugate
To understand the operation of division by complex numbers
To understand the modulus-argument form of a complex number and the basic
operations on complex numbers in that form
To understand the geometrical significance of multiplication and division of
complex numbers in the modulus-argument form
To be able to factorise quadratic polynomials over C
To be able to solve quadratic polynomials over C
A new set of numbers called Complex numbers is introduced in this chapter. The need for this
new set of numbers can be equated to the need for a solution of the equation x 2 + 1 = 0. A
geometric interpretation is also shown to be useful.
Complex numbers can be expressed in two ways, cartesian form and polar form. As a
preliminary to this, polar coordinates are introduced.
410
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Chapter 16 — Polar coordinates and complex numbers
16.1
411
Polar coordinates
E
In previous work the cartesian coordinate system has been used to represent points in
two-dimensional space. The point (x, y) is described in terms of its horizontal displacement (x)
and its vertical displacement (y) from a fixed point called the origin (O).
An alternative way of locating the point P is to describe it in terms of its polar coordinates
[r, ] where r specifies the distance from the origin or pole and specifies the angle of the line
OP relative to the line OZ which extends to the right from O and is called the polar axis.
P[r, θ]
Note: An angle in an anticlockwise direction from OZ is
considered to be positive.
r
O
Z
P[4, 60°]
PL
For example, the point P[4, 60◦ ] is located a
distance of 4 units along a line forming an angle
of 60◦ with the polar axis.
θ
4
O
60°
Z
SA
M
Using this system it is clear that any point can be specified in a number of different ways.
For example, the point [4, 60◦ ] may also be specified by [−4, −120◦ ].
The angle = 120◦ is measured in a clockwise direction from O.
The diagram below and to the left illustrates the point P [4, −120◦ ] and the diagram to the
right, [−4, −120◦ ].
P
O
4
Z
4
120°
O
60°
Z
120°
P'
P[4, 60◦ ] may also be specified by P[4, −300◦ ] or P[−4, 240◦ ].
Example 1
Plot the point P with coordinates [3, −30◦ ].
Solution
O
Z
30°
3
P
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412
Essential Advanced General Mathematics
The relationship between cartesian and
polar coordinates
y
If a set of cartesian axes is superimposed
over a polar axis, the relationship between
cartesian and polar coordinates can be
established.
From the diagram
1
y = r sin 2
r
r sin (θ)
θ
E
and
x = r cos P(x, y)
O
r cos (θ)
x(Z)
The angle can be found by finding a solution which satisfies both equations 1 and 2
Squaring both sides of equations 1 and 2 and adding yields
PL
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x 2 + y 2 = r 2 cos2 + r 2 sin2 = r 2 (cos2 + sin2 )
i.e. x 2 + y 2 = r 2
Using these relationships, coordinates can be converted from cartesian to polar and vice versa.
SA
M
Example 2
√
a Express ( 3, 1) in polar form.
√
b Express [ 2, 45◦ ] in cartesian form.
Solution
1
r 2 = x 2 + y2
sin =
a
2
√
= ( 3)2 + (1)2
=4
√
3
and cos =
2
∴ = 30◦
∴r =2
√
∴ ( 3, 1) specifies the same point as [2, 30◦ ]
√
◦
b r = √2 and = 45√
1
x = 2 cos 45◦ = 2 × √ = 1
2
√
√
1
y = 2 sin 45◦ = 2 × √ = 1
2
√
∴ [ 2, 45◦ ] specifies the same point as (1, 1)
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Chapter 16 — Polar coordinates and complex numbers
413
Curve sketching using polar coordinates
Using the TI-Nspire
Example 3
E
In the same way that graphs of relationships in cartesian form can be sketched, relationships
expressed in polar form can also be sketched. Some very interesting curves result from simple
polar equations.
It is recommended that to sketch these graphs a graphics calculator or a computer graphing
package be used. Graphs can of course be plotted using a table of values. A sheet of polar
graph paper is also useful although a sheet of blank paper will suffice as long as a ruler and a
protractor are used.
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Plot the graph of r = 3 (1 − cos ).
Solution
SA
M
Open a Graphs & Geometry application
(c 2) and select Polar from the Graph
Type menu (b 3 3).
Enter r 1 () = 3 (1 − cos ()) as shown
Note that the domain of ( as well as the
step size can be adjusted in this window.
The graph is shown using the Zoom,
Fit command from the Window menu
(b 4 ).
Note that every point on the graph satisfies
r = 3 (1 − cos ()) . For example, for
= 60◦
3
1
◦
r = 3 (1 − cos (60 )) = 3 1 −
=
2
2
For = 180◦ , r = 3 (1 − cos (180◦ ))
= 3 (1 − (−1)) = 6
◦
For = −90 , r = 3 (1 − cos (−90)) = 3◦
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Essential Advanced General Mathematics
E
Note that Trace (b 5 ) can be used to
show the coordinates of the points on the
graph in the form [r, ].
To go to the point where = , simply
type followed by enter. The cursor will
then move to the point [r, ] = [6, ] as
shown.
Example 4
PL
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Sketch the graph of r = .
Solution
SA
M
Open a Graphs & Geometry application
(c 2) and select Polar from the Graph
Type menu (b 3 3).
Enter r 1 () = .
The graph is shown.
If the domain of is extended, the graph
continues to spiral out. This can be
observed by extending the domain to
0 < < 6.
The resulting graph is shown using the
Zoom, Out command from the Window
menu (b 4 4).
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Chapter 16 — Polar coordinates and complex numbers
415
Using the Casio ClassPad
E
Plot the graph of r = 3 (1 − cos ).
Ensure that the mode is set to radians.
tap
and from the menu select
In
.
Enter the equation r 1 = 3 (1 − cos ()) and tap
$ to produce the graph.
PL
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In the screen shown, the window was selected by tapping Zoom, quick initialize.
Example 5
SA
M
Find the polar equation of the circle whose cartesian equation is
x 2 + y 2 = 4x
Solution
Let x = r cos and y = r sin Then
∴
∴
∴
∴
r 2 cos2 + r 2 sin2 = 4r cos r 2 (cos2 + sin2 ) = 4r cos r 2 − 4r cos = 0
r (r − 4 cos ) = 0
r = 0 or r = 4 cos ∴ r = 4 cos is the polar equation of the circle
Example 6
Find the cartesian equation corresponding to each of the following polar equations.
1
a r =3
b r=
c r = 3(1 − cos )
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416
Essential Advanced General Mathematics
Solution
r =3
2
x + y2 = 3
x 2 + y2 = 9
The circle with centre (0, 0)
and radius 3
b
1
1 + sin implies r (1 + sin ) = 1
i.e. r+ r sin = 1
∴ x 2 + y2 = 1 − y
x 2 + y 2 = 1 − 2y + y 2
1
2
∴ x = −2 y −
2
r=
E
a
∴ y=−
r = 3(1 − cos )
∴
r 2 = 3r− 3r cos (Multiplying both sides of equation by r )
∴
x 2 + y 2 = 3x 2 + y 2 − 3x
2
x + y 2 + 3x = 3 x 2 + y 2
PL
c
1
x2
+
2
2
Exercise 16A
Example
1
1 Plot each of the following points using a polar axis.
b B[3, 45◦ ]
f F[−5, −50◦ ]
SA
M
a A[2, 30◦ ]
e E[5, 50◦ ]
c C[−2, 60◦ ]
g G[−5, 130◦ ]
2 Plot each of the following points using a polar axis.
a A[1, ]
b B[−1, −]
c C 2,
2
d D[4, −30◦ ]
h H [5, −130◦ ]
3
d D 3,
4
Example
2a
3 Convert the following cartesian coordinates to polar coordinates. (Remember to note
which quadrant each point is in.)
√
√
a (4, 4)
b (1, − 3)
c (2 3, −2)
d (−5, 12)
√
g (−5, −12)
h (4, 3)
e (6, −5)
f ( 3, 1)
Example
2b
4 Convert the following polar coordinates to cartesian
coordinates.
5
◦
a [−2, 30 ]
b −4,
d [4, −2]
c −1,
2
4
7
g [2, 180◦ ]
h [1, −120◦ ]
f [5, 240◦ ]
e 2, −
6
Examples
3, 4
5 Plot each of the following polar graphs.
4
3
b r=
a r=
sin cos d r = 2, 0 ≤ ≤ 6
e r = , ≤ ≤ 4
6
g r = 5(1 + cos )
h r =
2(1 − sin )
√
, 0 ≤ ≤ 6
k r=
j r = ± cos 2
c r = 2 cos f r = cos 2
i r = 3 cos + 2
l r = 2 sin 2
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Chapter 16 — Polar coordinates and complex numbers
Example
5
6 Obtain the polar equations of each of the following.
a x 2 + y 2 = 16
Example
6
b x+y=6
c x2 = y
d
x2
+ y2 = 1
4
7 Obtain the cartesian equations of each of the following.
a r =2
d r = 2a(1 + sin 2)
b r = a(1 + cos )
a
e r=
1 + cos c r = a cos a
f r=
1 + sin The set of complex numbers
E
16.2
417
PL
In earlier work in mathematics it was assumed that an equation of the form x 2 = −1 had no
solutions. Mathematicians of the eighteenth century introduced the imaginary number i with
√
the property i 2 = −1. i is defined as i = −1 and the equation x 2 = −1 has two solutions, i
and −i. By considering i such that i 2 = −1 then the square roots of all negative numbers may
be found.
√
√
For example −4 = 4 × −1
√
√
= 4 × −1
= 2i
SA
M
Imaginary numbers led to the introduction of complex numbers, which further broadened the
scope of mathematical thinking. Today complex numbers are widely used in engineering, the
study of aerodynamics and many other branches of physics.
Consider the equation x 2 + 2x + 3 = 0. Using the quadratic formula to solve yields:
√
−2 ± 4 − 12
x=
2
√
−2 ± −8
=
2√
= −1 ± −2
This equation has no real solutions since the discriminant = b2 − 4ac is less than zero.
However, for complex numbers
√
x = −1 ± 2i
A complex number is an expression of the form a + bi, where a and b are real numbers. C
is the set of complex numbers, i.e. C = {a + bi : a, b ∈ R}. The letter often used to denote a
complex number is z.
Therefore z ∈ C implies z = a + bi where a, b ∈ R
If a = 0, z is said to be imaginary.
If b = 0, z is real.
Real numbers and imaginary numbers are subsets of C.
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Essential Advanced General Mathematics
Functions of complex numbers
Let z = a + bi
Re (z) is a function which defines the real component of z. Im (z) is a function which defines
the value of the imaginary component of z.
Re (z) = a and Im (z) = b
Re (z) and Im (z) are both real-valued functions of z, i.e. Re : C → R and Im : C → R.
So for the complex number z = 2 + 5i, Re (z) = 2 and Im (z) = 5.
E
Note:
Equality of complex numbers
Two complex numbers are equal if and only if both their real and imaginary parts are equal.
i.e.
x1 + y1 i = x2 + y2 i
if and only if
x1 = x2 and y1 = y2
Example 7
PL
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If 4 − 3i = 2a + bi find the values of a and b.
Solution
and
b = −3
SA
M
2a = 4
a=2
Example 8
Find the values of a and b such that (2a + 3b) + (a − 2b)i = −1 + 3i
Solution
2a + 3b
a − 2b
2 × 2 gives
2a − 4b
1 − 3 gives
7b
∴
= −1 1
=3 2
=6
3
= −7
b = −1 and a = 1
Operations with complex numbers
Addition and subtraction
If z 1 = a + bi and z 2 = c + di
Then z 1 ± z 2 = (a ± c) + i(b ± d)
(a, b, c, d ∈ R)
i.e. Re (z 1 ± z 2 ) = Re (z 1 ) ± Re (z 2 ) and Im (z 1 ± z 2 ) = Im (z 1 ) ± Im (z 2 )
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Chapter 16 — Polar coordinates and complex numbers
419
Example 9
If z 1 = 2 − 3i and z 2 = −4 + 5i find
a z1 + z2
b z1 − z2
Solution
Multiplication by a real constant
z = a + bi and
kz = k(a + bi)
= ka + kbi
If
then
k∈R
b z 1 − z 2 = (2 − −4) + (−3 − 5)i
= 6 − 8i
E
a z 1 + z 2 = 2(2 + −4) + (−3 + 5)i
= −2 + 2i
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For example, if z = 3 − 6i then 3z = 9 − 18i
Multiplication by powers of i
Successive multiplication by powers of i gives the following:
SA
M
i1 = i
i 2 = −1
i 3 = −i
i 4 = (−1)2 = 1
i5 = i
and so on
In general, for n = 0, 1, 2, 3, . . .
i 4n = 1
=i
i
4n+2
= −1
i
4n+3
= −i
i
4n+1
When multiplying by powers of i, the usual index laws apply.
Example 10
Simplify
a i 13
b 3i 4 × (−2i)3
Solution
a i 13 = i 4×3+1
=i
b 3i 4 × (−2i)3 = 3 × (−2)3 × i 4 × i 3
= −24i 7
= 24i
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Essential Advanced General Mathematics
Exercise 16B
1 State the values of Re (z) and Im (z) for each of the following.
1 3
a 2 + 3i
b 4 + 5i
− i
c
2 2 √
√
2 − 2 2i
d −4
e 3i
f
7, 8
2 Find the values of a and b in each of the following if
a 2a − 3bi = 4 + 6i
c 2a + bi = 10
9
3 Simplify the following.
a
c
e
g
(2 − 3i) + (4 − 5i)
(−3 − i) − (3 + i)
(1 − i) − (2i + 3)
4(2 − 3i) − (2 − 8i)
(4 + i) + (2 − 2i)
√
√
(2 − 2i) + (5 − 8i)
(2 + i) − (−2 − i)
−(5 − 4i) + (1 + 2i)
3
1
(4 − 3i) − (2 − i)
j
2
2
b
d
f
h
PL
Example
b a + b − 2abi = 5 − 12i
d 3a + (a − b)i = 2 + i
i 5(i + 4) + 3(2i − 7)
Example
10
4 Simplify
√
a
−16
20
f i
√
b 2 −9
g −2i × i 3
SA
M
5 Simplify
a i(2 − i)
16.3
E
Examples
√
c
−2
h 4i 4 × 3i 2
b i 2 (3 − 4i)
c
√
2i(i −
d i3
√ 5 √
i
8i × −2
2)
e i 14
√ √
√
d − 3( −3 + 2)
Multiplication and division of
complex numbers
Multiplication of complex numbers
If z 1 = a + bi and z 2 = c + di(a, b, c, d ∈ R)
Then
z 1 × z 2 = (a + bi) (c + di)
= ac + bci + adi + dbi 2
= (ac − bd) + (bc + ad)i
(bdi 2 = −bd)
Example 11
If z 1 = 3 − 2i and z 2 = 1 + i, find z 1 z 2 .
Solution
z 1 z 2 = (3 − 2i)(1 + i)
= 3 − 2i + 3i − 2i 2
= 5+i
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Chapter 16 — Polar coordinates and complex numbers
421
Conjugate of a complex number
If z = a + bi then the conjugate of z denoted by the symbol z is
z = a − bi
For example, the conjugate of −4 + 3i is −4 − 3i and vice versa.
E
Note that zz = (a + bi) (a − bi)
= a 2 + abi − abi − b2 i 2
= a 2 + b2 which is a real number
Using this result, a 2 + b2 can now be factorised over the set of complex numbers.
Example 12
If z 1 = 2 − 3i and z 2 = −1 + 2i find
a (z 1 + z 2 ) and z 1 + z 2
Solution
b z 1 z 2 and z 1 z 2
z 1 = 2 + 3i and z 2 = −1 − 2i
b z 1 z 2 = (2 − 3i)(−1 + 2i) = 4 + 7i
z 1 + z 2 = (2 − 3i) + (−1 + 2i)
z 1 z 2 = 4 − 7i
= 1−i
z 1 z 2 = (2 + 3i)(−1 − 2i) = 4 − 7i
(z 1 + z 2 ) = 1 + i
z 1 + z 2 = 2 + 3i + −1 − 2i
= 1+i
SA
M
a
PL
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In general it can be stated that
the conjugate of the sum of two complex numbers is equal to the sum of the conjugates
the conjugate of the product of two complex numbers is equal to the product of the
conjugates.
i.e. (z 1 + z 2 ) = z 1 + z 2 and (z 1 z 2 ) = z 1 z 2
Division of complex numbers
Division of one complex number by another relies on the fact that the product of a complex
number and its conjugate is a real number.
If
Then
z 1 = a + bi and z 2 = c + di
(a + bi)
z1
=
z2
(c + di)
(a, b, c, d ∈ R)
If the numerator and denominator are multiplied by the conjugate of z 2 then
(a + bi) (c − di)
z1
×
=
z2
(c + di) (c − di)
ac + bci − adi − bdi 2
=
c2 + d 2
(ac + bd) (bc − ad)i
= 2
+ 2
c + d2
c + d2
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Essential Advanced General Mathematics
Example 13
If z 1 = 2 − i and z 2 = 3 + 2i, find
z1
.
z2
Solution
6 − 3i − 4i + 2i 2
32 + 22
4 − 7i
=
13
1
=
(4 − 7i)
13
=
Example 14
E
3 − 2i
2−i
z1
×
=
z2
3 + 2i
3 − 2i
PL
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Solve for z the equation (2 + 3i) z = −1 − 2i
Solution
SA
M
(2 + 3i)z = −1 − 2i
−1 − 2i
z=
2 + 3i
−1 − 2i
2 − 3i
=
×
2 + 3i
2 − 3i
−8 − i
z=
13
There is an obvious similarity in the process of expressing a complex number with a real
denominator and the process of rationalising the denominator of a surd expression.
Example 15
If z = 2 − 5i find z −1 and express with a real denominator.
Solution
z −1 =
1
z
1
2 − 5i
2 + 5i
1
×
=
2 − 5i
2 + 5i
2 + 5i
=
29
1
=
(2 + 5i)
29
=
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Chapter 16 — Polar coordinates and complex numbers
423
Using the TI-Nspire
PL
E
The TI-Nspire can be used to deal with complex numbers. Select Rectangular form in
the Document Settings (c 8 1) to perform calculations on complex numbers in
the form a + bi.
The square root of a negative number can now
be performed as shown.
SA
M
√
The results of the operations +, −, × and ,
are illustrated using the two complex numbers
2 + 3i and 3 + 4i.
It is possible to perform arithmetic
operations with complex numbers as shown.
The Real Part command from the Complex
Number Tools submenu of the Number menu
(b 2 8 2) can be used as shown to find
the real part of a complex number.
The Magnitude command from the
Complex Number Tools submenu of the
Number menu (b 2 8 5 ) can be used as
shown to find the modulus of a complex number.
The Complex Conjugate command from the
Complex Number Tools submenu of the Number
menu (b 2 8 1) can be used as shown
to find the complex conjugate of a complex
number.
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424
Essential Advanced General Mathematics
E
There are also commands for factorising
polynomials over the complex numbers and for
solving polynomial equations over the complex
numbers. These are available from the Complex
submenu of the Algebra menu (b 3 ).
Using the Casio ClassPad
PL
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In
tap Real in the status bar at the bottom of the screen to enter Cplx mode. In this
√
√
to obtain the answer i. Similarly, −16 will return the
mode enter −1 and tap
answer 4i.
Operations
SA
M
i is found in
in the on-screen keyboard.
With the calculator set to Complex mode, a
number of arithmetic operations can be carried out,
as shown in the screen at right using options from
Interactive, Complex.
Polynomials can be factorised and solved over the
complex number field using Interactive, transformation
and Equation/inequality, solve.
Exercise 16C
Example
11
1 Expand and simplify
a (4 + i)2
d (−1 − i)2
b (2 − 2i)2
√
√
√
√
e ( 2 − 3i)( 2 + 3i)
c (3 + 2i)(2 + 4i)
f (5 − 2i)(−2 + 3i)
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Chapter 16 — Polar coordinates and complex numbers
425
2 Write down the conjugate of each of the following complex numbers.
√
a 2 − 5i
b −1 + 3i
c
5 − 2i
d −5i
12
3 If z 1 = 2 − i and z 2 = −3 + 2i find
a z1
e z1 z2
Example
15
b z2
f z1 + z2
c z1 z2
g z1 + z2
d z1 z2
h z1 + z2
4 If z = 2 − 4i express each of the following in the form x + yi.
a z
b zz
c z+z
e z−z
f i(z − z)
g z −1
d z(z + z)
z
h
i
E
Example
5 Find the values of a and b such that (a + bi)(2 + 5i) = 3 − i
13
6 Express in the form x + yi
3 + 2i
2−i
b
a
2 − 3i
4+i
2 − 2i
1
d
e
4i
2 − 3i
4 + 3i
1+i
i
f
2 + 6i
PL
Example
c
7 Find the values of a and b if (3 − i)(a + bi) = 6 − 7i
Example
14
8 Solve each of the following for z.
a (2 − i)z = 4 + 2i
d 2(4 − 7i)z = 5 + 2i
b (1 + 3i)z = −2 − i
e z(1 + i) = 4
c (3i + 5)z = 1 + i
Argand diagrams
SA
M
16.4
An Argand diagram is a geometrical representation of the set of complex numbers. In a
vector sense, a complex number has two dimensions; the real part and the imaginary part.
Therefore a plane is required to represent C.
Im(z)
An Argand diagram is drawn with two
3
perpendicular axes. The horizontal axis
represents Re(z), z ∈ C, and the vertical
2
axis represents Im(z), z ∈ C.
(3 + i)
(–2 + i)
1
Each point on an Argand diagram represents
a complex number. The complex number a + bi
0 1
–3 –2 –1
2
3 Re(z)
is situated at the point (a, b) on the equivalent
–1
cartesian axes as shown by the examples in this
–2
figure. The number written as a + bi is
called the cartesian form of the complex number.
(2 – 3i)
–3
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426
Essential Advanced General Mathematics
Example 16
Im(z)
5
4 B
Write down the complex number represented
by each of the points A to F on this Argand diagram.
A
D
5
C
0
E
Re(z)
E
–5
F
–5
Solution
PL
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A : 2 + 3i
C : −5
E : −5 − 2i
B : 4i
D : −1 + i
F : 1 − 3i
Geometrical representation of the basic
operations on complex numbers
SA
M
The addition of two complex numbers is similar to a vector sum and follows the triangle of
vectors rule.
The multiplication by a scalar follows vector properties of parallel position vectors.
z1 + z2
Im (z)
Im(z)
az
z2
z
0
z1
0
Re (z)
bz
Re(z)
cz
a>1
0<b<1
c<0
The subtraction z 1 − z 2 is represented by the sum z 1 + (−z 2 ).
Example 17
a Represent the following points on an Argand diagram.
i 2
ii −3i
iii 2 − i
iv −(2 + 3i) v −1 + 2i
b Let z 1 = 2 + i and z 2 = −1 + 3i.
Represent z 1 , z 2 , z 1 + z 2 and z 1 − z 2 on an Argand diagram and verify that the complex
number sum and difference follow the vector triangle properties.
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Chapter 16 — Polar coordinates and complex numbers
Solution
a
Im (z)
b
4
z2 3
2
1
Im(z)
3
–1 + 2i
2
1
2
–3 –2 –1 0 1
–1
z1 + z 2
z1
–4 –3 –2 –1 0 1 2 3 4 Re (z)
–1
z 1 – z2
–2
–z2
–3
–4
2 3 Re (z)
2–i
–2
–(2 + 3i)
427
–3 –3i
E
z 1 + z 2 = (2 + i) + (−1 + 3i)
= 1 + 4i
z 1 − z 2 = (2 + i) − (−1 + 3i)
= 3 − 2i
Rotation about the origin
PL
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When the complex number 2 + 3i is multiplied by –1 the result is −2 − 3i.
This can be considered to be achieved through a rotation of 180◦ about the origin. When the
complex number 2 + 3i is multiplied by i,
i(2 + 3i) = 2i + 3i 2
= 2i − 3
= −3 + 2i
i.e.
Im(z)
2 + 3i
–3 + 2i
SA
M
the result can be seen to be achieved through
a rotation of 90◦ in an anticlockwise direction
about the origin. If −3 + 2i is multiplied by
i the result is −2 − 3i. This is again achieved
through a rotation of 90◦ in an anticlockwise
direction about the origin.
0
Re(z)
–2 – 3i
Example 18
If z 1 = 1 − 4i and z 2 = −2 + 2i, find z 1 + z 2 algebraically and illustrate z 1 + z 2 on an
Argand diagram.
Solution
Im(z)
z 1 + z 2 = (1 − 4i) + (−2 + 2i)
= −1 − 2i
3
2
z2
1
0
–4
–3
–2
–1
z1 + z2
–1 1
2
3
4 Re(z)
–2
–3
–4
z1
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428
Essential Advanced General Mathematics
Exercise 16D
Example
16
Im(z)
1 Write down the complex numbers
represented on the following Argand diagram.
3
B 2
1
E
Example
17
18
3
4
5 Re(z)
D
2 Represent each of the following complex numbers as points on an Argand plane.
a 3 − 4i
Example
2
E
–5 –4 –3 –2 –1 0 –1 1
F
–2
–3
–4
C
A
b −4 + i
c 4+i
d −3 + 0i
e 0 − 2i
f −5 − 2i
PL
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3 If z 1 = 6 − 5i and z 2 = −3 + 4i, find algebraically and represent on an Argand diagram.
a z1 + z2
b z1 − z2
4 If z = 1 + 3i, represent on an Argand diagram
a z
b z
c z2
d −z
e
1
z
5 If z = 2 − 5i, represent on an Argand diagram
b zi
c zi 2
d zi 3
e zi 4
SA
M
a z
16.5
Solving equations over the complex field
Quadratic equations for which the discriminant is less than zero have no solutions for the real
numbers. The introduction of the complex number enables such quadratic equations to be
solved. Further solutions to higher degree polynomials may also be found using complex
numbers. Solution of higher degree polynomials appears in the Specialist Mathematics course.
In this chapter only quadratics will be considered.
Sum of two squares
Earlier it was seen that the product of a complex number a + bi and its conjugate a − bi
yielded the result
(a + bi)(a − bi) = a 2 + b2
Hence sums of two squares can be factorised enabling equations of the form z 2 + a 2 = 0 to
be solved.
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Chapter 16 — Polar coordinates and complex numbers
429
Example 19
Solve the equations
a z 2 + 16 = 0
b 2z 2 + 6 = 0
Solution
z 2 + 16 = 0
∴ z − 16i 2 = 0
(z + 4i)(z − 4i) = 0
∴ z = ±4i
2
b
2z 2 + 6 = 0
∴ 2(z 2 + 3) = 0
∴ 2(z 2 − 3i 2 ) = 0
√
√
∴ 2(z + 3i)(z − 3i) = 0
√
∴ z = ± 3i
E
a
Solution of quadratic equations
PL
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To solve quadratic equations where the discriminant is less than zero, still use the quadratic
formula in the usual way.
Example 20
Solve the equation 3z 2 + 5z + 3 = 0
Solution
SA
M
Using the quadratic formula
√
−5 ± 25 − 36
z=
√6
−5 ± −11
=
6
√
1
= (−5 ± 11i)
6
Using the TI-Nspire
Each of the expressions in the above examples can be factorised using cFactor from the
Complex submenu of Algebra, for example,
cfactor (z 2 + 16, z).
Each of the equations in the above examples can besolved using cSolve from the
Complex submenu of Algebra, for example
cSolve (3z 2 + 5z + 3 = 0, z).
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430
Essential Advanced General Mathematics
Using the Casio ClassPad
E
To factorise in the above examples, ensure the mode is set to Cplx.
Enter and highlight the expression z 2 + 16 then
tap Interactive, Transformation, rFactor.
PL
To solve in the above examples, the usual method for solving equations is used. For
example, enter and highlight 3z 2 + 5z + 3 = 0 then tap Interactive,
Equation/inequality, solve and ensure that the variable selected is z.
Exercise 16E
Examples
19, 20
1 Solve each of the following equations over C.
z2 + 4 = 0
(z − 2)2 + 16 = 0
z 2 + 3z + 3 = 0
2z = z 2 + 5
b
e
h
k
2z 2 + 18 = 0
(z + 1)2 = −49
2z 2 + 5z + 4 = 0
2z 2 − 6z = −10
c
f
i
l
3z 2 = −15
z 2 − 2z + 3 = 0
3z 2 = z − 2
z 2 − 6z = −14
SA
M
a
d
g
j
16.6
Polar form of a complex number
Earlier in this chapter it was shown that points on a cartesian plane (x, y) may be represented in
terms of polar coordinates [r, ]. Similarly, complex numbers may be represented in polar
form.
Im (z)
Recalling that x = r cos and y = r sin where
r 2 = x 2 + y 2 then the point P in the complex plane
P(x + iy)
corresponding to the complex number in cartesian form,
r
y
z = x + yi, may be represented as shown in the diagram.
θ
z = r cos + r sin i
x
0
Re (z)
= r (cos + sin i)
The polar form is abbreviated to z = r cis .
r = x 2 + y 2 is called the absolute value or modulus of z. It is denoted by mod z or |z|.
Remember that is measured in an anticlockwise direction from the horizontal axis.
The same point may be represented a number of ways in polar form, since
cos = cos( ± 2n) and sin = sin ( ± 2n), where n ∈ Z , the polar form of a complex
number is not unique.
Note:
i.e.
z = r cis = r cis ( + 2n), n ∈ Z
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Chapter 16 — Polar coordinates and complex numbers
431
Usually the interval − < ≤ is used. The corresponding value of is called the principal
value of the argument of z and is denoted by Arg z.
i.e.
− < Arg z ≤ Example 21
Express in polar form the following complex numbers
√
b z = 2 − 2i
a z = 1 + 3i
Solution
E
P1: FXS/ABE
√
3 i is a point in the 1st quadrant.
∴
0< <
2
√
Now
x = 1 and y = 3
√
Therefore r = 1 + 3
=2
√
1
3
)
also = (since cos = and sin =
3 √
2
2
∴ z = 1 + 3i
= 2 cis
3
PL
a First note that z = 1 +
SA
M
b z = 2 − 2i is a point in the 4th quadrant.
−
<<0
2
Now
x = 2 and y = 2
√
Therefore
r = 4+4
√
= 8
√
=2 2
1
−1
−
(since cos = √ and sin = √ )
Also
=
4
2
2
∴
z = 2 − 2i √
−
= 2 2 cis
4
∴
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Essential Advanced General Mathematics
Example 22
−2
Express in cartesian form z = 2 cis
3
Solution
−2
x = r cos = 2 cos
3
1
=2 −
2
= −1
−2
y = r sin = 2 sin
3
√ − 3
=2
2
√
=− 3
−2
∴ z = 2 cis
3
√
= −1 − 3i
PL
E
P1: FXS/ABE
SA
M
Multiplication and division in polar form
z 1 = r1 cis and z 2 = r2 cis 2
If
Then
and
z 1 z 2 = r1r2 cis (1 + 2 )
r1
z1
=
cis (1 − 2 )
z2
r2
These results may be proved using the addition formulas for sine and cosine established in
Chapter 11. This is left as an exercise for the reader.
Example 23
If z 1 = 2 cis 30◦ and z 2 = 4 cis 20◦ find the product z 1 z 2 and represent it on an Argand
diagram.
Solution
Im (z)
z 1 z 2 = r1r2 cis (1 + 2 )
= 2 × 4 cis (20◦ + 30◦ )
= 8 cis 50◦
z1z2
50°
z1
30°
z2
Re (z)
0
20°
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Chapter 16 — Polar coordinates and complex numbers
433
Example 24
5
, find the product z 1 z 2 .
If z 1 = 3 cis and z 2 = 2 cis
2
6
Solution
E
z 1 z 2 = r1r2 cis (1 + 2 )
5
= 6 cis
+
2
6
4
= 6 cis
3
−2
∴ z 1 z 2 = 6 cis
since − < Arg z ≤ 3
Example 25
PL
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√
√
z1
If z 1 = − 3 + i and z 2 = 2 3 + 2i, find the quotient
and express it in cartesian form.
z2
Solution
First express z 1 and z 1 in polar form.
3+1
SA
M
|z 1 | =
√
=2
√
|z 2 | = 12 + 4
=4
√
5
1
− 3
Arg z 1 =
, since sin 1 = and cos 1 =
6
2
2
where Arg z 1 = 1
√
1
3
Arg z 2 = , since sin 2 = and cos 2 =
6
2
2
where Arg z 2 = 2
5
and z 2 = 4 cis
z 1 = 2 cis
6
6
z1
r1
=
cis (1 − 2 )
z2
r2
2
5 = cis
−
4
6
6
1
2
= cis
2
3
∴
1
2
1
2
z1
= cos
+ sin
i
In cartesian form
z2
2
3
2
3
√ 1 −1
3
1
=
+
i
2 2
2
2
√
1
= − (1 − 3)i
4
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434
Essential Advanced General Mathematics
Exercise 16F
21
1 Express each of the following in the simplest polar form.
√
√
a 1 + 3i
b 1−i
c −2 3 + 2i
√
1 1
d −4 − 4i
e 12 − 12 3i
f − + i
2 2
Example
22
2 Express each of the following in the form x + yi.
√
3
c 2 cis
2 cis
b
a 3 cis
d 5 cis
6
3
2
4
5
√
4
−2
−
e 12 cis
g 5 cis
h 5 cis
f 3 2 cis
6
3
3
4
3 Simplify the following and express the answers in cartesian form.
. 3 cis
b 4 cis
. 3 cis
a 2 cis
12
4
6
12
5
2
−
c
cis
. 5 cis
. 3 cis
d
12 cis
4
12
3
3
√
√
5
−3
e
12 cis
. 3 cis
f ( 2 cis ).
3 cis
6
2
4
−
12 cis
10 cis
3
4
g h
2
5 cis
3 cis
12
3
√
3
−
12 8 cis
20 cis
4
6
i √
j
5
3 2 cis
8 cis
12
6
PL
23, 24, 25
SA
M
Examples
E
Example
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Chapter 16 — Polar coordinates and complex numbers
435
The polar coordinates [r, ] may be represented as follows.
is measured in an anticlockwise direction
from the polar axis.
P[r, θ]
θ
polar axis
E
O
pole
For conversion of coordinates from cartesian to polar and vice versa
PL
x = r cos , y = r sin and hence x 2 + y 2 = r 2 .
√
describes the same point as 2 cos , 2 sin
= (1, 3)
Therefore 2,
3
3
3
√
For (1, −1),
r= 2
√
√
1 = 2 cos and − 1 = 2 sin −1
1
Therefore
cos = √ and sin = √
2
2
−
and =
4
√ −
(1, −1) =
2,
4
SA
M
For conversion of an equation from cartesian to polar use
x = r cos , y = r sin and x 2 + y 2 = r 2
Therefore x + y = 1
becomes
r cos + r sin = 1
i.e. r (cos + sin ) = 1
Consider y = x 2
This becomes
∴
∴
r sin = r 2 cos2 r sin − r 2 cos2 = 0
r (sin − r cos2 ) = 0
r = 0 or sin = r cos2 r = 0 is the pole. The second equation becomes r =
tan cos i is an imaginary number with the property i 2 = −1.
C, the set of complex numbers, is defined by C = {a + bi : a, b ∈ R}.
Real numbers and Imaginary numbers are subsets of C.
Re (z) is the real component of z.
Im (z) is the value of the imaginary component of z.
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Review
Chapter summary
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Essential Advanced General Mathematics
z = r (cos + i sin ) where r = |z|
Im (z)
Re (z)
and sin =
cos =
|z|
|z|
E
z 1 = z 2 ⇔ Re (z 1 ) = Re (z 2 ) and Im (z 1 ) = Im (z 2 )
The Argand diagram is a geometrical representation of C.
Let z 1 = a + bi and z 2 = c + di, then z 1 z 2 = (ac − bd) + (ad + bc)i
The modulus of z, |z|, is the distance from the origin of the point represented by z.
The argument of z, arg z, is an angle measured anticlockwise about the origin from the
positive direction of the x axis to the line joining the origin to z.
The Argument of z, Arg z, is arg z expressed as an angle in the interval (−, ].
The modulus-argument form of the complex number z is given as:
PL
r (cos + i sin ) is usually written as r cis The complex conjugate of z is denoted by z, where z = Re (z) − Im (z)i; zz, (z + z) ∈ R
z1
z1 z2
The division of complex numbers:
=
z2
|z 2 |2
Multiplication and division of the modulus-argument form
Let z 1 = r1 cis 1 , z 2 = r2 cis 2
Then z 1 z 2 = r1r2 cis (1 + 2 ),
z1
r1
= cis (1 − 2 )
z2
r2
Multiple-choice questions
SA
M
Review
436
1 The polar coordinates [−3, 30◦ ] define a point that can also be described by
B [3, −30◦ ]
C [3, 150◦ ]
D [3, −150◦ ] E [−3, 150◦ ]
A [3, 30◦ ]
2 The polar coordinates of point A are
B [2, −40◦ ]
C [2, 140◦ ]
A [2, 40◦ ]
◦
◦
E [2, −140 ]
D [4, 40 ]
A
2
40°
O
√
3 The polar coordinates of the point with cartesian coordinates (−1, − 3) are
2
−4
C
2,
D
2,
E −2, −
A 2,
B −2,
3
3
3
3
3
4 The cartesian coordinates of the point with polar coordinates 3,
are
√ 6
√ √ √
3 3 3
3 3 3
1
3
3
,
C
,
D
B
3, √
A (3, 3)
,
E
2 2
2 2
2 2
3
5 The polar equation of the circle with centre 3,
and radius 3 is
2
√
A r = 3 sin B r = 3
C r =3
D r = 3 cos E r = 6 sin Z
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Chapter 16 — Polar coordinates and complex numbers
Review
6 The graph of r cos = 2 is
A
B
437
C
O
O
O
E
D
O
E
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PL
O
SA
M
7 The polar equation of the circle with cartesian equation x 2 + y 2 = 16 is
D r = 4 cos E r = 4
A r = 16
B r = 4 sin C r 2 + cos2 = 4
1
8 If u = 1 + i, then
=
2−u
1 1
1 2
1 1
1 1
A − − i
+ i
B
+ i
C
E 1 + 5i
D − + i
2 2
5 5
2 2
2 5
9 The point C on the Argand diagram represents the complex number z. Which point
represents the complex number i × z?
Im (z)
A A
B B
C C
D D
E E
1
10 If |z| = 5 then =
z
1
1
A √
B −√
5
5
C
B
C
A
Re (z)
D
1
5
D −
1
5
E
E
√
5
Short-answer questions (technology-free)
1 Graph each of the following.
a [3, ]
b 2,
3
◦
c [−2, 210 ]
d
11
−3,
6
2 Find the cartesian coordinates of the points in 1.
Cambridge University Press • Uncorrected Sample Pages • 978-0-521-61252-4
2008 © Evans, Lipson, Jones, Avery, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
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CUAU033-EVANS
September 12, 2008
10:50
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Essential Advanced General Mathematics
E
3 Graph each of the following.
a {[r, ] : r = 3}
b [r, ] : =
3
−5
c {[r, ] : r = −4} d [r, ] : =
4
4 Express each of the following in polar form.
√
√ √
√
−5 −5 3
3 −1
a (3, 3)
c
,
d (4 2, −4 2)
,
b
2
2
2
2
5 Transform each of the following equations from cartesian to polar form.
b x 2 + y2 = 9
c y 2 = 8x
a x 2 + y 2 = 16
e x 2 + 4y 2 = 64
f 2x − y + 2 = 0
d x 2 = 4y
PL
6 Transform each of the following equations from polar to cartesian form.
a r =5
b r = 3 sin c r 2 cos 2 = 9
d r (1 − 2 cos ) = 8
e r (2 − cos ) = 7
f r (1 − sin ) = 1
7 For z 1 = m + in and z 2 = p + iq, express each of the following in the form a + ib.
b z2
c z1 z2
a 2z 1 + 3z 2
z1
e z1 + z1
f (z 1 + z 2 )(z 1 − z 2 )
d
z2
z2
1
3z 1
h
g
i
z1
z1
z2
√
8 In the following, z = 1 − 3i. Express each in the form a + ib and mark each of the
following on an Argand diagram.
1
1
c z3
d
f
e z
a z
b z2
z
z
9 Write each of the following in polar form.
√
√
c 2 3+i
a 1+i
b 1 − 3i
√
√
√
√
√
e −3 2 − 3 2i
f
3−i
d 3 2 + 3 2i
SA
M
Review
438
10 Write each of the following in cartesian form.
a −2 cis
b 3 cis
3
4
−3
−5
d −3 cis
e 3 cis
4
6
3
4
√
−
2 cis
f
4
c 3 cis
Extended-response questions
1 a Plot the graphs of r = 2 sin and r = 2 cos .
b Write the corresponding cartesian equation for each of these relations.
c Describe the curves you obtain from the polar equations r = 2a sin or r = 2a cos where a is a non-zero constant.
Cambridge University Press • Uncorrected Sample Pages • 978-0-521-61252-4
2008 © Evans, Lipson, Jones, Avery, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
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Chapter 16 — Polar coordinates and complex numbers
439
E
SA
M
PL
4 Let z be a complex number with |z| = 6. Let A be the point representing z. Let B be the
point representing (1 + i)z.
Find
a |(1 + i)z|
b |(1 + i)z − z|
c Prove that OAB is an isosceles right-angled triangle.
1
1
5 Let z = √ + √ i
2
2
a On an Argand diagram O, A, Z , P, Q represent the complex numbers 0, 1, z, 1 + z and
1 − z respectively. Show these points on a diagram.
|OP|
b Prove that the magnitude of ∠POQ = . Find the ratio
.
|OQ|
2
Cambridge University Press • Uncorrected Sample Pages • 978-0-521-61252-4
2008 © Evans, Lipson, Jones, Avery, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
Review
2 Investigate each of the families of graphs defined by:
a r = a + b sin or r = a + b cos where a and b are non-zero constants
b r 2 = a 2 sin 2, r 2 = −a 2 sin 2, r 2 = a 2 cos 2, r 2 = −a 2 cos 2 where a is a positive
constant
c r = a
d r = a sin n and r = a cos n where a is a non-zero constant
√
3 a Find the exact solutions in C for the equation z 2 − 2 3z + 4 = 0.
b i Plot the two solutions from a on an Argand diagram.
ii Find the equation of the circle, with centre the origin, which passes through these
two points.
iii Find the value of a ∈ Z such that the circle passes through (0, ±a)