1911 Encyclopædia Britannica/Number/Complex Numbers

19. Complex Numbers.—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

$a_{0}z^{n}+a_{1}z^{n-1}+\dots +a_{n-1}z+a_{n}=0$, 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 ${\mathfrak {N}}$.