“JUST THE MATHS”
UNIT NUMBER
6.1
COMPLEX NUMBERS 1
(Definitions and algebra)
by
A.J.Hobson
6.1.1
6.1.2
6.1.3
6.1.4
The definition of a complex number
The algebra of complex numbers
Exercises
Answers to exercises
UNIT 6.1 - COMPLEX NUMBERS 1 - DEFINITIONS AND ALGEBRA
6.1.1 THE DEFINITION OF A COMPLEX NUMBER
Students who are already familiar with the Differential Calculus may appreciate that equations of the form
a
d2 y
dy
+b
+ cy = f (x),
2
dx
dx
which are called “Differential Equations”, have wide-reaching applications in science and
engineering. They are particularly applicable to problems involving either electrical circuits
or mechanical vibrations.
It is possible to show that, in order to determine a formula (without derivatives) giving the
variable y in terms of the variable x, one method is to solve, first, the quadratic equation
whose coefficents are a, b and c and whose solutions are therefore
−b ±
√
b2 − 4ac
.
2a
Note:
Students who are not already familiar with the Differential Calculus should consider only
the quadratic equation whose coefficients are a, b and c, ignoring references to differential
equations.
ILLUSTRATION
One method of solving the differential equation
d2 y
dy
−6
+ 13 = 2 sin x
2
dx
dx
would be to solve, first, the quadratic equation whose coefficients are 1, −6 and 13.
Its solutions are
√
36 − 52
6 ± −16
=
2
2
which clearly do not exist since we cannot find the square root of a negative number in
elementary arithmetic.
6±
√
1
However, if we assume that the differential equation represents a genuine scientific problem
with a genuine scientific solution, we cannot simply dismiss the result obtained from the
quadratic formula.
The difficulty seems to be, not so much with the −16 but with the minus sign in front of the
16. We shall therefore write the solutions in the form
√
√
6 ± 4 −1
= 3 ± 2 −1.
2
Notes:
√
(i) The symbol −1 will be regarded as an “imaginary” number.
√
(ii) In mathematical work, −1 is normally denoted by i but, in scientific work it is denoted
by j in order to avoid confusion with other quantities (eg. electric current) which could be
denoted by the same symbol.
√
(iii) Whenever the imaginary quantity j = −1 occurs in the solutions of a quadratic
equation, those solutions will always be of the form a + bj (or a + jb), where a and b are
ordinary numbers of elementary arithmetic.
DEFINITIONS
1. The term “complex number” is used to denote any expression of the form a + bj or
a + jb where a and b are ordinary numbers
of elementary arithmetic (including zero)
√
and j denotes the imaginary number −1; i.e. j 2 = −1.
2. If the value a happens to be zero, then the complex number a + bj or a + jb is called
“purely imaginary” and is written bj or jb.
3. If the value b happens to be zero, then the complex number a + bj or a + jb is defined
to be the same as the number a and is called “real”. That is a + j0 = a + 0j = a.
4. For the complex number a + bj or a + jb, the value a is called the “real part” and the
value b is called the “imaginary part”. Notice that the imaginary part is b and not
jb.
5. The complex numbers a ± bj are said to form a pair of “complex conjugates” and
similarly a ± jb form a pair of complex conjugates. Alternatively, we may say, for
instance, that a − jb is the complex conjugate of a + jb and a + jb is the complex
conjugate of a − jb.
Note:
In some work on complex numbers, especially where many complex numbers may be under
2
discussion at the same time, it is convenient to denote real and imaginary parts by the
symbols x and y respectively, rather than a and b. It is also convenient, on some occasions,
to denote the whole complex number x + jy by the symbol z in which case the conjugate,
x − jy, will be denoted by z.
6.1.2 THE ALGEBRA OF COMPLEX NUMBERS
INTRODUCTION
An “Algebra” (coming from the Arabic word AL-JABR) refers to any mathematical system
which uses the concepts of equality, addition, subtraction, multiplication and division. For
example, the algebra of real numbers is what we normally call “arithmetic”; but algebraical
concepts can be applied to other mathematical systems of which the system of complex
numbers is one.
In meeting a new mathematical system for the first time, the concepts of equality, addition,
subtraction, multiplication and division need to be properly defined, and that is the purpose
of the present section. In some cases, the definitions are fairly obvious, but need to be made
without contradicting ideas already established in the system of real numbers which complex
numbers include.
(a) EQUALITY
Unlike a real number, a complex number does not have a “value”; and so the word “equality”
must take on a meaning, here, which is different from that used in elementary arithmetic.
In fact two complex numbers are defined to be equal if they have the same real part and the
same imaginary part.
That is
a + jb = c + jd if and only if a = c and b = d
EXAMPLE
Determine x and y such that
(2x − 3y) + j(x + 5y) = 11 − j14.
3
Solution
From the definition of equality, we may
EQUATE REAL AND IMAGINARY PARTS.
Thus,
2x − 3y = 11,
x + 5y = −14
These simultaneous linear equations are satisfied by x = 1 and y = −3.
(b) ADDITION AND SUBTRACTION
These two concepts are very easily defined. We simply add (or subtract) the real parts and
the imaginary parts of the two complex numbers whose sum (or difference) is required.
That is,
(a + jb) + (c + jd) = (a + c) + j(b + d)
and
(a + jb) − (c + jd) = (a − c) + j(b − d).
EXAMPLE
(−7 + j2) + (10 − j5) = 3 − j3 = 3(1 − j)
and
(−7 + j2) − (10 − j5) = −17 + j7.
4
(c) MULTIPLICATION
The definition of multiplication essentially treats j in the same way as any other algebraic
symbol, but uses the fact that j 2 = −1.
Thus,
(a + jb)(c + jd) = (ac − bd) + j(bc + ad);
but this is not so much a formula to be learned off-by-heart as a technique to be applied in
future examples.
EXAMPLES
1.
(5 + j9)(2 + j6) = (10 − 54) + j(18 + 30) = −44 + j48.
2.
(3 − j8)(1 + j4) = (3 + 32) + j(−8 + 12) = 35 + j4.
3.
(a + jb)(a − jb) = a2 + b2 .
Note:
The third example above will be useful in the next section. It shows that the product of
a complex number and its complex conjugate is always a real number consisting
of the sum of the squares of the real and imaginary parts.
(d) DIVISION
The objective here is to make a definition which provides the real and imaginary parts of
the complex expression
a + jb
.
c + jd
Once again, we make this definition in accordance with what would be obtained algebraically
by treating j in the same way as any other algebraic symbol, but using the fact that j 2 = −1.
5
The method is to multiply both the numerator and the denominator of the complex ratio
by the conjugate of the denominator giving
a + jb c − jd
(ac + bd) + j(bc − ad)
.
=
.
c + jd c − jd
c2 + d2
The required definition is thus
a + jb
(ac + bd) + j(bc − ad)
=
,
c + jd
c2 + d2
which, again, is not so much a formula to be learned off-by-heart as a technique to be applied
in future examples.
EXAMPLES
1.
5 + j3
5 + j3 2 − j7
=
.
2 + j7
2 + j7 2 − j7
=
Hence, the real part is
31
53
(10 + 21) + j(6 − 35)
31 − j29
=
.
2
2
2 +7
53
and the imaginary part is − 29
.
53
2.
6+j
6 + j −j2 − 4
=
.
j2 − 4
j2 − 4 −j2 − 4
=
(−24 + 2) + j(−4 − 12)
−22 − j16
=
.
2
2
(−2) + (−4)
20
Hence, the real part is − 22
= − 11
and the imaginary part is − 16
= − 45 .
20
10
20
6
6.1.3 EXERCISES
1. Simplify the following:
(a) j 3 ; (b) j 4 ; (c) j 5 ; (d) j 15 ; (e) j 22 .
2. If z1 = 2 − j5, z2 = 1 + j7 and z3 = −3 − j4, determine the following in the form a + jb:
(a)
z1 − z2 + z3 ;
(b)
2z1 + z2 − z3 ;
(c)
z1 − (4z2 − z3 );
(d)
z1
;
z2
(e)
z2
;
z3
(f)
z3
.
z1
3. Determine the values of x and y such that
(3x − 5y) + j(x + 3y) = 20 + j2.
4. Determine the real and imaginary parts of the expression
(1 − j3)2 + j(2 + j5) −
3(4 − j)
.
1−j
5. If z ≡ x + jy and z ≡ x − jy are conjugate complex numbers, determine the values of
x and y such that
4zz − 3(z − z) = 2 + j.
7
6.1.4 ANSWERS TO EXERCISES
1. (a) −j; (b) 1; (c) j; (d) −j; (e) −1.
2. (a)
4 − j16;
(b)
8 + j;
(c)
−5 − j37;
(d)
−0.66 − j0.38;
(e)
−1.24 − j0.68;
(f)
0.48 − j0.79
3.
x = 5 and y = −1.
4. The real part = −20.5; the imaginary part = −8.5
5.
x=±
1
1
and y = − .
2
2
8
“JUST THE MATHS”
UNIT NUMBER
6.2
COMPLEX NUMBERS 2
(The Argand Diagram)
by
A.J.Hobson
6.2.1
6.2.2
6.2.3
6.2.4
6.2.5
6.2.6
Introduction
Graphical addition and subtraction
Multiplication by j
Modulus and argument
Exercises
Answers to exercises
UNIT 6.2 - COMPLEX NUMBERS 2
THE ARGAND DIAGRAM
6.2.1 INTRODUCTION
It may be observed that a complex number x + jy is completely specified if we know the
values of x and y in the correct order. But the same is true for the cartesian co-ordinates,
(x, y), of a point in two dimesions. There is therefore a “one-to-one correspondence”
between the complex number x + jy and the point with co-ordinates (x, y).
Hence it is possible to represent the complex number x+jy by the point (x, y) in a geometrical
diagram called the Argand Diagram:
y
6
......................... (x, y)
..
.
..
.
..
. -x
O
DEFINITIONS:
1. The x-axis is called the “real axis” since the points on it represent real numbers.
2. The y-axis is called the “imaginary axis” since the points on it represent purely
imaginary numbers.
6.2.2 GRAPHICAL ADDITON AND SUBTRACTION
If two complex numbers, z1 = x1 + jy1 and z2 = x2 + jy2 , are represented in the Argand
Diagram by the points P1 (x1 , y1 ) and P2 (x2 , y2 ) respectively, then the sum, z1 + z2 , of the
complex numbers will be represented by the point Q(x1 + x2 , y1 + y2 ).
If O is the origin, it is possible to show that Q is the fourth vertex of the parallelogram
having OP1 and OP2 as adjacent sides.
1
y
Q
P1
S
1
P2
- x
O
6
R
In the diagram, the triangle ORP1 has exactly the same shape as the triangle P2 SQ. Hence,
the co-ordinates of Q must be (x1 + x2 , y1 + y2 ).
Note:
The difference z1 − z2 of the two complex numbers may similarly be found by completing the
parallelogram of which two adjacent sides are the straight line segments joining the origin
to the points with co-ordinates (x1 , y1 ) and (−x2 , −y2 ).
6.2.3 MULTIPLICATION BY j OF A COMPLEX NUMBER
Given any complex number z = x + jy, we observe that
jz = j(x + jy) = −y + jx.
Thus, if z is represented in the Argand Diagram by the point with co-ordinates A(x, y), then
jz is represented by the point with co-ordinates B(−y, x).
2
B
y
6
A
A
A
A
A
A
AO
-x
But OB is in the position which would be occupied by OA if it were rotated through 90◦ in
a counter-clockwise direction.
We conclude that, in the Argand Diagram, multiplication by j of a complex number rotates,
through 90◦ in a counter-clockwise direction, the straight line segment joining the origin to
the point representing the complex number.
6.2.4 MODULUS AND ARGUMENT
y
P(x, y)
*
r θ
-x
O
6
(a) Modulus
If a complex number, z = x + jy is represented in the Argand Diagram by the point, P,
3
with cartesian co-ordinates (x, y) then the distance, r, of P from the origin is called the
“modulus” of z and is denoted by either |z| or |x + jy|.
Using the theorem of Pythagoras in the diagram, we conclude that
r = |z| = |x + jy| =
q
x2 + y 2 .
Note:
This definition of modulus is consistent with the definition of modulus for real numbers
(which are included in the system of complex numbers). For any real number x, we may say
that
|x| = |x + j0| =
√
√
x2 + 02 =
x2 ,
giving the usual numerical value of x.
ILLUSTRATIONS
1.
|3 − j4| =
q
32 + (−4)2 =
√
25 = 5.
2.
|1 + j| =
√
12 + 12 =
√
2.
3.
|j7| = |0 + j7| =
√
02 + 72 =
√
49 = 7.
Note:
The result of the last example above is obvious from the Argand Diagram since the point on
the y-axis representing j7 is a distance of exactly 7 units from the origin. In the same way,
a real number is represented by a point on the x-axis whose distance from the origin is the
numerical value of the real number.
(b) Argument
The “argument” (or “amplitude”) of a complex number, z, is defined to be the angle, θ,
which the straight line segment OP makes with the positive real axis (measuring θ positively
from this axis in a counter-clockwise sense).
4
In the diagram,
tan θ =
y
y
; that is, θ = tan−1 .
x
x
Note:
For a given complex number, there will be infinitely many possible values of the argument,
any two of which will differ by a whole multiple of 360◦ . The complete set of possible values
is denoted by Argz, using an upper-case A.
The particular value of the argument which lies in the interval −180◦ < θ ≤ 180◦ is called
the “principal value” of the argument and is denoted by arg z using a lower-case a. The
particular value, 180◦ , in preference to −180◦ , represents the principal value of the argument
of a negative real number.
ILLUSTRATIONS
1.
√
1
Arg( 3 + j) = tan−1 √
3
!
= 30◦ + k360◦ ,
where k may be any integer. But we note that
√
arg( 3 + j) = 30◦ only.
2.
Arg(−1 + j) = tan−1 (−1) = 135◦ + k360◦
but not −45◦ + k360◦ , since the complex number −1 + j is represented by a point in
the second quadrant of the Argand Diagram.
We note also that
arg(−1 + j) = 135◦ only.
3.
Arg(−1 − j) = tan−1 (1) = 225◦ + k360◦ or − 135◦ + k360◦
but not 45◦ + k360◦ since the complex number −1 − j is represented by a point in the
third quadrant of the Argand Diagram.
We note also that
arg(−1 − j) = −135◦ only.
5
Note:
It is worth mentioning here that, in the Argand Diagram, the directed straight line segment
described from the point P1 (representing the complex number z1 = x1 + jy1 ) to the point
P2 (representing the complex number z2 = x2 + jy2 ) has length, r, equal to |z2 − z1 |, and is
inclined to the positive direction of the real axis at an angle, θ, equal to arg(z2 − z1 ). This
follows from the relationship
z2 − z1 = (x2 − x1 ) + j(y2 − y1 )
in which x2 − x1 and y2 − y1 are the distances separating the two points, parallel to the real
axis and the imaginary axis respectively.
y
P2
6
y2 − y1
r
-x
O
P1 θ
x2 − x1
6.2.5 EXERCISES
1. Determine the modulus (in decimals, where appropriate, correct to three significant
figures) and the principal value of the argument (in degrees, correct to the nearest
degree) of the following complex numbers:
(a)
1 − j;
(b)
−3 + j4;
(c)
√
√
− 2 − j 2;
6
(d)
√
1
3
−j
;
2
2
(e)
−7 − j9.
2. If z = 4 − j5, verify that jz has the same modulus as z but that the principal value of
the argument of jz is greater, by 90◦ than the principal value of the argument of z.
3. Illustrate the following statements in the Argand Diagram:
(a)
(6 − j11) + (5 + j3) = 11 − j8;
(b)
(6 − j11) − (5 + j3) = −1 − j14.
6.2.6 ANSWERS TO EXERCISES
1. (a) 1.41 and −45◦ ;
(b) 5 and 127◦ ;
(c) 2 and −135◦ ;
(d) 1 and −60◦ ;
(e) 11.4 and −128◦ .
√
2. 4 − j5 has modulus 41 and argument
−51◦ ;
√
j(4 − j5) = 5 + j4 has modulus 41 and argument 39◦ = −51◦ + 90◦ .
3. Construct the graphical sum and difference of the two complex numbers.
7
“JUST THE MATHS”
UNIT NUMBER
6.3
COMPLEX NUMBERS 3
(The polar & exponential forms)
by
A.J.Hobson
6.3.1
6.3.2
6.3.3
6.3.4
6.3.5
The polar form
The exponential form
Products and quotients in polar form
Exercises
Answers to exercises
UNIT 6.3 - COMPLEX NUMBERS 3
THE POLAR AND EXPONENTIAL FORMS
6.3.1 THE POLAR FORM
y
P(x, y)
*
6
r θ
O
-x
From the above diagram, we may observe that
x
y
= cos θ and
= sin θ.
r
r
Hence, the relationship between x, y, r and θ may also be stated in the form
x = r cos θ,
y = r sin θ,
which means that the complex number x + jy may be written as r cos θ + jr sin θ. In other
words,
x + jy = r(cos θ + j sin θ).
The left-hand-side of this relationship is called the “rectangular form” or “cartesian
form” of the complex number while the right-hand-side is called the “polar form”.
1
Note:
For convenience, the polar form may be abbreviated to r6 θ, where θ may be positive, negative
or zero and may be expressed in either degrees or radians.
EXAMPLES
1. Express the complex number z =
√
3 + j in polar form.
Solution
|z| = r =
√
3+1=2
and
1
Argz = θ = tan−1 √ = 30◦ + k360◦ ,
3
where k may be any integer.
Alternatively, using radians,
Argz =
π
+ k2π,
6
where k may be any integer.
Hence, in polar form,
z = 2(cos[30◦ + k360◦ ] + j sin[30◦ + k360◦ ]) = 26 [30◦ + k360◦ ]
or
z = 2 cos
π
π
+ k2π + j sin
+ k2π
6
6
= 26
π
+ k2π .
6
2. Express the complex number z = −1 − j in polar form.
Solution
√
√
|z| = r = 1 + 1 = 2
and
Argz = θ = tan−1 (1) = −135◦ + k360◦ ,
where k may be any integer.
2
Alternatively,
Argz = −
3π
+ k2π,
4
where k may be any integer.
Hence, in polar form,
z=
√
2(cos[−135◦ + k360◦ ] + j sin[−135◦ + k360◦ ]) =
√
26 [−135◦ + k360◦ ]
or
√ 3π
√ 3π
3π
6
z = 2 cos −
+ k2π + j sin −
+ k2π = 2 −
+ k2π .
4
4
4
Note:
If it is required that the polar form should contain only the principal value of the argument,
θ, then, provided −180◦ < θ ≤ 180◦ or −π < θ ≤ π, the component k360◦ or k2π of the
result is simply omitted.
6.3.2 THE EXPONENTIAL FORM
Using some theory from the differential calculus of complex variables (not included here) it
is possible to show that, for any complex number, z,
ez = 1 +
z
z2 z3 z4
+
+
+
+ ...,
1! 2!
3!
4!
sin z = z −
z3 z5 z7
+
−
+ ...
3!
5!
7!
cos z = 1 −
z2 z4 z6
+
−
+ ...
2!
4!
6!
and
These are, in fact, taken as the definitions of the functions ez , sin z and cos z.
Students who are already familiar with the differential calculus of a real variable, x, may
recognise similarities between the above formulae and the “MacLaurin Series” for the functions ex , sin x and cos x. In the case of the series for sin x and cos x, the value, x, must be
expressed in radians and not degrees.
A useful deduction can be made from the three formulae if we make the substitution z = jθ
into the first one, obtaining:
3
ejθ = 1 +
jθ (jθ)2 (jθ)3 (jθ)4
+
+
+
+ ...
1!
2!
3!
4!
and, since j 2 = −1, this gives
θ
θ2
θ3 θ4
e =1+j −
−j +
+ ...
1! 2!
3!
4!
jθ
On regrouping this into real and imaginary parts, then using the sine and cosine series, we
obtain
ejθ = cos θ + j sin θ,
provided θ is expressed in radians and not degrees.
The complex number x+jy, having modulus r and argument θ +k2π, may thus be expressed
not only in polar form but also in
the exponential form, rejθ .
ILLUSTRATIONS
Using the examples of the previous section
1.
√
π
3 + j = 2ej ( 6 +k2π) .
2.
√
−1 + j =
3π
2ej ( 4 +k2π) .
3.
−1 − j =
√
3π
2e−j ( 4 +k2π) .
Note:
If it is required that the exponential form should contain only the principal value of the
argument, θ, then, provided −π < θ ≤ π, the component k2π of the result is simply omitted.
4
6.3.3 PRODUCTS AND QUOTIENTS IN POLAR FORM
Let us suppose that two complex numbers z1 and z2 have already been expressed in polar
form, so that
z1 = r1 (cos θ1 + j sin θ1 ) = r1 6 θ1
and
z2 = r2 (cos θ2 + j sin θ2 ) = r2 6 θ2 .
It is then possible to establish very simple rules for determining both the product and the
quotient of the two complex numbers. The explanation is as follows:
(a) The Product
z1 .z2 = r1 .r2 (cos θ1 + j sin θ1 ).(cos θ2 + j sin θ2 ).
That is,
z1 .z2 = r1 .r2 ([cos θ1 . cos θ2 − sin θ1 . sin θ2 ] + j[sin θ1 . cos θ2 + cos θ1 . sin θ2 ]).
Using trigonometric identities, this reduces to
z1 .z2 = r1 .r2 (cos[θ1 + θ2 ] + j sin[θ1 + θ2 ]) = r1 .r2 6 [θ1 + θ2 ].
We have shown that, to determine the product of two complex numbers in polar form, we
construct the product of their modulus values and the sum of their argument values.
(b) The Quotient
z1
r1 (cos θ1 + j sin θ1 )
=
.
z2
r2 (cos θ2 + j sin θ2 )
On multiplying the numerator and denominator by cos θ2 − j sin θ2 , we obtain
5
r1
z1
= ([cos θ1 . cos θ2 + sin θ1 . sin θ2 ] + j[sin θ1 . cos θ2 − cos θ1 . sin θ2 ]).
z2
r2
Using trigonometric identities, this reduces to
z1
r1
r1
= (cos[θ1 − θ2 ] + j sin[θ1 − θ2 ]) = 6 [θ1 − θ2 ].
z2
r2
r2
We have shown that, to determine the quotient of two complex numbers in polar form, we
construct the quotient of their modulus values and the difference of their argument values.
ILLUSTRATIONS
Using results from earlier examples:
1.
√
√
√
( 3 + j).(−1 − j) = 26 30◦ . 26 (−135◦ ) = 2 26 (−105◦ ).
We notice that, for all of the complex numbers in this example, including the result,
the argument appears as the principal value.
2.
√
√
3+j
26 30◦
=√
=
26 165◦ .
−1 − j
26 (−135◦ )
Again, for all of the complex numbers in this example, including the result, the argument
appears as the principal value.
Note:
It will not always turn out that the argument of a product or quotient of two complex
numbers appears as the principal value. For instance,
3.
√
√
√
(−1 − j).(− 3 − j) = 26 (−135◦ ).26 (−150◦ ) = 2 26 (−285◦ ),
√
which must be converted to 2 26 (75◦ ) if the principal value of the argument is required.
6
6.3.4 EXERCISES
In the following cases, express the complex numbers z1 and z2 in
(a) the polar form, r6 θ
and
(b) the exponential form, rejθ
using only the principal value of θ.
(c) For each case, determine also the product, z1 .z2 , and the quotient,
only the principal value of the argument.
1.
z1 = 1 + j, z2 =
√
3 − j.
2.
√
√
z1 = − 2 + j 2, z2 = −3 − j4.
3.
z1 = −4 − j5, z2 = 7 − j9.
6.3.5 ANSWERS TO EXERCISES
1. (a)
z1 =
√
26 45◦ z2 = 26 (−30◦ );
(b)
z1 =
√
π
π
2ej 4 z2 = 2e−j 6 ;
(c)
√
√
z1
2
◦
6 75◦ .
z1 .z2 = 2 26 15
=
z2
2
2. (a)
z1 = 26 (135◦ ) z2 = 56 (−127◦ );
7
z1
,
z2
in polar form using
(b)
z1 = 2ej
3π
4
z2 = 5e−j2.22 ;
(c)
z1 .z2 = 106 8◦
z1
2
= 6 (−98◦ ).
z2
5
3. (a)
z1 = 6.406 (−128.66◦ ) z2 = 11.406 (−55.13◦ );
(b)
z1 = 6.40e−j2.25 z2 = 11.40e−j0.96 ;
(c)
z1 .z2 = 72.966 176.21◦
8
z1
= 0.566 (−73.53◦ ).
z2
“JUST THE MATHS”
UNIT NUMBER
6.4
COMPLEX NUMBERS 4
(Powers of complex numbers)
by
A.J.Hobson
6.4.1
6.4.2
6.4.3
6.4.4
6.4.5
Positive whole number powers
Negative whole number powers
Fractional powers & De Moivre’s Theorem
Exercises
Answers to exercises
UNIT 6.4 - COMPLEX NUMBERS 4
POWERS OF COMPLEX NUMBERS
6.4.1 POSITIVE WHOLE NUMBER POWERS
As an application of the rule for multiplying together complex numbers in polar form, it is
a simple matter to multiply a complex number by itself any desired number of times.
Suppose that
z = r6 θ.
Then,
z 2 = r.r6 (θ + θ) = r2 6 2θ;
z 3 = z.z 2 = r.r2 6 (θ + 2θ) = r3 6 3θ;
and, by continuing this process,
z n = rn 6 nθ.
This result is due to De Moivre, but other aspects of it will need to be discussed before we
may formalise what is called “De Moivre’s Theorem”.
EXAMPLE
1
1
√ + j√
2
2
!19
π 19
) = 16
4
=
(16
19π
= 16
4
6.4.2 NEGATIVE WHOLE NUMBER POWERS
If n is a negative whole number, we shall suppose that
n = −m,
where m is a positive whole number.
Thus, if z = r6 θ,
1
3π
1
1
= −√ + j √ .
4
2
2
z n = z −m =
1
1
= m
.
m
6
z
r mθ
In more detail,
zn =
1
,
rm (cos mθ + j sin mθ)
giving
zn =
1 (cos mθ − j sin mθ)
.
= r−m (cos[−mθ] + j sin[−mθ]).
rm cos2 mθ + sin2 mθ
But −m = n, and so
z n = rn (cos nθ + j sin nθ) = rn 6 nθ,
showing that the result of the previous section remains true for negative whole number
powers.
EXAMPLE
√
1
j
( 3 + j)−3 = (26 30◦ )−3 = 6 (−90◦ ) = − .
8
8
6.4.3 FRACTIONAL POWERS AND DE MOIVRE’S THEOREM
To begin with, here, we consider the complex number
1
zn,
where n is a positive whole number and z = r6 θ.
1
We define z n to be any complex number which gives z itself when raised to the power n.
Such a complex number is called “an n-th root of z”.
2
Certainly one such possibility is
θ
,
n
1
rn6
by virtue of the paragraph dealing with positive whole number powers.
But the general expression for z is given by
z = r6 (θ + k360◦ ),
1
where k may be any integer; and this suggests other possibilities for z n , namely
r
1
n6
θ + k360◦
.
n
However, this set of n-th roots is not an infinite set because the roots which are given by
k = 0, 1, 2, 3............n − 1 are also given by k = n, n + 1, n + 2, n + 3, ......., 2n − 1, 2n,
2n + 1, 2n + 2, 2n + 3, ..... and so on, respectively.
We conclude that there are precisely n n-th roots given by k = 0, 1, 2, 3........., n − 1.
EXAMPLE
Determine the cube roots (i.e. 3rd roots) of the complex number j8.
Solution
We first write
j8 = 86 (90◦ + k360◦ ).
Hence,
1
3
(j8) = 8
1
36
(90◦ + k360◦ )
,
3
where k = 0, 1, 2
3
The three distinct cube roots are therefore
26 30◦ , 26 150◦ and 26 270◦ = 26 (−90◦ ).
They all have the same modulus of 2 but their arguments are spaced around the Argand
◦
Diagram at regular intervals of 360
= 120◦ .
3
Notes:
(i) In general, the n-th roots of a complex number will all have the same modulus, but their
◦
arguments will be spaced at regular intervals of 360
.
n
(ii) Assuming that −180◦ < θ ≤ 180◦ ; that is, assuming that the polar form of z uses the
principal value of the argument, then the particular n-th root of z which is given by k = 0
is called the “principal n-th root”.
(iii) If
m
n
is a fraction in its lowest terms, we define
m
zn
1
to be either z n
m
1
or (z m ) n both of which turn out to give the same set of n distinct results.
The discussion, so far, on powers of complex numbers leads us to the following statement:
DE MOIVRE’S THEOREM
If z = r6 θ, then, for any rational number n, one value of z n is rn 6 nθ.
6.4.4 EXERCISES
1. Determine the following in the form a + jb, expressing a and b in decimals correct to
four significant figures:
(a)
√
(1 + j 3)10 ;
(b)
(2 − j5)−4 .
2. Determine the fourth roots of j81 in exponential form rejθ where r > 0 and −π < θ ≤ π.
3. Determine the fifth roots of the complex number −4 + j4 in the form a + jb expressing
a and b in decimals, where appropriate, correct to two places. State also which root is
the principal root.
4
4. Determine all the values of
3
(3 + j4) 2
in polar form.
6.4.5 ANSWERS TO EXERCISES
1. (a)
√
(1 + j 3)10 = −512.0 − j886.8;
(b)
(2 − j5)−4 = 5.796 − j1.188
2. The fourth roots are
π
3π
7π
5π
3e− 8 , 3e 8 , 3e 8 , 3e− 8 .
3. The fifth roots are
1.26 + j0.64,
− 0.22 + j1.40,
− 1.40 + j0.22,
− 0.64 − j1.26, 1 − j.
The principal root is 1.26 + j0.64.
4. There are two values, namely
11.186 79.695◦ and 11.186 (−100.305◦ ).
5
“JUST THE MATHS”
UNIT NUMBER
6.5
COMPLEX NUMBERS 5
(Applications to trigonometric identities)
by
A.J.Hobson
6.5.1 Introduction
6.5.2 Expressions for cos nθ, sin nθ in terms of cos θ, sin θ
6.5.3 Expressions for cosn θ and sinn θ in terms of sines and
cosines of whole multiples of x
6.5.4 Exercises
6.5.5 Answers to exercises
UNIT 6.5 - COMPLEX NUMBERS 5
APPLICATIONS TO TRIGONOMETRIC IDENTITIES
6.5.1 INTRODUCTION
It will be useful for the purposes of this section to restate the result known as “Pascal’s
Triangle” previously discussed in Unit 2.2.
If n is a positive whole number, the diagram
1
1
1
1
1
2
3
4
1
3
6
1
4
1
provides the coefficients in the expansion of (A + B)n which contains the sequence of terms
An , An−1 B, An−2 B 2 , An−3 B 3 , ........., B n .
6.5.2 EXPRESSIONS FOR cos nθ AND sin nθ IN TERMS OF cos θ AND sin θ.
From De Moivre’s Theorem
(cos θ + j sin θ)n ≡ cos nθ + j sin nθ,
from which we may deduce that, in the expansion of the left-hand-side, using Pascal’s Triangle, the real part will coincide with cos nθ and the imaginary part will coincide with sin nθ.
EXAMPLE
(cos θ + j sin θ)3 ≡ cos3 θ + 3cos2 θ.(j sin θ) + 3 cos θ.(j sin θ)2 + (j sin θ)3 .
That is,
cos 3θ ≡ cos3 θ − 3 cos θ.sin2 θ or 4cos3 θ − 3 cos θ,
1
using sin2 θ ≡ 1 − cos2 θ;
and
sin 3θ ≡ 3cos2 θ. sin θ − sin3 θ or 3 sin θ − 4sin3 θ,
using cos2 θ ≡ 1 − sin2 θ.
6.5.3 EXPRESSIONS FOR cosn θ AND sinn θ IN TERMS OF SINES AND COSINES
OF WHOLE MULTIPLES OF θ.
The technique described here is particularly useful in calculus problems when we are required
to integrate an integer power of a sine function or a cosine function. It does stand, however,
as a self-contained application to trigonometry of complex numbers.
Suppose
z ≡ cos θ + j sin θ
−
(1)
Then, by De Moivre’s Theorem, or by direct manipulation,
1
≡ cos θ − j sin θ
z
−
(2).
Adding (1) and (2) together, then subtracting (2) from (1), we obtain
z+
1
z
≡ 2 cos θ
z−
1
z
≡ j2 sin θ
Also, by De Moivre’s Theorem,
z n ≡ cos nθ + j sin nθ
−
(3)
1
≡ cos nθ − j sin nθ
zn
−
(4).
and
2
Adding (3) and (4) together, then subtracting (4) from (3), we obtain
zn +
1
zn
zn −
≡ 2 cos nθ
1
zn
≡ j2 sin nθ
We are now in a position to discuss some examples on finding trigonometric identities for
whole number powers of sin θ or cos θ.
EXAMPLES
1. Determine an identity for sin3 θ.
Solution
We use the result
j 3 23 sin3 θ ≡ z −
1
z
3
,
where z ≡ cos θ + j sin θ.
That is,
2
1
1
−j8sin3 θ ≡ z 3 − 3z 2 . + 3z.
z
z
−
1
z3
or, after cancelling common factors,
3
1
1
1
−j8sin θ ≡ z − 3z + − 3 ≡ z 3 − 3 − 3 z −
,
z z
z
z
3
3
which gives
−j8sin3 θ ≡ j2 sin 3θ − j6 sin θ.
Hence,
sin3 θ ≡
1
(3 sin θ − sin 3θ) .
4
3
2. Determine an identity for cos4 θ.
Solution
We use the result
1
2 cos θ ≡ z +
z
4
4
4
,
where z ≡ cos θ + j sin θ.
That is,
2
1
1
16cos4 θ ≡ z 4 + 4z 3 . + 6z 2 .
z
z
3
+ 4z.
1
z
4
+
1
z
or, after cancelling common factors,
4
1
1
1
16cos θ ≡ z + 4z + 6 + 2 + 4 ≡ z 4 + 4 + 4 z 2 + 2 + 6,
z
z
z
z
4
4
2
which gives
16cos4 θ ≡ 2 cos 4θ + 8 cos 2θ + 6.
Hence,
cos4 θ ≡
1
(cos 4θ + 4 cos 2θ + 3)
8
6.5.4 EXERCISES
1. Use a complex number method to determine identities for cos 4θ and sin 4θ in terms of
sin θ and cos θ.
2. Use a complex number method to determine an identity for sin5 θ in terms of sines of
whole multiples of θ.
3. Use a complex number method to deternine an identity for cos6 θ in terms of cosines of
whole multiples of θ.
4
6.5.5 ANSWERS TO EXERCISES
1.
cos 4θ ≡ cos4 θ − 6cos2 θ.sin2 θ
and
sin 4θ ≡ 4cos3 θ. sin θ − 4 cos θ.sin3 θ.
2.
sin5 θ ≡
1
(sin 5θ − 5 sin 3θ + 10 sin θ].
16
3.
cos6 θ ≡
1
(cos 6θ + 6 cos 4θ + 15 cos 2θ + 10).
32
5
“JUST THE MATHS”
UNIT NUMBER
6.6
COMPLEX NUMBERS 6
(Complex loci)
by
A.J.Hobson
6.6.1
6.6.2
6.6.3
6.6.4
6.6.5
6.6.6
Introduction
The circle
The half-straight-line
More general loci
Exercises
Answers to exercises
UNIT 6.6 - COMPLEX NUMBERS 6
COMPLEX LOCI
6.6.1 INTRODUCTION
In Unit 6.2, it was mentioned that the directed line segment joining the point representing
a complex number z1 to the point representing a complex number z2 is of length equal
to |z2 − z1 | and is inclined to the positive direction of the real axis at an angle equal to
arg(z2 − z1 ).
This observation now has significance when discussing variable complex numbers which are
constrained to move along a certain path (or “locus”) in the Argand Diagram. For many
practical applications, such paths (or “loci”) will normally be either straight lines or circles
and two standard types of example appear in what follows.
In both types, we shall assume that z = x + jy denotes a variable complex number (represented by the point (x, y) in the Argand Diagram), while z0 = x0 + jy0 denotes a fixed
complex number (represented by the point (x0 , y0 ) in the Argand Diagram).
6.6.2 THE CIRCLE
Suppose that the moving point representing z moves on a circle, with radius a, whose centre
is at the fixed point representing z0 .
y
z
6
a z0
O
-
1
x
Then the distance between these two points will always be equal to a. In other words,
|z − z0 | = a
and this is the standard equation of the circle in terms of complex numbers.
Note:
By substituting z = x + jy and z0 = x0 + jy0 in the above equation, we may obtain the
equivalent equation in terms of cartesian co-ordinates, namely,
|(x − x0 ) + j(y − y0 )| = a.
That is,
(x − x0 )2 + (y − y0 )2 = a2 .
ILLUSTRATION
The equation
|z − 3 + j4| = 7
represents a circle, with radius 7, whose centre is the point representing the complex number
3 − j4.
In cartesian co-ordinates, it is the circle with equation
(x − 3)2 + (y + 4)2 = 49.
6.6.3 THE HALF-STRAIGHT-LINE
Suppose now that the “directed” straight line segment described from the fixed point
representing z0 to the moving point representing z is inclined at an angle θ to the positive
direction of the real axis.
2
Then,
arg(z − z0 ) = θ
and this equation is satisfied by all of the values of z for which the inclination of the directed
line segment is genuinely θ and not 180◦ − θ. The latter angle would correspond to points
on the other half of the straight line joining the two points.
y
z
6
θ
*
-x
O
z0
Note:
If we substitute z = x + jy and z0 = x0 + jy0 , we obtain
arg([x − x0 ] + j[y − y0 ]) = θ.
That is,
tan−1
y − y0
=θ
x − x0
or
y − y0 = tan θ(x − x0 ),
which is certainly the equation of a straight line with gradient tan θ passing through the
point (x0 , y0 ); but it represents only that half of the straight line for which x − x0 and y − y0
correspond, in sign as well as value, to the real and imaginary parts of a complex number
whose argument is genuinely θ and not 180◦ − θ.
3
ILLUSTRATION
The equation
arg(z + 1 − j5) = −
π
6
represents the half-straight-line described from the point representing z0 = −1 + j5 to the
point representing z = x + jy and inclined to the positive direction of the real axis at an
angle of − π6 .
y
z0P
6
PP
PP
P
O
PP
q
P
-x
In terms of cartesian co-ordinates,
π
arg([x + 1] + j[y − 5]) = − ,
6
in which it must be true that x + 1 > 0 and y − 5 < 0 in order that the argument of
[x + 1] + j[y − 5] may be a negative acute angle.
We thus have the half-straight-line with equation
y − 5 = tan −
π
1
(x + 1) = − √ (x + 1)
6
3
which lies to the right of, and below the point (−1, 5).
4
6.6.4 MORE GENERAL LOCI
Certain types of locus problem may be encountered which cannot be identified with either
of the two standard types discussed above. The secret, in such problems is to substitute
z = x + jy in order to obtain the cartesian equation of the locus. We have already seen that
this method is applicable to the two standard types anyway.
ILLUSTRATIONS
1. The equation
− 1 √
= 3
z + 2
z
may be written
|z − 1| =
√
3 |z + 2|.
That is,
(x − 1)2 + y 2 = 3[(x + 2)2 + y 2 ],
which simplifies to
2x2 + 2y 2 + 14x + 13 = 0
or
7
x+
2
2
+ y2 =
23
,
4
representing a circle with centre − 27 , 0 and radius
q
23
.
4
2. The equation
arg
π
z−3
=
z
4
may be written
arg(z − 3) − arg z =
5
π
.
4
That is,
arg([x − 3] + jy) − arg(x + jy) =
π
.
4
Taking tangents of both sides and using the trigonometric identity for tan(A − B), we
obtain
y
− xy
x−3
y y = 1.
1 + x−3
x
On simplification, the equation becomes
x2 + y 2 − 3x − 3y = 0
or
3
x−
2
the equation of a circle with centre
2
3
+ y−
2
3 3
,
2 2
2
9
= ,
2
and radius
√3 .
2
However, we observe that the original complex number,
cannot have an argument of
In fact,
π
4
z−3
,
z
unless its real and imaginary parts are both positive.
(x − 3) + jy x − jy
x(x − 3) + y 2 + j3
z−3
=
.
=
z
x + jy
x − jy
x2 + y 2
which requires, therefore, that
x(x − 3) + y 2 > 0.
That is,
x2 + y 2 − 3x > 0
or
3
x−
2
2
9
+ y2 > .
4
Conclusion
The locus is that part of the circle with centre 32 , 23 and radius
the circle with centre
3
,0
2
and radius
3
.
2
6
√3
2
which lies outside
y
6
this region
O
-
6.6.5 EXERCISES
1. Identify the loci whose equations are
(a)
|z − 3| = 4;
(b)
|z − 4 + j7| = 2.
2. Identify the loci whose equations are
(a)
arg(z + 1) =
π
;
3
(b)
arg(z − 2 − j3) =
3π
.
2
3. Identify the loci whose equations are
(a)
z
z
+ j2 = 1;
− j3 (b)
arg
z+j
π
=− .
z−1
4
7
x
6.6.6 ANSWERS TO EXERCISES
1. (a) A circle with centre (3, 0) and radius 4;
(b) A circle with centre 4, −7) and radius 2.
2. (a) A half-straight-line to the right of, and above the point (−1, 0) inclined at an angle
of π3 to the positive direction of the real axis;
(b) A half-straight-line below the point (2, 3) and perpendicular to the real axis.
3. (a) The straight line y = 12 ;
(b) That part of the circle x2 + y 2 = 1 which lies outside the circle with centre
and radius √12 and above the straight line whose equation is y = x − 1.
1
, − 12
2
Note:
Examples like No. 3(b) are often quite difficult and will not normally be included in the
more elementary first year courses in mathematics.
8