1911 Encyclopædia Britannica/Number/Complex Numbers

19. Complex Numbers.&mdash;If $$a$$ is an assigned number, rational or irrational, and $$n$$ a natural number, it can be proved that there is a real number satisfying the equation $$x^n = a$$, except when $$n$$ is even and $$a$$ is negative: in this case the equation is not satisfied by any real number whatever. To remove the difficulty we construct an aggregate of polar couples $$\{x, y\}$$, where $$x, y$$ are any two real numbers, and define the addition and multiplication of such couples by the rules
 * $$\{x, y\} + \{x', y'\} = \{x+x', y+y'\}$$;
 * $$\{x, y\} \times \{x', y'\} = \{xx'-yy', xy'+x'y\}$$.

We also agree that $$\{x, y\} <\{x', y'\},$$ if $$x < x'$$ or if $$x = x'$$ and $$y < y'$$. It follows that the aggregate has the ground element $$\{1, 0\}$$, which we may denote by $$\sigma$$; and that, if we write $$\tau$$ for the element $$\{0, 1\}$$,
 * $$\tau^2 =\{-1, 0\}=-\sigma$$.

Whenever $$m, n$$ are rational, $$\{m, n\} = m\sigma+n\tau$$, and we are thus justified in writing, if we like, $$x\sigma+y\tau$$ for $$\{x, y\}$$, in all circumstances. A further simplification is gained by writing $$x$$ instead of $$x\sigma$$, and regarding $$\tau$$ as a symbol which is such that $$\tau^2= -1$$, but in other respects obeys the ordinary laws of operation. It is usual to write $$i$$ instead of $$\tau$$; we thus have an aggregate $$\mathfrak{I}$$ of complex numbers $$x+yi$$. In this aggregate, which includes the real continuum as part of itself, not only the four rational operations (excluding division by $$\{0, 0\}$$, the zero element), but also the extraction of roots, may be effected without any restriction. Moreover (as first proved by Gauss and Cauchy), if $$a_0, a_1, \dots a_n$$ are any assigned real or complex numbers, the equation
 * $$a_0z^n + a_1z^{n-1} + \dots + a_{n-1}z + a_n = 0$$, is always satisfied by precisely $$n$$ real or complex values of $$z$$, with a proper convention as to multiple roots. Thus any algebraic function of any finite number of elements of

$$\mathfrak{I}$$ is also contained in $$\mathfrak{I}$$, which is, in this sense, a closed arithmetical field, just as $$\mathfrak{N}$$ is when we restrict ourselves to rational operations. The power of $$\mathfrak{I}$$ is the same as that of $$\mathfrak{N}$$.