“JUST THE MATHS”

UNIT NUMBER

6.1

COMPLEX NUMBERS 1(Definitions and algebra)

by

A.J.Hobson

6.1.1 The definition of a complex number6.1.2 The algebra of complex numbers6.1.3 Exercises6.1.4 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 equa-tions of the form

ad2y

dx2+ b

dy

dx+ cy = f(x),

which are called “Differential Equations”, have wide-reaching applications in science andengineering. They are particularly applicable to problems involving either electrical circuitsor mechanical vibrations.

It is possible to show that, in order to determine a formula (without derivatives) giving thevariable y in terms of the variable x, one method is to solve, first, the quadratic equationwhose 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 onlythe quadratic equation whose coefficients are a, b and c, ignoring references to differentialequations.

ILLUSTRATION

One method of solving the differential equation

d2y

dx2− 6

dy

dx+ 13 = 2 sin x

would be to solve, first, the quadratic equation whose coefficients are 1, −6 and 13.

Its solutions are6±

√36− 52

2=

6±√−16

2

which clearly do not exist since we cannot find the square root of a negative number inelementary arithmetic.

1

However, if we assume that the differential equation represents a genuine scientific problemwith a genuine scientific solution, we cannot simply dismiss the result obtained from thequadratic formula.

The difficulty seems to be, not so much with the −16 but with the minus sign in front of the16. We shall therefore write the solutions in the form

6± 4√−1

2= 3± 2

√−1.

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 bedenoted 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 areordinary numbers of elementary arithmetic.

DEFINITIONS

1. The term “complex number” is used to denote any expression of the form a + bj ora + jb where a and b are ordinary numbers of elementary arithmetic (including zero)and j denotes the imaginary number

√−1; i.e. j2 = −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 definedto 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 thevalue b is called the “imaginary part”. Notice that the imaginary part is b and notjb.

5. The complex numbers a ± bj are said to form a pair of “complex conjugates” andsimilarly a ± jb form a pair of complex conjugates. Alternatively, we may say, forinstance, that a − jb is the complex conjugate of a + jb and a + jb is the complexconjugate 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 thesymbols 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 systemwhich uses the concepts of equality, addition, subtraction, multiplication and division. Forexample, the algebra of real numbers is what we normally call “arithmetic”; but algebraicalconcepts can be applied to other mathematical systems of which the system of complexnumbers 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 purposeof the present section. In some cases, the definitions are fairly obvious, but need to be madewithout contradicting ideas already established in the system of real numbers which complexnumbers 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 thesame 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 andthe 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 algebraicsymbol, but uses the fact that j2 = −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 infuture 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 ofa complex number and its complex conjugate is always a real number consistingof 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 ofthe complex expression

a + jb

c + jd.

Once again, we make this definition in accordance with what would be obtained algebraicallyby treating j in the same way as any other algebraic symbol, but using the fact that j2 = −1.

5

The method is to multiply both the numerator and the denominator of the complex ratioby the conjugate of the denominator giving

a + jb

c + jd.c− jd

c− jd=

(ac + bd) + j(bc− ad)

c2 + d2.

The required definition is thus

a + jb

c + jd=

(ac + bd) + j(bc− ad)

c2 + d2,

which, again, is not so much a formula to be learned off-by-heart as a technique to be appliedin future examples.

EXAMPLES

1.

5 + j3

2 + j7=

5 + j3

2 + j7.2− j7

2− j7

=(10 + 21) + j(6− 35)

22 + 72=

31− j29

53.

Hence, the real part is 3153

and the imaginary part is −2953

.

2.

6 + j

j2− 4=

6 + j

j2− 4.−j2− 4

−j2− 4

=(−24 + 2) + j(−4− 12)

(−2)2 + (−4)2=−22− j16

20.

Hence, the real part is −2220

= −1110

and the imaginary part is −1620

= −45.

6

6.1.3 EXERCISES

1. Simplify the following:

(a) j3; (b) j4; (c) j5; (d) j15; (e) j22.

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 ofx 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

2and y = −1

2.

8

“JUST THE MATHS”

UNIT NUMBER

6.2

COMPLEX NUMBERS 2(The Argand Diagram)

by

A.J.Hobson

6.2.1 Introduction6.2.2 Graphical addition and subtraction6.2.3 Multiplication by j6.2.4 Modulus and argument6.2.5 Exercises6.2.6 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 thevalues 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 geometricaldiagram called the Argand Diagram:

-

6

x

y.........................

...

...

...(x, y)

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 purelyimaginary numbers.

6.2.2 GRAPHICAL ADDITON AND SUBTRACTION

If two complex numbers, z1 = x1 + jy1 and z2 = x2 + jy2, are represented in the ArgandDiagram by the points P1(x1, y1) and P2(x2, y2) respectively, then the sum, z1 + z2, of thecomplex 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 parallelogramhaving OP1 and OP2 as adjacent sides.

1

6

y

- x�������

P1

���������1P2

��������� Q

R

S������

O

In the diagram, the triangle ORP1 has exactly the same shape as the triangle P2SQ. 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 theparallelogram of which two adjacent sides are the straight line segments joining the originto 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), thenjz is represented by the point with co-ordinates B(−y, x).

2

-

6

x

y

������

AA

AA

AA

A

B

O

But OB is in the position which would be occupied by OA if it were rotated through 90◦ ina 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 tothe point representing the complex number.

6.2.4 MODULUS AND ARGUMENT

-

6

������

������*

P(x, y)

x

y

θ

r

O

(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| =√

x2 + y2.

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 saythat

|x| = |x + j0| =√

x2 + 02 =√

x2,

giving the usual numerical value of x.

ILLUSTRATIONS

1.

|3− j4| =√

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 onthe 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 thenumerical 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 positivelyfrom this axis in a counter-clockwise sense).

4

In the diagram,

tan θ =y

x; that is, θ = tan−1 y

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 valuesis denoted by Argz, using an upper-case A.

The particular value of the argument which lies in the interval −180◦ < θ ≤ 180◦ is calledthe “principal value” of the argument and is denoted by arg z using a lower-case a. Theparticular value, 180◦, in preference to −180◦, represents the principal value of the argumentof a negative real number.

ILLUSTRATIONS

1.

Arg(√

3 + j) = tan−1

(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 inthe second quadrant of the Argand Diagram.

We note also thatarg(−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 thethird quadrant of the Argand Diagram.

We note also thatarg(−1− j) = −135◦ only.

5

Note:It is worth mentioning here that, in the Argand Diagram, the directed straight line segmentdescribed from the point P1 (representing the complex number z1 = x1 + jy1) to the pointP2 (representing the complex number z2 = x2 + jy2) has length, r, equal to |z2 − z1|, and isinclined to the positive direction of the real axis at an angle, θ, equal to arg(z2 − z1). Thisfollows 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 realaxis and the imaginary axis respectively.

-

6

x

y

��

��

��

���

P1

P2

x2 − x1

y2 − y1r

θ

O

6.2.5 EXERCISES

1. Determine the modulus (in decimals, where appropriate, correct to three significantfigures) and the principal value of the argument (in degrees, correct to the nearestdegree) of the following complex numbers:

(a)

1− j;

(b)

−3 + j4;

(c)

−√

2− j√

2;

6

(d)

1

2− j

√3

2;

(e)

−7− j9.

2. If z = 4− j5, verify that jz has the same modulus as z but that the principal value ofthe 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 The polar form6.3.2 The exponential form6.3.3 Products and quotients in polar form6.3.4 Exercises6.3.5 Answers to exercises

UNIT 6.3 - COMPLEX NUMBERS 3

THE POLAR AND EXPONENTIAL FORMS

6.3.1 THE POLAR FORM

-

6

�����

�������*

P(x, y)

x

y

θ

r

O

From the above diagram, we may observe that

x

r= cos θ and

y

r= sin θ.

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 otherwords,

x + jy = r(cos θ + j sin θ).

The left-hand-side of this relationship is called the “rectangular form” or “cartesianform” 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 r 6 θ, where θ may be positive, negativeor 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

Argz = θ = tan−1 1√3

= 30◦ + k360◦,

where k may be any integer.

Alternatively, using radians,

Argz =π

6+ k2π,

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

6+ k2π

]+ j sin

[π

6+ k2π

])= 26

[π

6+ k2π

].

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

4+ k2π,

where k may be any integer.

Hence, in polar form,

z =√

2(cos[−135◦ + k360◦] + j sin[−135◦ + k360◦]) =√

26 [−135◦ + k360◦]

or

z =√

2(cos

[−3π

4+ k2π

]+ j sin

[−3π

4+ k2π

])=√

26[−3π

4+ k2π

].

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 theresult is simply omitted.

6.3.2 THE EXPONENTIAL FORM

Using some theory from the differential calculus of complex variables (not included here) itis possible to show that, for any complex number, z,

ez = 1 +z

1!+

z2

2!+

z3

3!+

z4

4!+ ...,

sin z = z − z3

3!+

z5

5!− z7

7!+ ...

and

cos z = 1− z2

2!+

z4

4!− z6

6!+ ...

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, mayrecognise similarities between the above formulae and the “MacLaurin Series” for the func-tions ex, sin x and cos x. In the case of the series for sin x and cos x, the value, x, must beexpressed in radians and not degrees.

A useful deduction can be made from the three formulae if we make the substitution z = jθinto the first one, obtaining:

3

ejθ = 1 +jθ

1!+

(jθ)2

2!+

(jθ)3

3!+

(jθ)4

4!+ ...

and, since j2 = −1, this gives

ejθ = 1 + jθ

1!− θ2

2!− j

θ3

3!+

θ4

4!+ ...

On regrouping this into real and imaginary parts, then using the sine and cosine series, weobtain

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 expressednot only in polar form but also in

the exponential form, rejθ.

ILLUSTRATIONS

Using the examples of the previous section

1.√

3 + j = 2ej(π6+k2π).

2.

−1 + j =√

2ej( 3π4

+k2π).

3.

−1− j =√

2e−j( 3π4

+k2π).

Note:If it is required that the exponential form should contain only the principal value of theargument, 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 polarform, 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 thequotient 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, weconstruct the product of their modulus values and the sum of their argument values.

(b) The Quotient

z1

z2

=r1

r2

(cos θ1 + j sin θ1)

(cos θ2 + j sin θ2).

On multiplying the numerator and denominator by cos θ2 − j sin θ2, we obtain

5

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 quotient of two complex numbers in polar form, weconstruct 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

−1− j=

26 30◦√26 (−135◦)

=√

26 165◦.

Again, for all of the complex numbers in this example, including the result, the argumentappears as the principal value.

Note:It will not always turn out that the argument of a product or quotient of two complexnumbers 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, r 6 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, z1

z2, in polar form using

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 =√

2ej π4 z2 = 2e−j π

6 ;

(c)

z1.z2 = 2√

26 15◦z1

z2

=

√2

26 75◦.

2. (a)

z1 = 26 (135◦) z2 = 56 (−127◦);

7

(b)

z1 = 2ej 3π4 z2 = 5e−j2.22;

(c)

z1.z2 = 106 8◦z1

z2

=2

56 (−98◦).

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◦z1

z2

= 0.566 (−73.53◦).

8

“JUST THE MATHS”

UNIT NUMBER

6.4

COMPLEX NUMBERS 4(Powers of complex numbers)

by

A.J.Hobson

6.4.1 Positive whole number powers6.4.2 Negative whole number powers6.4.3 Fractional powers & De Moivre’s Theorem6.4.4 Exercises6.4.5 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 isa simple matter to multiply a complex number by itself any desired number of times.

Suppose thatz = r 6 θ.

Then,

z2 = r.r 6 (θ + θ) = r2 6 2θ;

z3 = z.z2 = r.r2 6 (θ + 2θ) = r3 6 3θ;

and, by continuing this process,

zn = rn 6 nθ.

This result is due to De Moivre, but other aspects of it will need to be discussed before wemay formalise what is called “De Moivre’s Theorem”.

EXAMPLE

(1√2

+ j1√2

)19

= (16[π

4

])19 = 16

[19π

4

]= 16

[3π

4

]= − 1√

2+ j

1√2.

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 = r 6 θ,

1

zn = z−m =1

zm=

1

rm 6 mθ.

In more detail,

zn =1

rm(cos mθ + j sin mθ),

giving

zn =1

rm.(cos mθ − j sin mθ)

cos2mθ + sin2mθ= r−m(cos[−mθ] + j sin[−mθ]).

But −m = n, and so

zn = rn(cos nθ + j sin nθ) = rn 6 nθ,

showing that the result of the previous section remains true for negative whole numberpowers.

EXAMPLE

(√

3 + j)−3 = (26 30◦)−3 =1

86 (−90◦) = −j

8.

6.4.3 FRACTIONAL POWERS AND DE MOIVRE’S THEOREM

To begin with, here, we consider the complex number

z1n ,

where n is a positive whole number and z = r 6 θ.

We define z1n 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

r1n 6

θ

n,

by virtue of the paragraph dealing with positive whole number powers.

But the general expression for z is given by

z = r 6 (θ + k360◦),

where k may be any integer; and this suggests other possibilities for z1n , namely

r1n 6

θ + k360◦

n.

However, this set of n-th roots is not an infinite set because the roots which are given byk = 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 writej8 = 86 (90◦ + k360◦).

Hence,

(j8)13 = 8

13 6

(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 ArgandDiagram at regular intervals of 360◦

3= 120◦.

Notes:

(i) In general, the n-th roots of a complex number will all have the same modulus, but theirarguments 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 theprincipal value of the argument, then the particular n-th root of z which is given by k = 0is called the “principal n-th root”.

(iii) If mn

is a fraction in its lowest terms, we define

zmn

to be either(z

1n

)mor (zm)

1n 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 = r 6 θ, then, for any rational number n, one value of zn is rn 6 nθ.

6.4.4 EXERCISES

1. Determine the following in the form a + jb, expressing a and b in decimals correct tofour 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 expressinga and b in decimals, where appropriate, correct to two places. State also which root isthe principal root.

4

4. Determine all the values of

(3 + j4)32

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

3e−π8 , 3e

3π8 , 3e

7π8 , 3e−

5π8 .

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 Introduction6.5.2 Expressions for cos nθ, sin nθ in terms of cos θ, sin θ

6.5.3 Expressions for cosnθ and sinnθ in terms of sines andcosines of whole multiples of x

6.5.4 Exercises6.5.5 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’sTriangle” previously discussed in Unit 2.2.

If n is a positive whole number, the diagram

1 1

1 2 1

1 3 3 1

1 4 6 4 1

provides the coefficients in the expansion of (A + B)n which contains the sequence of terms

An, An−1B, An−2B2, An−3B3, ........., Bn.

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 Trian-gle, the real part will coincide with cos nθ and the imaginary part will coincide with sin nθ.

EXAMPLE

(cos θ + j sin θ)3 ≡ cos3θ + 3cos2θ.(j sin θ) + 3 cos θ.(j sin θ)2 + (j sin θ)3.

That is,

cos 3θ ≡ cos3θ − 3 cos θ.sin2θ or 4cos3θ − 3 cos θ,

1

using sin2θ ≡ 1− cos2θ;

and

sin 3θ ≡ 3cos2θ. sin θ − sin3θ or 3 sin θ − 4sin3θ,

using cos2θ ≡ 1− sin2θ.

6.5.3 EXPRESSIONS FOR cosnθ AND sinnθ IN TERMS OF SINES AND COSINESOF WHOLE MULTIPLES OF θ.

The technique described here is particularly useful in calculus problems when we are requiredto 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

z≡ cos θ − j sin θ − (2).

Adding (1) and (2) together, then subtracting (2) from (1), we obtain

z + 1z≡ 2 cos θ z − 1

z≡ j2 sin θ

Also, by De Moivre’s Theorem,

zn ≡ cos nθ + j sin nθ − (3)

and

1

zn≡ cos nθ − j sin nθ − (4).

2

Adding (3) and (4) together, then subtracting (4) from (3), we obtain

zn + 1zn ≡ 2 cos nθ zn − 1

zn ≡ j2 sin nθ

We are now in a position to discuss some examples on finding trigonometric identities forwhole number powers of sin θ or cos θ.

EXAMPLES

1. Determine an identity for sin3θ.

Solution

We use the result

j323sin3θ ≡(z − 1

z

)3

,

where z ≡ cos θ + j sin θ.

That is,

−j8sin3θ ≡ z3 − 3z2.1

z+ 3z.

(1

z

)2

− 1

z3

or, after cancelling common factors,

−j8sin3θ ≡ z3 − 3z +3

z− 1

z3≡

(z3 − 1

z3

)− 3

(z − 1

z

),

which gives

−j8sin3θ ≡ j2 sin 3θ − j6 sin θ.

Hence,

sin3θ ≡ 1

4(3 sin θ − sin 3θ) .

3

2. Determine an identity for cos4θ.

Solution

We use the result

24cos4θ ≡(z +

1

z

)4

,

where z ≡ cos θ + j sin θ.

That is,

16cos4θ ≡ z4 + 4z3.1

z+ 6z2.

(1

z

)2

+ 4z.(

1

z

)3

+(

1

z

)4

or, after cancelling common factors,

16cos4θ ≡ z4 + 4z2 + 6 +4

z2+

1

z4≡ z4 +

1

z4+ 4

(z2 +

1

z2

)+ 6,

which gives

16cos4θ ≡ 2 cos 4θ + 8 cos 2θ + 6.

Hence,

cos4θ ≡ 1

8(cos 4θ + 4 cos 2θ + 3)

6.5.4 EXERCISES

1. Use a complex number method to determine identities for cos 4θ and sin 4θ in terms ofsin θ and cos θ.

2. Use a complex number method to determine an identity for sin5θ in terms of sines ofwhole multiples of θ.

3. Use a complex number method to deternine an identity for cos6θ in terms of cosines ofwhole multiples of θ.

4

6.5.5 ANSWERS TO EXERCISES

1.

cos 4θ ≡ cos4θ − 6cos2θ.sin2θ

and

sin 4θ ≡ 4cos3θ. sin θ − 4 cos θ.sin3θ.

2.

sin5θ ≡ 1

16(sin 5θ − 5 sin 3θ + 10 sin θ].

3.

cos6θ ≡ 1

32(cos 6θ + 6 cos 4θ + 15 cos 2θ + 10).

5

“JUST THE MATHS”

UNIT NUMBER

6.6

COMPLEX NUMBERS 6(Complex loci)

by

A.J.Hobson

6.6.1 Introduction6.6.2 The circle6.6.3 The half-straight-line6.6.4 More general loci6.6.5 Exercises6.6.6 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 representinga complex number z1 to the point representing a complex number z2 is of length equalto |z2 − z1| and is inclined to the positive direction of the real axis at an angle equal toarg(z2 − z1).

This observation now has significance when discussing variable complex numbers which areconstrained to move along a certain path (or “locus”) in the Argand Diagram. For manypractical applications, such paths (or “loci”) will normally be either straight lines or circlesand two standard types of example appear in what follows.

In both types, we shall assume that z = x + jy denotes a variable complex number (rep-resented by the point (x, y) in the Argand Diagram), while z0 = x0 + jy0 denotes a fixedcomplex 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 centreis at the fixed point representing z0.

z0

z

a

O - x

6

y

1

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 theequivalent 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 number3− 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 pointrepresenting z0 to the moving point representing z is inclined at an angle θ to the positivedirection 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 directedline segment is genuinely θ and not 180◦ − θ. The latter angle would correspond to pointson the other half of the straight line joining the two points.

-

6

x

y

����

���

���

���

�����*

θz0

z

O

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

x− x0

= θ

or

y − y0 = tan θ(x− x0),

which is certainly the equation of a straight line with gradient tan θ passing through thepoint (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 numberwhose argument is genuinely θ and not 180◦ − θ.

3

ILLUSTRATION

The equation

arg(z + 1− j5) = −π6

represents the half-straight-line described from the point representing z0 = −1 + j5 to thepoint representing z = x + jy and inclined to the positive direction of the real axis at anangle of −π

6.

-

6

x

y

PPPPPPPPPq

z0

O

In terms of cartesian co-ordinates,

arg([x+ 1] + j[y − 5]) = −π6,

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

6

)(x+ 1) = − 1√

3(x+ 1)

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 eitherof the two standard types discussed above. The secret, in such problems is to substitutez = x+ jy in order to obtain the cartesian equation of the locus. We have already seen thatthis method is applicable to the two standard types anyway.

ILLUSTRATIONS

1. The equation ∣∣∣∣z − 1

z + 2

∣∣∣∣ =√

3

may be written

|z − 1| =√

3 |z + 2|.

That is,

(x− 1)2 + y2 = 3[(x+ 2)2 + y2],

which simplifies to

2x2 + 2y2 + 14x+ 13 = 0

or (x+

7

2

)2

+ y2 =23

4,

representing a circle with centre(−7

2, 0)

and radius√

234

.

2. The equation

arg(z − 3

z

)=π

4

may be written

arg(z − 3)− arg z =π

4.

5

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), weobtain

yx−3− y

x

1 + yx−3

yx

= 1.

On simplification, the equation becomes

x2 + y2 − 3x− 3y = 0

or (x− 3

2

)2

+(y − 3

2

)2

=9

2,

the equation of a circle with centre(

32, 3

2

)and radius 3√

2.

However, we observe that the original complex number,

z − 3

z,

cannot have an argument of π4

unless its real and imaginary parts are both positive.

In fact,

z − 3

z=

(x− 3) + jy

x+ jy.x− jyx− jy

=x(x− 3) + y2 + j3

x2 + y2

which requires, therefore, that

x(x− 3) + y2 > 0.

That is,

x2 + y2 − 3x > 0

or (x− 3

2

)2

+ y2 >9

4.

Conclusion

The locus is that part of the circle with centre(

32, 3

2

)and radius 3√

2which lies outside

the circle with centre(

32, 0)

and radius 32.

6

-

6y

xO

this region

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 + j2

z − j3

∣∣∣∣∣ = 1;

(b)

arg(z + j

z − 1

)= −π

4.

7

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 angleof π

3to 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 + y2 = 1 which lies outside the circle with centre(

12,−1

2

)and radius 1√

2and above the straight line whose equation is y = x− 1.

Note:Examples like No. 3(b) are often quite difficult and will not normally be included in themore elementary first year courses in mathematics.

8