43. The quadratic equation with real coefficients

The equations $z^{2} + 1 = 0$ , $a z^{2} + 2 b z + c = 0$ . There is no real number $z$ such that $z^{2} + 1 = 0$ ; 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 $z^{2} = - 1$ , it is sometimes written in the form $\sqrt{- 1}$ .

Complex numbers are sometimes called imaginary.¹ 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 $x + y i$ . Such a pair of numbers is no less ‘real’ than any ordinary number such as $\frac{1}{2}$ , or than the paper on which this is printed, or than the Solar System. Thus $i = 0 + 1 i$ 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 $z^{2} + 1 = 0$ , 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 $a z^{2} + 2 b z + c = 0,$ where $a$ , $b$ , $c$ are real numbers. If $b^{2} > a c$ , the ordinary method of solution gives two real roots ${- b \pm \sqrt{b^{2} - a c}} / a .$ If $b^{2} < a c$ , the equation has no real roots. It may be written in the form ${z + (b / a)}^{2} = - (a c - b^{2}) / a^{2},$ an equation which is evidently satisfied if we substitute for $z + (b / a)$ either of the complex numbers $\pm i \sqrt{a c - b^{2}} / a$ .² We express this by saying that the equation has the two complex roots ${- b \pm i \sqrt{a c - b^{2}}} / a .$

If we agree as a matter of convention to say that when $b^{2} = a c$ (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 $n$ th degree cannot have more than $n$ real roots. Let us denote the complex number $x + y i$ by the single letter $z$ , a convention which we may express by writing $z = x + y i$ . 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 $n$ th degree, and $f (z) = 0$ has the $n$ roots $a_{1}$ , $a_{2}$ , …, $a_{n}$ , then $f (z) = A (z - a_{1}) (z - a_{2}) \dots (z - a_{n}),$ where $A$ is a constant, real or complex, in fact the coefficient of $z^{n}$ 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 $n$ th 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.³ 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 $\frac{3}{7}$ 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 $a z^{2} + 2 b z + c = 0,$ then $α + β = - (2 b / a), α β = (c / a) .$

Similarly, if $α$ , $β$ , $γ$ are the roots of $a z^{3} + 3 b z^{2} + 3 c z + d = 0,$ then $α + β + γ = - (3 b / a), β γ + γ α + α β = (3 c / a), α β γ = - (d / a) .$ All such theorems as these are true whether $a$ , $b$ , … $α$ , $β$ , … are real or complex.

The phrase ‘real number’ was introduced as an antithesis to ‘imaginary number’.↩︎
We shall sometimes write $x + i y$ instead of $x + y i$ for convenience in printing.↩︎
See Appendix I.↩︎

\leftarrow

39-42. Complex numbers

Main Page

44. Argand’s diagram

\to