Chapter III
Complex Numbers

34. Displacements along a line and in a plane. The ‘real number’ x, with which we have been concerned in the two preceding chapters, may be regarded from many different points of view. It may be regarded as a pure number, destitute of geometrical significance, or a geometrical significance may be attached to it in at least three different ways. It may be regarded as the measure of a length, viz. the length A0P along the line Λ of Chap. I. It may be regarded as the mark of a point, viz. the point P whose distance from A0 is x. Or it may be regarded as the measure of a displacement or change of position on the line Λ. It is on this last point of view that we shall now concentrate our attention.

Imagine a small particle placed at P on the line Λ and then displaced to Q. We shall call the displacement or change of position which is needed to transfer the particle from P to Q the displacement PQ¯. To specify a displacement completely three things are needed, its magnitude, its sense forwards or backwards along the line, and what may be called its point of application, i.e. the original position P of the particle. But, when we are thinking merely of the change of position produced by the displacement, it is natural to disregard the point of application and to consider all displacements as equivalent whose lengths and senses are the same. Then the displacement is completely specified by the length PQ = x, the sense of the displacement being fixed by the sign of x. We may therefore, without ambiguity, speak of the displacement [x],26 and we may write PQ¯ = [x].

We use the square bracket to distinguish the displacement [x] from the length or number x.27 If the coordinate of P is a, that of Q will be a + x; the displacement [x] therefore transfers a particle from the point a to the point a + x.

We come now to consider displacements in a plane. We may define the displacement PQ¯ as before. But now more data are required in order to specify it completely. We require to know: (i) the magnitude of the displacement, i.e. the length of the straight line PQ; (ii) the direction of the displacement, which is determined by the angle which PQ makes with some fixed line in the plane; (iii) the sense of the displacement; and (iv) its point of application. Of these requirements we may disregard the fourth, if we consider two displacements as equivalent if they are


pict

Fig. 19.

the same in magnitude, direction, and sense. In other words, if PQ and RS are equal and parallel, and the sense of motion from P to Q is the same as that of motion from R to S, we regard the displacements PQ¯ and RS¯ as equivalent, and write

PQ¯ = RS¯.

Now let us take any pair of coordinate axes in the plane (such as OXOY in Fig. 19). Draw a line OA equal and parallel to PQ, the sense of motion from O to A being the same as that from P to Q. Then PQ¯ and OA¯ are equivalent displacements. Let x and y be the coordinates of A. Then it is evident that OA¯ is completely specified if x and y are given. We call OA¯ the displacement [x,y] and write

OA¯ = PQ¯ = RS¯ = [x,y].

35. Equivalence of displacements. Multiplication of displacements by numbers. If ξ and η are the coordinates of P, and ξ and η those of Q, it is evident that

x = ξ ξ,y = η η.
The displacement from (ξ,η) to (ξ,η) is therefore
[ξ ξ,η η].

It is clear that two displacements [x,y], [x,y] are equivalent if, and only if, x = x, y = y. Thus [x,y] = [x,y] if and only if

x = x,y = y. (1)

The reverse displacement QP¯ would be [ξ ξ,η η], and it is natural to agree that

[ξ ξ,η η] = [ξ ξ,η η], QP¯ = PQ¯,

these equations being really definitions of the meaning of the symbols [ξ ξ,η η], PQ¯. Having thus agreed that

[x,y] = [x,y],
it is natural to agree further that

α[x,y] = [αx,αy], (2)

where α is any real number, positive or negative. Thus (Fig. 19) if OB = 1 2OA then

OB¯ = 1 2OA¯ = 1 2[x,y] = [1 2x,1 2y].

The equations (1) and (2) define the first two important ideas connected with displacements, viz. equivalence of displacements, and multiplication of displacements by numbers.

36. Addition of displacements. We have not yet given any definition which enables us to attach any meaning to the expressions

PQ¯ + PQ¯,[x,y] + [x,y].
Common sense at once suggests that we should define the sum of two displacements as the displacement which is the result of the successive application of the two given displacements. In other words, it suggests that if QQ1 be drawn equal and parallel to PQ, so that the result of successive displacements PQ¯PQ¯ on a particle at P is to transfer it first to Q and then to Q1 then we should define the sum of PQ¯ and PQ¯ as being PQ1¯. If then we draw OA equal and parallel to PQ, and OB equal and parallel to PQ, and complete the parallelogram OACB, we have


pict

Fig. 20.

PQ¯ + PQ¯ = PQ1¯ = OA¯ + OB¯ = OC¯.

Let us consider the consequences of adopting this definition. If the coordinates of B are xy, then those of the middle point of AB are 1 2(x + x), 1 2(y + y), and those of C are x + x, y + y. Hence

[x,y] + [x,y] = [x + x,y + y], (3)

which may be regarded as the symbolic definition of addition of displacements. We observe that

[x,y] + [x,y] = [x + x,y + y] = [x + x,y + y] = [x,y] + [x,y]

In other words, addition of displacements obeys the commutative law expressed in ordinary algebra by the equation a + b = b + a. This law expresses the obvious geometrical fact that if we move from P first through a distance PQ2 equal and parallel to PQ, and then through a distance equal and parallel to PQ, we shall arrive at the same point Q1 as before.

In particular

[x,y] = [x, 0] + [0,y]. (4)
Here [x, 0] denotes a displacement through a distance x in a direction parallel to OX. It is in fact what we previously denoted by [x], when we were considering only displacements along a line. We call [x, 0] and [0,y] the components of [x,y], and [x,y] their resultant.

When we have once defined addition of two displacements, there is no further difficulty in the way of defining addition of any number. Thus, by definition,

[x,y] + [x,y] + [x,y] = ([x,y] + [x,y]) + [x,y] = [x + x,y + y] + [x,y] = [x + x + x,y + y + y].

We define subtraction of displacements by the equation

[x,y] [x,y] = [x,y] + ([x,y]), (4)

which is the same thing as [x,y] + [x,y] or as [x x,y y]. In particular

[x,y] [x,y] = [0, 0].

The displacement [0, 0] leaves the particle where it was; it is the zero displacement, and we agree to write [0, 0] = 0.

Examples XX. 1. Prove that

(i) α[βx,βy] = β[αx,αy] = [αβx,αβy],

(ii) ([x,y] + [x,y]) + [x,y] = [x,y] + ([x,y] + [x,y]),

(iii) [x,y] + [x,y] = [x,y] + [x,y],

(iv) (α + β)[x,y] = α[x,y] + β[x,y],

(v) α{[x,y] + [x,y]}= α[x,y] + α[x,y].

[We have already proved (iii). The remaining equations follow with equal ease from the definitions. The reader should in each case consider the geometrical significance of the equation, as we did above in the case of (iii).]

2. If M is the middle point of PQ, then OM¯ = 1 2(OP¯ + OQ¯). More generally, if M divides PQ in the ratio μ : λ, then

OM¯ = λ λ + μOP¯ + μ λ + μOQ¯.

3. If G is the centre of mass of equal particles at P1, P2, …, Pn, then

OG¯ = (OP1¯ + OP2¯ + + OPn¯)/n.

4. If PQR are collinear points in the plane, then it is possible to find real numbers αβγ, not all zero, and such that

α OP¯ + β OQ¯ + γ OR¯ = 0;
and conversely. [This is really only another way of stating Ex. 2.]

5. If AB¯ and AC¯ are two displacements not in the same straight line, and

α AB¯ + β AC¯ = γ AB¯ + δ AC¯,
then α = γ and β = δ.

[Take AB1 = α AB, AC1 = β AC. Complete the parallelogram AB1P1C1. Then AP1¯ = α AB¯ + β AC¯. It is evident that AP1¯ can only be expressed in this form in one way, whence the theorem follows.]

6. ABCD is a parallelogram. Through Q, a point inside the parallelogram, RQS and TQU are drawn parallel to the sides. Show that RUTS intersect on AC.


pict

Fig. 21.

[Let the ratios AT : AB, AR : AD be denoted by αβ. Then

AT¯ = α AB¯,AR¯ = β AD¯, AU¯ = α AB¯ + AD¯,AS¯ = AB¯ + β AD¯.

Let RU meet AC in P. Then, since RUP are collinear,

AP¯ = λ λ + μAR¯ + μ λ + μAU¯,
where μ/λ is the ratio in which P divides RU. That is to say
AP¯ = αμ λ + μAB¯ + βλ + μ λ + μ AD¯.

But since P lies on AC, AP¯ is a numerical multiple of AC¯; say

AP¯ = k AC¯ = k AB¯ + k AD¯.
Hence (Ex. 5) αμ = βλ + μ = (λ + μ)k, from which we deduce
k = αβ α + β 1.
The symmetry of this result shows that a similar argument would also give
AP¯ = αβ α + β 1AC¯,
if P is the point where TS meets AC. Hence P and P are the same point.]

7. ABCD is a parallelogram, and M the middle point of AB. Show that DM trisects and is trisected by AC.28

37. Multiplication of displacements. So far we have made no attempt to attach any meaning whatever to the notion of the product of two displacements. The only kind of multiplication which we have considered is that in which a displacement is multiplied by a number. The expression

[x,y] × [x,y]
so far means nothing, and we are at liberty to define it to mean anything we like. It is, however, fairly clear that if any definition of such a product is to be of any use, the product of two displacements must itself be a displacement.

We might, for example, define it as being equal to

[x + x,y + y];
in other words, we might agree that the product of two displacements was to be always equal to their sum. But there would be two serious objections to such a definition. In the first place our definition would be futile. We should only be introducing a new method of expressing something which we can perfectly well express without it. In the second place our definition would be inconvenient and misleading for the following reasons. If α is a real number, we have already defined α[x,y] as [αx,αy]. Now, as we saw in § 34, the real number α may itself from one point of view be regarded as a displacement, viz. the displacement [α] along the axis OX, or, in our later notation, the displacement [α, 0]. It is therefore, if not absolutely necessary, at any rate most desirable, that our definition should be such that
[α, 0][x,y] = [αx,αy],
and the suggested definition does not give this result.

A more reasonable definition might appear to be

[x,y][x,y] = [xx,yy].
But this would give
[α, 0][x,y] = [αx, 0];
and so this definition also would be open to the second objection.

In fact, it is by no means obvious what is the best meaning to attach to the product [x,y][x,y]. All that is clear is (1) that, if our definition is to be of any use, this product must itself be a displacement whose coordinates depend on x and y, or in other words that we must have

[x,y][x,y] = [X,Y ],
where X and Y are functions of xyx, and y; (2) that the definition must be such as to agree with the equation
[x, 0][x,y] = [xx,xy];
and (3) that the definition must obey the ordinary commutative, distributive, and associative laws of multiplication, so that

[x,y][x,y] = [x,y][x,y], ([x,y] + [x,y])[x,y] = [x,y][x,y] + [x,y][x,y], [x,y]([x,y] + [x,y]) = [x,y][x,y] + [x,y][x,y],  and [x,y]([x,y][x,y]) = ([x,y][x,y])[x,y].

38. The right definition to take is suggested as follows. We know that, if OABOCD are two similar triangles, the angles corresponding in the order in which they are written, then

OB/OA = OD/OC,
or OB OC = OA OD. This suggests that we should try to define multiplication and division of displacements in such a way that
OB¯/OA¯ = OD¯/OC¯,OB¯ OC¯ = OA¯ OD¯.


pict

Fig. 22.

Now let

OB¯ = [x,y],OC¯ = [x,y],OD¯ = [X,Y ],
and suppose that A is the point (1, 0), so that OA¯ = [1, 0]. Then
OA¯ OD¯ = [1, 0][X,Y ] = [X,Y ],
and so
[x,y][x,y] = [X,Y ].
The product OB¯ OC¯ is therefore to be defined as OD¯, D being obtained by constructing on OC a triangle similar to OAB. In order to free this definition from ambiguity, it should be observed that on OC we can describe two such triangles, OCD and OCD. We choose that for which the angle COD is equal to AOB in sign as well as in magnitude. We say that the two triangles are then similar in the same sense.

If the polar coordinates of B and C are (ρ,θ) and (σ,φ), so that

x = ρ cos θ,y = ρ sin θ,x = σ cos φ,y = σ sin φ,
then the polar coordinates of D are evidently ρσ and θ + φ. Hence

X = ρσ cos(θ + φ) = xx yy, Y = ρσ sin(θ + φ) = xy + yx.

The required definition is therefore

[x,y][x,y] = [xx yy,xy + yx]. (5)

We observe (1) that if y = 0, then X = xx, Y = xy, as we desired; (2) that the right-hand side is not altered if we interchange x and x, and y and y, so that

[x,y][x,y] = [x,y][x,y];
and (3) that

{[x,y] + [x,y]}[x,y] = [x + x,y + y][x,y] = [(x + x)x (y + y)y, (x + x)y + (y + y)x] = [xx yy,xy + yx] + [xx yy,xy + yx] = [x,y][x,y] + [x,y][x,y].

Similarly we can verify that all the equations at the end of § 37 are satisfied. Thus the definition (6) fulfils all the requirements which we made of it in § 37.

Example. Show directly from the geometrical definition given above that multiplication of displacements obeys the commutative and distributive laws. [Take the commutative law for example. The product OB¯ OC¯ is OD¯ (Fig. 22), COD being similar to AOB. To construct the product OC¯ OB¯ we should have to construct on OB a triangle BOD1 similar to AOC; and so what we want to prove is that D and D1 coincide, or that BOD is similar to AOC. This is an easy piece of elementary geometry.]

39. Complex numbers. Just as to a displacement [x] along OX correspond a point (x) and a real number x, so to a displacement [x,y] in the plane correspond a point (x,y) and a pair of real numbers xy.

We shall find it convenient to denote this pair of real numbers xy by the symbol

x + yi.
The reason for the choice of this notation will appear later. For the present the reader must regard x + yi as simply another way of writing [x,y]. The expression x + yi is called a complex number.

We proceed next to define equivalence, addition, and multiplication of complex numbers. To every complex number corresponds a displacement. Two complex numbers are equivalent if the corresponding displacements are equivalent. The sum or product of two complex numbers is the complex number which corresponds to the sum or product of the two corresponding displacements. Thus

x + yi = x + yi, (1)

if and only if x = x, y = y;

(x + yi) + (x + yi) = (x + x) + (y + y)i;  (2) (x + yi)(x + yi) = xx yy + (xy + yx)i.  (3)

In particular we have, as special cases of (2) and (3),

x + yi = (x + 0i) + (0 + yi), (x + 0i)(x + yi) = xx + xyi;

and these equations suggest that there will be no danger of confusion if, when dealing with complex numbers, we write x for x + 0i and yi for 0 + yi, as we shall henceforth.

Positive integral powers and polynomials of complex numbers are then defined as in ordinary algebra. Thus, by putting x = x, y = y in (3), we obtain

(x + yi)2 = (x + yi)(x + yi) = x2 y2 + 2xyi.

The reader will easily verify for himself that addition and multiplication of complex numbers obey the laws of algebra expressed by the equations

(x + yi) + (x + yi) = (x + yi) + (x + yi), {(x + yi) + (x + yi)} + (x + yi) = (x + yi) + {(x + yi) + (x + yi)}, (x + yi)(x + yi) = (x + yi)(x + yi), (x + yi){(x + yi) + (x + yi)} = (x + yi)(x + yi) + (x + yi)(x + yi), {(x + yi) + (x + yi)}(x + yi) = (x + yi)(x + yi) + (x + yi)(x + yi), (x + yi){(x + yi)(x + yi)} = {(x + yi)(x + yi)}(x + yi),

the proofs of these equations being practically the same as those of the corresponding equations for the corresponding displacements.

Subtraction and division of complex numbers are defined as in ordinary algebra. Thus we may define (x + yi) (x + yi) as

(x + yi) + {(x + yi)} = x + yi + (x yi) = (x x) + (y y)i;
or again, as the number ξ + ηi such that
(x + yi) + (ξ + ηi) = x + yi,
which leads to the same result. And (x + yi)/(x + yi) is defined as being the complex number ξ + ηi such that
(x + yi)(ξ + ηi) = x + yi,
or
xξ yη + (xη + yξ)i = x + yi,
or

xξ yη = x,xη + yξ = y. (4)

Solving these equations for ξ and η, we obtain

ξ = xx + yy x2 + y2 ,η = yx xy x2 + y2 .
This solution fails if x and y are both zero, i.e. if x + yi = 0. Thus subtraction is always possible; division is always possible unless the divisor is zero.

Examples. (1) From a geometrical point of view, the problem of the division of the displacement OB¯ by OC¯ is that of finding D so that the triangles COBAOD are similar, and this is evidently possible (and the solution unique) unless C coincides with 0, or OC¯ = 0.


pict

Fig. 23.

(2) The numbers x + yi, x yi are said to be conjugate. Verify that

(x + yi)(x yi) = x2 + y2,
so that the product of two conjugate numbers is real, and that
x + yi x+ yi = (x + yi)(xyi) (x+ yi)(xyi) = xx+ yy+ (xy xy)i x2 + y2 .

40. One most important property of real numbers is that known as the factor theorem, which asserts that the product of two numbers cannot be zero unless one of the two is itself zero. To prove that this is also true of complex numbers we put x = 0, y = 0 in the equations (4) of the preceding section. Then

xξ yη = 0,xη + yξ = 0.
These equations give ξ = 0, η = 0, i.e.
ξ + ηi = 0,
unless x = 0 and y = 0, or x + yi = 0. Thus x + yi cannot vanish unless either x + yi or ξ + ηi vanishes.

41. The equation i2 = 1. We agreed to simplify our notation by writing x instead of x + 0i and yi instead of 0 + yi. The particular complex number 1i we shall denote simply by i. It is the number which corresponds to a unit displacement along OY . Also

i2 = ii = (0 + 1i)(0 + 1i) = (0 0 1 1) + (0 1 + 1 0)i = 1.
Similarly (i)2 = 1. Thus the complex numbers i and  i satisfy the equation x2 = 1.

The reader will now easily satisfy himself that the upshot of the rules for addition and multiplication of complex numbers is this, that we operate with complex numbers in exactly the same way as with real numbers, treating the symbol i as itself a number, but replacing the product ii = i2 by  1 whenever it occurs. Thus, for example,

(x + yi)(x + yi) = xx + xyi + yxi + yyi2 = (xx yy) + (xy + yx)i.

42. The geometrical interpretation of multiplication by i. Since

(x + yi)i = y + xi,
it follows that if x + yi corresponds to OP¯, and OQ is drawn equal to OP and so that POQ is a positive right angle, then (x + yi)i corresponds to OQ¯. In other words, multiplication of a complex number by i turns the corresponding displacement through a right angle.

We might have developed the whole theory of complex numbers from this point of view. Starting with the ideas of x as representing a displacement along OX, and of i as a symbol of operation equivalent to turning x through a right angle, we should have been led to regard yi as a displacement of magnitude y along OY . It would then have been natural to define x + yi as in §§ 37 and 40, and (x + yi)i would have represented the displacement obtained by turning x + yi through a right angle, i.e.  y + xi. Finally, we should naturally have defined (x + yi)x as xx + yxi, (x + yi)yi as yy + xyi, and (x + yi)(x + yi) as the sum of these displacements, i.e. as

xx yy + (xy + yx)i.

43. The equations z2 + 1 = 0, az2 + 2bz + c = 0. There is no real number z such that z2 + 1 = 0; this is expressed by saying that the equation has no real roots. But, as we have just seen, the two complex numbers i and  i satisfy this equation. We express this by saying that the equation has the two complex roots i and  i. Since i satisfies z2 = 1, it is sometimes written in the form  1.

Complex numbers are sometimes called imaginary.29 The expression is by no means a happily chosen one, but it is firmly established and has to be accepted. It cannot, however, be too strongly impressed upon the reader that an ‘imaginary number’ is no more ‘imaginary’, in any ordinary sense of the word, than a ‘real’ number; and that it is not a number at all, in the sense in which the ‘real’ numbers are numbers, but, as should be clear from the preceding discussion, a pair of numbers (x,y), united symbolically, for purposes of technical convenience, in the form x + yi. Such a pair of numbers is no less ‘real’ than any ordinary number such as 1 2, or than the paper on which this is printed, or than the Solar System. Thus

i = 0 + 1i
stands for the pair of numbers (0, 1), and may be represented geometrically by a point or by the displacement [0, 1]. And when we say that i is a root of the equation z2 + 1 = 0, what we mean is simply that we have defined a method of combining such pairs of numbers (or displacements) which we call ‘multiplication’, and which, when we so combine (0, 1) with itself, gives the result (1, 0).

Now let us consider the more general equation

az2 + 2bz + c = 0,
where abc are real numbers. If b2 > ac, the ordinary method of solution gives two real roots
{b ±b2 ac}/a.
If b2 < ac, the equation has no real roots. It may be written in the form
{z + (b/a)}2 = (ac b2)/a2,
an equation which is evidently satisfied if we substitute for z + (b/a) either of the complex numbers ± iac b2/a.30 We express this by saying that the equation has the two complex roots
{b ± iac b2}/a.

If we agree as a matter of convention to say that when b2 = ac (in which case the equation is satisfied by one value of x only, viz.  b/a), the equation has two equal roots, we can say that a quadratic equation with real coefficients has two roots in all cases, either two distinct real roots, or two equal real roots, or two distinct complex roots.

The question is naturally suggested whether a quadratic equation may not, when complex roots are once admitted, have more than two roots. It is easy to see that this is not possible. Its impossibility may in fact be proved by precisely the same chain of reasoning as is used in elementary algebra to prove that an equation of the nth degree cannot have more than n real roots. Let us denote the complex number x + yi by the single letter z, a convention which we may express by writing z = x + yi. Let f(z) denote any polynomial in z, with real or complex coefficients. Then we prove in succession:

(1) that the remainder, when f(z) is divided by z a, a being any real or complex number, is f(a);

(2) that if a is a root of the equation f(z) = 0, then f(z) is divisible by z a;

(3) that if f(z) is of the nth degree, and f(z) = 0 has the n roots a1, a2, …, an, then

f(z) = A(z a1)(z a2)(z an),
where A is a constant, real or complex, in fact the coefficient of zn in f(z). From the last result, and the theorem of § 40, it follows that f(z) cannot have more than n roots.

We conclude that a quadratic equation with real coefficients has exactly two roots. We shall see later on that a similar theorem is true for an equation of any degree and with either real or complex coefficients: an equation of the nth degree has exactly n roots. The only point in the proof which presents any difficulty is the first, viz. the proof that any equation must have at least one root. This we must postpone for the present.31 We may, however, at once call attention to one very interesting result of this theorem. In the theory of number we start from the positive integers and from the ideas of addition and multiplication and the converse operations of subtraction and division. We find that these operations are not always possible unless we admit new kinds of numbers. We can only attach a meaning to 3 7 if we admit negative numbers, or to 3 7 if we admit rational fractions. When we extend our list of arithmetical operations so as to include root extraction and the solution of equations, we find that some of them, such as that of the extraction of the square root of a number which (like 2) is not a perfect square, are not possible unless we widen our conception of a number, and admit the irrational numbers of Chap. I.

Others, such as the extraction of the square root of  1, are not possible unless we go still further, and admit the complex numbers of this chapter. And it would not be unnatural to suppose that, when we come to consider equations of higher degree, some might prove to be insoluble even by the aid of complex numbers, and that thus we might be led to the considerations of higher and higher types of, so to say, hyper-complex numbers. The fact that the roots of any algebraical equation whatever are ordinary complex numbers shows that this is not the case. The application of any of the ordinary algebraical operations to complex numbers will yield only complex numbers. In technical language ‘the field of the complex numbers is closed for algebraical operations’.

Before we pass on to other matters, let us add that all theorems of elementary algebra which are proved merely by the application of the rules of addition and multiplication are true whether the numbers which occur in them are real or complex, since the rules referred to apply to complex as well as real numbers. For example, we know that, if α and β are the roots of

az2 + 2bz + c = 0,
then
α + β = (2b/a),αβ = (c/a).

Similarly, if αβγ are the roots of

az3 + 3bz2 + 3cz + d = 0,
then
α + β + γ = (3b/a),βγ + γα + αβ = (3c/a),αβγ = (d/a).
All such theorems as these are true whether ab, …, αβ, … are real or complex.

44. Argand’s diagram. Let P (Fig. 24) be the point (x,y), r the length OP, and θ the angle XOP, so that

x = r cos θ,y = r sin θ,r = x2 + y2, cos θ : sin θ : 1 :: x : y : r.

We denote the complex number x + yi by z, as in § 43, and we call z the complex variable.


pict

Fig. 24.

We call P the point z, or the point corresponding to z; z the argument of P, x the real part, y the imaginary part, r the modulus, and θ the amplitude of z; and we write

x = R(z),y = I(z),r = z,θ = am z.

When y = 0 we say that z is real, when x = 0 that z is purely imaginary. Two numbers x + yi, x yi which differ only in the signs of their imaginary parts, we call conjugate. It will be observed that the sum 2x of two conjugate numbers and their product x2 + y2 are both real, that they have the same modulus x2 + y2 and that their product is equal to the square of the modulus of either. The roots of a quadratic with real coefficients, for example, are conjugate, when not real.

It must be observed that θ or am z is a many-valued function of x and y, having an infinity of values, which are angles differing by multiples of 2π.32 A line originally lying along OX will, if turned through any of these angles, come to lie along OP. We shall describe that one of these angles which lies between π and π as the principal value of the amplitude of z. This definition is unambiguous except when one of the values is π, in which case π is also a value. In this case we must make some special provision as to which value is to be regarded as the principal value. In general, when we speak of the amplitude of z we shall, unless the contrary is stated, mean the principal value of the amplitude.

Fig. 24 is usually known as Argand’s diagram.

45. De Moivre’s Theorem. The following statements follow immediately from the definitions of addition and multiplication.

(1) The real (or imaginary) part of the sum of two complex numbers is equal to the sum of their real (or imaginary) parts.

(2) The modulus of the product of two complex numbers is equal to the product of their moduli.

(3) The amplitude of the product of two complex numbers is either equal to the sum of their amplitudes, or differs from it by 2π.

It should be observed that it is not always true that the principal value of am(zz) is the sum of the principal values of amz and amz. For example, if z = z= 1 + i, then the principal values of the amplitudes of z and z are each 3 4π. But zz= 2i, and the principal value of am(zz) is  1 2π and not 3 2π.

The two last theorems may be expressed in the equation

r(cos θ + i sin θ) × ρ(cos φ + i sin φ) = rρ{cos(θ + φ) + i sin(θ + φ)},
which may be proved at once by multiplying out and using the ordinary trigonometrical formulae for cos(θ + φ) and sin(θ + φ). More generally

r1(cos θ1 + i sin θ1) × r2(cos θ2 + i sin θ2) × × rn(cos θn + i sin θn) = r1r2rn{cos(θ1 + θ2 + + θn) + i sin(θ1 + θ2 + + θn)}.

A particularly interesting case is that in which

r1 = r2 = = rn = 1,θ1 = θ2 = = θn = θ.

We then obtain the equation

(cos θ + i sin θ)n = cos nθ + i sin nθ,
where n is any positive integer: a result known as De Moivre’s Theorem.33

Again, if

z = r(cos θ + i sin θ)
then
1/z = (cos θ i sin θ)/r.
Thus the modulus of the reciprocal of z is the reciprocal of the modulus of z, and the amplitude of the reciprocal is the negative of the amplitude of z. We can now state the theorems for quotients which correspond to (2) and (3).

(4) The modulus of the quotient of two complex numbers is equal to the quotient of their moduli.

(5) The amplitude of the quotient of two complex numbers either is equal to the difference of their amplitudes, or differs from it by 2π.

Again

(cos θ + i sin θ)n = (cos θ i sin θ)n = {cos(θ) + i sin(θ)}n = cos(nθ) + i sin(nθ).

Hence De Moivre’s Theorem holds for all integral values of n, positive or negative.

To the theorems (1)–(5) we may add the following theorem, which is also of very great importance.

(6) The modulus of the sum of any number of complex numbers is not greater than the sum of their moduli.


pict

Fig. 25.

Let OP¯, OP¯, … be the displacements corresponding to the various complex numbers. Draw PQ equal and parallel to OP, QR equal and parallel to OP, and so on. Finally we reach a point U, such that

OU¯ = OP¯ + OP¯ + OP¯ + .
The length OU is the modulus of the sum of the complex numbers, whereas the sum of their moduli is the total length of the broken line OPQRU, which is not less than OU.

A purely arithmetical proof of this theorem is outlined in Exs. XXI. 1.

46. We add some theorems concerning rational functions of complex numbers. A rational function of the complex variable z is defined exactly as is a rational function of a real variable x, viz. as the quotient of two polynomials in z.

THEOREM 1. Any rational function R(z) can be reduced to the form X + Y i, where X and Y are rational functions of x and y with real coefficients.

In the first place it is evident that any polynomial P(x + yi) can be reduced, in virtue of the definitions of addition and multiplication, to the form A + Bi, where A and B are polynomials in x and y with real coefficients. Similarly Q(x + yi) can be reduced to the form C + Di. Hence

R(x + yi) = P(x + yi)/Q(x + yi)
can be expressed in the form

(A + Bi)/(C + Di) = (A + Bi)(C Di)/(C + Di)(C Di) = AC + BD C2 + D2 + BC AD C2 + D2 i,

which proves the theorem.

THEOREM 2. If R(x + yi) = X + Y i, R denoting a rational function as before, but with real coefficients, then R(x yi) = X Y i.

In the first place this is easily verified for a power (x + yi)n by actual expansion. It follows by addition that the theorem is true for any polynomial with real coefficients. Hence, in the notation used above,

R(x yi) = A Bi C Di = AC + BD C2 + D2 BC AD C2 + D2 i,
the reduction being the same as before except that the sign of i is changed throughout. It is evident that results similar to those of Theorems 1 and 2 hold for functions of any number of complex variables.

THEOREM 3. The roots of an equation

a0zn + a 1zn1 + + a n = 0,
whose coefficients are real, may, in so far as they are not themselves real, be arranged in conjugate pairs.

For it follows from Theorem 2 that if x + yi is a root then so is x yi. A particular case of this theorem is the result (§ 43) that the roots of a quadratic equation with real coefficients are either real or conjugate.

This theorem is sometimes stated as follows: in an equation with real coefficients complex roots occur in conjugate pairs. It should be compared with the result of Exs. VIII. 7, which may be stated as follows: in an equation with rational coefficients irrational roots occur in conjugate pairs.34

Examples XXI. 1. Prove theorem (6) of §45 directly from the definitions and without the aid of geometrical considerations.

[First, to prove that z + zz+ z is to prove that

(x + x)2 + (y + y)2 {x2 + y2 + x2 + y2}2.
The theorem is then easily extended to the general case.]

2. The one and only case in which

z+ z+ = z + z+ ,
is that in which the numbers z, z, … have all the same amplitude. Prove this both geometrically and analytically.

3. The modulus of the sum of any number of complex numbers is not less than the sum of their real (or imaginary) parts.

4. If the sum and product of two complex numbers are both real, then the two numbers must either be real or conjugate.

5. If

a + b2 + (c + d2)i = A + B2 + (C + D2)i,
where abcd, ABCD are real rational numbers, then
a = A,b = B,c = C,d = D.

6. Express the following numbers in the form A + Bi, where A and B are real numbers:

(1 + i)2, 1 + i 1 i2, 1 i 1 + i2,λ + μi λ μi, λ + μi λ μi2 λ μi λ + μi2,
where λ and μ are real numbers.

7. Express the following functions of z = x + yi in the form X + Y i, where X and Y are real functions of x and y: z2z3zn, 1/z, z + (1/z), (α + βz)/(γ + δz), where αβγδ are real numbers.

8. Find the moduli of the numbers and functions in the two preceding examples.

9. The two lines joining the points z = a, z = b and z = c, z = d will be perpendicular if

am a b c d = ±1 2π,
i.e. if (a b)/(c d) is purely imaginary. What is the condition that the lines should be parallel?

10. The three angular points of a triangle are given by z = α, z = β, z = γ, where αβγ are complex numbers. Establish the following propositions:

(i) the centre of gravity is given by z = 1 3(α + β + γ);

(ii) the circum-centre is given by z α= z β= z γ;

(iii) the three perpendiculars from the angular points on the opposite sides meet in a point given by

R z α β γ = R z β γ α = R z γ α β = 0;

(iv) there is a point P inside the triangle such that

CBP = ACP = BAP = ω,
and
cotω = cotA + cotB + cotC.

[To prove (iii) we observe that if ABC are the vertices, and P any point z, then the condition that AP should be perpendicular to BC is (Ex. 9) that (z α)/(β γ) should be purely imaginary, or that

R(z α)R(β γ) + I(z α)I(β γ) = 0.
This equation, and the two similar equations obtained by permuting αβγ cyclically, are satisfied by the same value of z, as appears from the fact that the sum of the three left-hand sides is zero.

To prove (iv), take BC parallel to the positive direction of the axis of x. Then35

γ β = a,α γ = bCis(C),β α = cCisB.

We have to determine z and ω from the equations

(z α)(β0 α0) (z0 α0)(β α) = (z β)(γ0 β0) (z0 β0)(γ β) = (z γ)(α0 γ0) (z0 γ0)(α γ) = Cis2ω,
where z0, α0β0γ0 denote the conjugates of z, αβγ.

Adding the numerators and denominators of the three equal fractions, and using the equation

icotω = (1 + Cis2ω)/(1 Cis2ω),
we find that
icotω = (β γ)(β0 γ0) + (γ α)(γ0 α0) + (α β)(α0 β0) βγ0 β0γ + γα0 γ0α + αβ0 α0β .
From this it is easily deduced that the value of cotω is (a2 + b2 + c2)/4Δ, where Δ is the area of the triangle; and this is equivalent to the result given.

To determine z, we multiply the numerators and denominators of the equal fractions by (γ0 β0)/(β α), (α0 γ0)/(γ β), (β0 α0)/(α γ), and add to form a new fraction. It will be found that

z = aαCisA + bβCisB + cγCisC aCisA + bCisB + cCisC .

11. The two triangles whose vertices are the points abc and xyz respectively will be similar if

111 a b c xyz = 0

[The condition required is that AB¯/AC¯ = XY ¯/XZ¯ (large letters denoting the points whose arguments are the corresponding small letters), or (b a)/(c a) = (y x)/(z x), which is the same as the condition given.]

12. Deduce from the last example that if the points xyz are collinear then we can find real numbers αβγ such that α + β + γ = 0 and αx + βy + γz = 0, and conversely (cf. Exs. XX. 4). [Use the fact that in this case the triangle formed by xyz is similar to a certain line-triangle on the axis OX, and apply the result of the last example.]

13. The general linear equation with complex coefficients. The equation αz + β = 0 has the one solution z = (β/α), unless α = 0. If we put

α = a + Ai,β = b + Bi,z = x + yi,
and equate real and imaginary parts, we obtain two equations to determine the two real numbers x and y. The equation will have a real root if y = 0, which gives ax + b = 0, Ax + B = 0, and the condition that these equations should be consistent is aB bA = 0.

14. The general quadratic equation with complex coefficients. This equation is

(a + Ai)z2 + 2(b + Bi)z + (c + Ci) = 0.

Unless a and A are both zero we can divide through by a + iA. Hence we may consider

z2 + 2(b + Bi)z + (c + Ci) = 0 (1)

as the standard form of our equation. Putting z = x + yi and equating real and imaginary parts, we obtain a pair of simultaneous equations for x and y, viz.

x2 y2 + 2(bx By) + c = 0,2xy + 2(by + Bx) + C = 0.

If we put

x + b = ξ,y + B = η,b2 B2 c = h,2bB C = k,
these equations become
ξ2 η2 = h,2ξη = k.

Squaring and adding we obtain

ξ2 + η2 = h2 + k2,ξ = ±1 2{h2 + k2 + h},η = ±1 2{h2 + k2 h}.
We must choose the signs so that ξη has the sign of k: i.e. if k is positive we must take like signs, if k is negative unlike signs.

Conditions for equal roots. The two roots can only be equal if both the square roots above vanish, i.e. if h = 0, k = 0, or if c = b2 B2, C = 2bB. These conditions are equivalent to the single condition c + Ci = (b + Bi)2, which obviously expresses the fact that the left-hand side of (1) is a perfect square.

Condition for a real root. If x2 + 2(b + Bi)x + (c + Ci) = 0, where x is real, then x2 + 2bx + c = 0, 2Bx + C = 0. Eliminating x we find that the required condition is

C2 4bBC + 4cB2 = 0.

Condition for a purely imaginary root. This is easily found to be

C2 4bBC 4b2c = 0.

Conditions for a pair of conjugate complex roots. Since the sum and the product of two conjugate complex numbers are both real, b + Bi and c + Ci must both be real, i.e. B = 0, C = 0. Thus the equation (1) can have a pair of conjugate complex roots only if its coefficients are real. The reader should verify this conclusion by means of the explicit expressions of the roots. Moreover, if b2 c, the roots will be real even in this case. Hence for a pair of conjugate roots we must have B = 0, C = 0, b2 < c.

15. The Cubic equation. Consider the cubic equation

z3 + 3Hz + G = 0,
where G and H are complex numbers, it being given that the equation has (a) a real root, (b) a purely imaginary root, (c) a pair of conjugate roots. If H = λ + μi, G = ρ + σi, we arrive at the following conclusions.

(a) Conditions for a real root. If μ is not zero, then the real root is  σ/3μ, and σ3 + 27λμ2σ 27μ3ρ = 0. On the other hand, if μ = 0 then we must also have σ = 0, so that the coefficients of the equation are real. In this case there may be three real roots.

(b) Conditions for a purely imaginary root. If μ is not zero then the purely imaginary root is (ρ/3μ)i, and ρ3 27λμ2ρ 27μ3σ = 0. If μ = 0 then also ρ = 0, and the root is yi, where y is given by the equation y3 3λy σ = 0, which has real coefficients. In this case there may be three purely imaginary roots.

(c) Conditions for a pair of conjugate complex roots. Let these be x + yi and x yi. Then since the sum of the three roots is zero the third root must be  2x. From the relations between the coefficients and the roots of an equation we deduce

y2 3x2 = 3H,2x(x2 + y2) = G.
Hence G and H must both be real.

In each case we can either find a root (in which case the equation can be reduced to a quadratic by dividing by a known factor) or we can reduce the solution of the equation to the solution of a cubic equation with real coefficients.

16. The cubic equation x3 + a1x2 + a2x + a3 = 0, where a1 = A1 + Ai, …, has a pair of conjugate complex roots. Prove that the remaining root is Aa3/A, unless A= 0. Examine the case in which A= 0.

17. Prove that if z3 + 3Hz + G = 0 has two complex roots then the equation

8α3 + 6αH G = 0
has one real root which is the real part α of the complex roots of the original equation; and show that α has the same sign as G.

18. An equation of any order with complex coefficients will in general have no real roots nor pairs of conjugate complex roots. How many conditions must be satisfied by the coefficients in order that the equation should have (a) a real root, (b) a pair of conjugate roots?

19. Coaxal circles. In Fig. 26, let abz be the arguments of ABP. Then

am z b z a = APB,
if the principal value of the amplitude is chosen. If the two circles shown in the figure are equal, and zz1z are the arguments of PP1P, and APB = θ, it is easy to see that
am zb za = π θ,am z1 b z1 a = θ,
and
am zb za = π + θ.

The locus defined by the equation

am z b z a = θ,
where θ is constant, is the arc APB. By writing π θ θ π + θ for θ, we obtain the other three arcs shown.


pict

Fig. 26.

The system of equations obtained by supposing that θ is a parameter, varying from  π to  + π, represents the system of circles which can be drawn through the points AB. It should however be observed that each circle has to be divided into two parts to which correspond different values of θ.

20. Now let us consider the equation

z b z a = λ, (1)

where λ is a constant.

Let K be the point in which the tangent to the circle ABP at P meets AB. Then the triangles KPAKBP are similar, and so

AP/PB = PK/BK = KA/KP = λ.
Hence KA/KB = λ2, and therefore K is a fixed point for all positions of P which satisfy the equation (1). Also KP2 = KA KB, and so is constant. Hence the locus of P is a circle whose centre is K.

The system of equations obtained by varying λ represents a system of circles, and every circle of this system cuts at right angles every circle of the system of Ex. 19.

The system of Ex. 19 is called a system of coaxal circles of the common point kind. The system of Ex. 20 is called a system of coaxal circles of the limiting point kind, A and B being the limiting points of the system. If λ is very large or very small then the circle is a very small circle containing A or B in its interior.

21. Bilinear Transformations. Consider the equation

z = Z + a, (1)

where z = x + yi and Z = X + Y i are two complex variables which we may suppose to be represented in two planes xoyXOY . To every value of z corresponds one of Z, and conversely. If a = α + βi then

x = X + α,y = Y + β,
and to the point (x,y) corresponds the point (X,Y ). If (x,y) describes a curve of any kind in its plane, (X,Y ) describes a curve in its plane. Thus to any figure in one plane corresponds a figure in the other. A passage of this kind from a figure in the plane xoy to a figure in the plane XOY by means of a relation such as (1) between z and Z is called a transformation. In this particular case the relation between corresponding figures is very easily defined. The (X,Y ) figure is the same in size, shape, and orientation as the (x,y) figure, but is shifted a distance α to the left, and a distance β downwards. Such a transformation is called a translation.

Now consider the equation

z = ρZ, (2)

where ρ is real. This gives x = ρX, y = ρY . The two figures are similar and similarly situated about their respective origins, but the scale of the (x,y) figure is ρ times that of the (X,Y ) figure. Such a transformation is called a magnification.

Finally consider the equation

z = (cosφ + isinφ)Z. (3)

It is clear that z= Z and that one value of amz is amZ + φ, and that the two figures differ only in that the (x,y) figure is the (X,Y ) figure turned about the origin through an angle φ in the positive direction. Such a transformation is called a rotation.

The general linear transformation

z = aZ + b (4)

is a combination of the three transformations (1), (2), (3). For, if a= ρ and ama = φ, we can replace (4) by the three equations

z = z+ b,z= ρZ,Z= (cosφ + isinφ)Z.
Thus the general linear transformation is equivalent to the combination of a translation, a magnification, and a rotation.

Next let us consider the transformation

z = 1/Z. (5)

If Z= R and amZ = Θ, then z= 1/R and amz = Θ, and to pass from the (x,y) figure to the (X,Y ) figure we invert the former with respect to o, with unit radius of inversion, and then construct the image of the new figure in the axis ox (i.e. the symmetrical figure on the other side of ox).

Finally consider the transformation

z = aZ + b cZ + d. (6)

This is equivalent to the combination of the transformations

z = (a/c) + (bc ad)(z/c),z= 1/Z,Z= cZ + d,
i.e. to a certain combination of transformations of the types already considered.

The transformation (6) is called the general bilinear transformation. Solving for Z we obtain

Z = dz b cz a.

The general bilinear transformation is the most general type of transformation for which one and only one value of z corresponds to each value of Z, and conversely.

22. The general bilinear transformation transforms circles into circles. This may be proved in a variety of ways. We may assume the well-known theorem in pure geometry, that inversion transforms circles into circles (which may of course in particular cases be straight lines). Or we may use the results of Exs. 19 and 20. If, e.g., the (x,y) circle is

(z σ)/(z ρ)= λ,
and we substitute for z in terms of Z, we obtain
(Z σ)/(Z ρ)= λ,
where
σ= b σd a σc,ρ= b ρd a ρc,λ= a ρc a σcλ.

23. Consider the transformations z = 1/Z, z = (1 + Z)/(1 Z), and draw the (X,Y ) curves which correspond to (1) circles whose centre is the origin, (2) straight lines through the origin.

24. The condition that the transformation z = (aZ + b)/(cZ + d) should make the circle x2 + y2 = 1 correspond to a straight line in the (X,Y ) plane is a= c.

25. Cross ratios. The cross ratio (z1,z2;z3,z4) is defined to be

(z1 z3)(z2 z4) (z1 z4)(z2 z3).

If the four points z1z2z3z4 are on the same line, this definition agrees with that adopted in elementary geometry. There are 24 cross ratios which can be formed from z1z2z3z4 by permuting the suffixes. These consist of six groups of four equal cross ratios. If one ratio is λ, then the six distinct cross ratios are λ, 1 λ, 1/λ, 1/(1 λ), (λ 1)/λ, λ/(λ 1). The four points are said to be harmonic or harmonically related if any one of these is equal to  1. In this case the six ratios are 1, 2, 1, 1 2, 21 2.

If any cross ratio is real then all are real and the four points lie on a circle. For in this case

am (z1 z3)(z2 z4) (z1 z4)(z2 z3)
must have one of the three values π0π, so that am{(z1 z3)/(z1 z4)}and am{(z2 z3)/(z2 z4)} must either be equal or differ by π (cf. Ex. 19).

If (z1,z2;z3,z4) = 1, we have the two equations

am z1 z3 z1 z4 = ±π + am z2 z3 z2 z4, z1 z3 z1 z4 = z2 z3 z2 z4 .
The four points A1, A2, A3A4 lie on a circle, A1 and A2 being separated by A3 and A4. Also A1A3/A1A4 = A2A3/A2A4. Let O be the middle point of A3A4. The equation
(z1 z3)(z2 z4) (z1 z4)(z2 z3) = 1
may be put in the form
(z1 + z2)(z3 + z4) = 2(z1z2 + z3z4),
or, what is the same thing,
{z1 1 2(z3 + z4)}{z2 1 2(z3 + z4)}= {1 2(z3 z4)}2.
But this is equivalent to OA1¯ OA2¯ = OA3¯2 = OA4¯2. Hence OA1 and OA2 make equal angles with A3A4, and OA1 OA2 = OA32 = OA42. It will be observed that the relation between the pairs A1A2 and A3A4 is symmetrical. Hence, if O is the middle point of A1A2, OA3 and OA4 are equally inclined to A1A2, and OA3 OA4 = OA12 = OA22.

26. If the points A1A2 are given by az2 + 2bz + c = 0, and the points A3A4 by az2 + 2bz + c= 0, and O is the middle point of A3A4, and ac+ ac 2bb= 0, then OA1OA2 are equally inclined to A3A4 and OA1 OA2 = OA32 = OA42.

(Math. Trip. 1901.)

27. ABCD are two intersecting lines in Argand’s diagram, and PQ their middle points. Prove that, if AB bisects the angle CPD and PA2 = PB2 = PC PD, then CD bisects the angle AQB and QC2 = QD2 = QA QB.

(Math. Trip. 1909.)

28. The condition that four points should lie on a circle. A sufficient condition is that one (and therefore all) of the cross ratios should be real (Ex. 25); this condition is also necessary. Another form of the condition is that it should be possible to choose real numbers αβγ such that

1 1 1 α β γ z1z4 + z2z3z2z4 + z3z1z3z4 + z1z2 = 0.

[To prove this we observe that the transformation Z = 1/(z z4) is equivalent to an inversion with respect to the point z4, coupled with a certain reflexion (Ex. 21). If z1z2z3 lie on a circle through z4, the corresponding points Z1 = 1/(z1 z4), Z2 = 1/(z2 z4), Z3 = 1/(z3 z4) lie on a straight line. Hence (Ex. 12) we can find real numbers αβγ such that α+ β+ γ= 0 and α/(z1 z4) + β/(z2 z4) + γ/(z3 z4) = 0, and it is easy to prove that this is equivalent to the given condition.]

29. Prove the following analogue of De Moivre’s Theorem for real numbers: if φ1φ2, φ3, … is a series of positive acute angles such that

tanφm+1 = tanφm secφ1 + secφm tanφ1,  then tanφm+n = tanφm secφn + secφm tanφn, secφm+n = secφm secφn + tanφm tanφn,

and

tanφm + secφm = (tanφ1 + secφ1)m.

[Use the method of mathematical induction.]

30. The transformation z = Zm. In this case r = Rm, and θ and mΘ differ by a multiple of 2π. If Z describes a circle round the origin then z describes a circle round the origin m times.

The whole (x,y) plane corresponds to any one of m sectors in the (X,Y ) plane, each of angle 2π/m. To each point in the (x,y) plane correspond m points in the (X,Y ) plane.

31. Complex functions of a real variable. If f(t)φ(t) are two real functions of a real variable t defined for a certain range of values of t, we call

z = f(t) + iφ(t) (1)

a complex function of t. We can represent it graphically by drawing the curve

x = f(t),y = φ(t);
the equation of the curve may be obtained by eliminating t between these equations. If z is a polynomial in t, or rational function of t, with complex coefficients, we can express it in the form (1) and so determine the curve represented by the function.

(i) Let

z = a + (b a)t,
where a and b are complex numbers. If a = α + αi, b = β + βi, then
x = α + (β α)t,y = α+ (βα)t.
The curve is the straight line joining the points z = a and z = b. The segment between the points corresponds to the range of values of t from 0 to 1. Find the values of t which correspond to the two produced segments of the line.

(ii) If

z = c + ρ 1 + ti 1 ti,
where ρ is positive, then the curve is the circle of centre c and radius ρ. As t varies through all real values z describes the circle once.

(iii) In general the equation z = (a + bt)/(c + dt) represents a circle. This can be proved by calculating x and y and eliminating: but this process is rather cumbrous. A simpler method is obtained by using the result of Ex. 22. Let z = (a + bZ)/(c + dZ), Z = t. As t varies Z describes a straight line, viz. the axis of X. Hence z describes a circle.

(iv) The equation

z = a + 2bt + ct2
represents a parabola generally, a straight line if b/c is real.

(v) The equation z = (a + 2bt + ct2)/(α + 2βt + γt2), where αβγ are real, represents a conic section.

[Eliminate t from

x = (A + 2Bt + Ct2)/(α + 2βt + γt2),y = (A+ 2Bt + Ct2)/(α + 2βt + γt2),
where A + Ai = a, B + Bi = b, C + Ci = c.]

47. Roots of complex numbers. We have not, up to the present, attributed any meaning to symbols such as anam/n, when a is a complex number, and m and n integers. It is, however, natural to adopt the definitions which are given in elementary algebra for real values of a. Thus we define an or a1/n, where n is a positive integer, as a number z which satisfies the equation zn = a; and am/n, where m is an integer, as (a1/n)m. These definitions do not prejudge the question as to whether there are or are not more than one (or any) roots of the equation.

48. Solution of the equation zn = a. Let

a = ρ(cos φ + i sin φ),
where ρ is positive and φ is an angle such that π < φ π. If we put z = r(cos θ + i sin θ), the equation takes the form
rn(cos nθ + i sin nθ) = ρ(cos φ + i sin φ);
so that

rn = ρ, cos nθ = cos φ, sin nθ = sin φ. (1)

The only possible value of r is ρn, the ordinary arithmetical nth root of ρ; and in order that the last two equations should be satisfied it is necessary and sufficient that nθ = φ + 2kπ, where k is an integer, or

θ = (φ + 2kπ)/n.
If k = pn + q, where p and q are integers, and 0 q < n, the value of θ is 2pπ + (φ + 2qπ)/n, and in this the value of p is a matter of indifference. Hence the equation
zn = a = ρ(cos φ + i sin φ)
has n roots and n only, given by z = r(cos θ + i sin θ), where
r = ρn,θ = (φ + 2qπ)/n,(q = 0,1,2,,n 1).

That these n roots are in reality all distinct is easily seen by plotting them on Argand’s diagram. The particular root

ρn{cos(φ/n) + i sin(φ/n)}
is called the principal value of an.

The case in which a = 1, ρ = 1, φ = 0 is of particular interest. The n roots of the equation xn = 1 are

cos(2qπ/n) + i sin(2qπ/n),(q = 0,1,,n 1).
These numbers are called the nth roots of unity; the principal value is unity itself. If we write ωn for cos(2π/n) + i sin(2π/n), we see that the nth roots of unity are
1,ωn,ωn2,,ω nn1.

Examples XXII. 1. The two square roots of 1 are 1 1; the three cube roots are 1, 1 2(1 + i3), 1 2(1 i3); the four fourth roots are 1, i, 1, i; and the five fifth roots are

1, 1 4[5 1 + i10 + 25],1 4[ 5 1 + i10 25], 1 4[ 5 1 i10 25],1 4[5 1 i10 + 25].

2. Prove that

1 + ωn + ωn2 + + ω nn1 = 0.

3. Prove that

(x + yω3 + zω32)(x + yω 32 + zω 3) = x2 + y2 + z2 yz zx xy.

4. The nth roots of a are the products of the nth roots of unity by the principal value of an.

5. It follows from Exs. XXI. 14 that the roots of

z2 = α + βi
are
±1 2{α2 + β2 + α}±i1 2{α2 + β2 α},
like or unlike signs being chosen according as β is positive or negative. Show that this result agrees with the result of §48.

6. Show that (x2m a2m)/(x2 a2) is equal to

(x2 2axcos π m + a2)(x2 2axcos 2π m + a2)(x2 2axcos (m 1)π m + a2).

[The factors of x2m a2m are

(x a),(x aω2m),(x aω2m2),(x aω 2m2m1).
The factor x aω2mm is x + a. The factors (x aω2ms), (x aω2m2ms) taken together give a factor x2 2axcos(sπ/m) + a2.]

7. Resolve x2m+1 a2m+1, x2m + a2m, and x2m+1 + a2m+1 into factors in a similar way.

8. Show that x2n 2xnan cosθ + a2n is equal to

x2 2xacos θ n + a2 x2 2xacos θ + 2π n + a2 x2 2xacos θ + 2(n 1)π n + a2 .

[Use the formula

x2n 2xnan cosθ + a2n = {xn an(cosθ + isinθ)}{xn an(cosθ isinθ)},
and split up each of the last two expressions into n factors.]

9. Find all the roots of the equation x6 2x3 + 2 = 0.

(Math. Trip. 1910.)

10. The problem of finding the accurate value of ωn in a numerical form involving square roots only, as in the formula ω3 = 1 2(1 + i3), is the algebraical equivalent of the geometrical problem of inscribing a regular polygon of n sides in a circle of unit radius by Euclidean methods, i.e. by ruler and compasses. For this construction will be possible if and only if we can construct lengths measured by cos(2π/n) and sin(2π/n); and this is possible (Ch. II, Misc. Exs. 22) if and only if these numbers are expressible in a form involving square roots only.

Euclid gives constructions for n = 3, 4, 5, 6, 8, 10, 12, and 15. It is evident that the construction is possible for any value of n which can be found from these by multiplication by any power of 2. There are other special values of n for which such constructions are possible, the most interesting being n = 17.

49. The general form of De Moivre’s Theorem. It follows from the results of the last section that if q is a positive integer then one of the values of (cos θ + i sin θ)1/q is

cos(θ/q) + i sin(θ/q).
Raising each of these expressions to the power p (where p is any integer positive or negative), we obtain the theorem that one of the values of (cos θ + i sin θ)p/q is cos(pθ/q) + i sin(pθ/q), or that if α is any rational number then one of the values of (cos θ + i sin θ)α is
cos αθ + i sin αθ.
This is a generalised form of De Moivre’s Theorem (§ 45).

Miscellaneous Examples on Chapter III.

1. The condition that a triangle (xyz) should be equilateral is that

x2 + y2 + z2 yz zx xy = 0.

[Let XY Z be the triangle. The displacement ZX¯ is Y Z¯ turned through an angle 2 3π in the positive or negative direction. Since Cis 2 3π = ω3, Cis(2 3π) = 1/ω3 = ω32, we have x z = (z y)ω3 or x z = (z y)ω32. Hence x + yω3 + zω32 = 0 or x + yω32 + zω3 = 0. The result follows from Exs. XXII. 3.]

2. If XY Z, XY Z are two triangles, and

Y Z¯ Y Z¯= ZX¯ ZX¯= XY ¯ XY ¯,
then both triangles are equilateral. [From the equations
(y z)(yz) = (z x)(zx) = (x y)(xy) = κ2,
say, we deduce 1/(yz) = 0, or x2 yz= 0. Now apply the result of the last example.]

3. Similar triangles BCX, CAY , ABZ are described on the sides of a triangle ABC. Show that the centres of gravity of ABCXY Z are coincident.

[We have (x c)/(b c) = (y a)/(c a) = (z b)/(a b) = λ, say. Express 1 3(x + y + z) in terms of abc.]

4. If XY Z are points on the sides of the triangle ABC, such that

BX/XC = CY/Y A = AZ/ZB = r,
and if ABC, XY Z are similar, then either r = 1 or both triangles are equilateral.

5. If ABCD are four points in a plane, then

AD BC BD CA + CD AB.

[Let z1z2z3z4 be the complex numbers corresponding to ABCD. Then we have identically

(x1 x4)(x2 x3) + (x2 x4)(x3 x1) + (x3 x4)(x1 x2) = 0.
Hence

(x1 x4)(x2 x3) = (x2 x4)(x3 x1) + (x3 x4)(x1 x2) (x2 x4)(x3 x1)+ (x3 x4)(x1 x2).]

6. Deduce Ptolemy’s Theorem concerning cyclic quadrilaterals from the fact that the cross ratios of four concyclic points are real. [Use the same identity as in the last example.]

7. If z2 + z2 = 1, then the points zz are ends of conjugate diameters of an ellipse whose foci are the points 1 1. [If CPCD are conjugate semi-diameters of an ellipse and SH its foci, then CD is parallel to the external bisector of the angle SPH, and SP HP = CD2.]

8. Prove that a + b2 + a b2 = 2{a2 + b2}. [This is the analytical equivalent of the geometrical theorem that, if M is the middle point of PQ, then OP2 + OQ2 = 2OM2 + 2MP2.]

9. Deduce from Ex. 8 that

a + a2 b2+ a a2 b2= a + b+ a b.

[If a + a2 b2 = z1, a a2 b2 = z2, we have

z12 + z 22 = 1 2z1 + z22 + 1 2z1 z22 = 2a2 + 2a2 b2,
and so
(z1+ z2)2 = 2{a2 + a2 b2+ b2}= a + b2 + a b2 + 2a2 b2.

Another way of stating the result is: if z1 and z2 are the roots of αz2 + 2βz + γ = 0, then

z1+ z2= (1/α){(β + αγ) + (β αγ)}.]

10. Show that the necessary and sufficient conditions that both the roots of the equation z2 + az + b = 0 should be of unit modulus are

a2,b= 1,amb = 2ama.

[The amplitudes have not necessarily their principal values.]

11. If x4 + 4a1x3 + 6a2x2 + 4a3x + a4 = 0 is an equation with real coefficients and has two real and two complex roots, concyclic in the Argand diagram, then

a32 + a 12a 4 + a23 a 2a4 2a1a2a3 = 0.

12. The four roots of a0x4 + 4a1x3 + 6a2x2 + 4a3x + a4 = 0 will be harmonically related if

a0a32 + a 12a 4 + a23 a 0a2a4 2a1a2a3 = 0.

[Express Z23,14Z31,24Z12,34, where Z23,14 = (z1 z2)(z3 z4) + (z1 z3)(z2 z4) and z1z2, z3z4 are the roots of the equation, in terms of the coefficients.]

13. Imaginary points and straight lines. Let ax + by + c = 0 be an equation with complex coefficients (which of course may be real in special cases).

If we give x any particular real or complex value, we can find the corresponding value of y. The aggregate of pairs of real or complex values of x and y which satisfy the equation is called an imaginary straight line; the pairs of values are called imaginary points, and are said to lie on the line. The values of x and y are called the coordinates of the point (x,y). When x and y are real, the point is called a real point: when abc are all real (or can be made all real by division by a common factor), the line is called a real line. The points x = α + βi, y = γ + δi and x = α βi, y = γ δi are said to be conjugate; and so are the lines

(A + Ai)x + (B + Bi)y + C + Ci = 0,(A Ai)x + (B Bi)y + C Ci = 0.

Verify the following assertions:—every real line contains infinitely many pairs of conjugate imaginary points; an imaginary line in general contains one and only one real point; an imaginary line cannot contain a pair of conjugate imaginary points:—and find the conditions (a) that the line joining two given imaginary points should be real, and (b) that the point of intersection of two imaginary lines should be real.

14. Prove the identities

(x + y + z)(x + yω3 + zω32)(x + yω 32 + zω 3) = x3 + y3 + z3 3xyz, (x + y + z)(x + yω5 + zω54)(x + yω 52 + zω 53)(x + yω 53 + zω 52)(x + yω 54 + zω 5) = x5 + y5 + z5 5x3yz + 5xy2z2.

15. Solve the equations

x3 3ax + (a3 + 1) = 0,x5 5ax3 + 5a2x + (a5 + 1) = 0.

16. If f(x) = a0 + a1x + + akxk, then

{f(x) + f(ωx) + + f(ωn1x)}/n = a 0 + anxn + a 2nx2n + + a λnxλn,
ω being any root of xn = 1 (except x = 1), and λn the greatest multiple of n contained in k. Find a similar formula for aμ + aμ+nxn + aμ+2nx2n + .

17. If

(1 + x)n = p 0 + p1x + p2x2 + ,
n being a positive integer, then
p0 p2 + p4 = 21 2n cos 1 4nπ,p1 p3 + p5 = 21 2n sin 1 4nπ.

18. Sum the series

x 2!(n 2)! + x2 5!(n 5)! + x3 8!(n 8)! + + xn/3 (n 1)!,
n being a multiple of 3.
(Math. Trip. 1899.)

19. If t is a complex number such that t= 1, then the point x = (at + b)/(t c) describes a circle as t varies, unless c= 1, when it describes a straight line.

20. If t varies as in the last example then the point x = 1 2{at + (b/t)} in general describes an ellipse whose foci are given by x2 = ab, and whose axes are a+ b and ab. But if a= bthen x describes the finite straight line joining the points ab, ab.

21. Prove that if t is real and z = t2 1 + t4 t2, then, when t2 < 1, z is represented by a point which lies on the circle x2 + y2 + x = 0. Assuming that, when t2 > 1, t4 t2 denotes the positive square root of t4 t2, discuss the motion of the point which represents z, as t diminishes from a large positive value to a large negative value.

(Math. Trip. 1912.)

22. The coefficients of the transformation z = (aZ + b)/(cZ + d) are subject to the condition ad bc = 1. Show that, if c0, there are two fixed points αβ, i.e. points unaltered by the transformation, except when (a + d)2 = 4, when there is only one fixed point α; and that in these two cases the transformation may be expressed in the forms

z α z β = KZ α Z β, 1 z α = 1 Z α + K.

Show further that, if c = 0, there will be one fixed point α unless a = d, and that in these two cases the transformation may be expressed in the forms

z α = K(Z α),z = Z + K.

Finally, if abcd are further restricted to positive integral values (including zero), show that the only transformations with less than two fixed points are of the forms (1/z) = (1/Z) + K, z = Z + K.

(Math. Trip. 1911.)

23. Prove that the relation z = (1 + Zi)/(Z + i) transforms the part of the axis of x between the points z = 1 and z = 1 into a semicircle passing through the points Z = 1 and Z = 1. Find all the figures that can be obtained from the originally selected part of the axis of x by successive applications of the transformation.

(Math. Trip. 1912.)

24. If z = 2Z + Z2 then the circle Z= 1 corresponds to a cardioid in the plane of z.

25. Discuss the transformation z = 1 2{Z + (1/Z)}, showing in particular that to the circles X2 + Y 2 = α2 correspond the confocal ellipses

x2 1 2 α + 1 α2 + y2 1 2 α 1 α2 = 1.

26. If (z + 1)2 = 4/Z then the unit circle in the z-plane corresponds to the parabola Rcos21 2Θ = 1 in the Z-plane, and the inside of the circle to the outside of the parabola.

27. Show that, by means of the transformation z = {(Z ci)/(Z + ci)}2, the upper half of the z-plane may be made to correspond to the interior of a certain semicircle in the Z-plane.

28. If z = Z2 1, then as z describes the circle z= κ, the two corresponding positions of Z each describe the Cassinian oval ρ1ρ2 = κ, where ρ1ρ2 are the distances of Z from the points 11. Trace the ovals for different values of κ.

29. Consider the relation az2 + 2hzZ + bZ2 + 2gz + 2fZ + c = 0. Show that there are two values of Z for which the corresponding values of z are equal, and vice versa. We call these the branch points in the Z and z-planes respectively. Show that, if z describes an ellipse whose foci are the branch points, then so does Z.

[We can, without loss of generality, take the given relation in the form

z2 + 2zZcosω + Z2 = 1 :
the reader should satisfy himself that this is the case. The branch points in either plane are cosecω and cosecω. An ellipse of the form specified is given by
z + cosecω+ z cosecω= C,
where C is a constant. This is equivalent (Ex. 9) to
z + z2 cosec2ω+ z z2 cosec2ω= C.
Express this in terms of Z.]

30. If z = aZm + bZn, where mn are positive integers and ab real, then as Z describes the unit circle, z describes a hypo- or epi-cycloid.

31. Show that the transformation

z = (a + di)Z0 + b cZ0 (a di),
where abcd are real and a2 + d2 + bc > 0, and Z0 denotes the conjugate of Z, is equivalent to an inversion with respect to the circle
c(x2 + y2) 2ax 2dy b = 0.
What is the geometrical interpretation of the transformation when
a2 + d2 + bc < 0?

32. The transformation

1 z 1 + z = 1 Z 1 + Zc,
where c is rational and 0 < c < 1, transforms the circle z= 1 into the boundary of a circular lune of angle π/c.