## 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 $$x$$$$y$$.

We shall find it convenient to denote this pair of real numbers $$x$$$$y$$ 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 $\begin{equation*}x + yi = x’ + y’i, \tag{1}\end{equation*}$ if and only if $$x = x’$$, $$y = y’$$; $\begin{equation*} (x + yi) + (x’ + y’i) = (x + x’) + (y + y’)i; \tag{2}\end{equation*}$$\begin{equation*} (x + yi) (x’ + y’i) = xx’ – yy’ + (xy’ + yx’)i. \tag{3}\end{equation*}$

In particular we have, as special cases of (2) and (3), $\begin{gathered} x + yi = (x + 0i) + (0 + yi),\\ (x + 0i) (x’ + y’i) = xx’ + xy’i;\end{gathered}$ 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 $(x + yi)^{2} = (x + yi) (x + yi) = x^{2} – y^{2} + 2xyi.$

The reader will easily verify for himself that addition and multiplication of complex numbers obey the laws of algebra expressed by the equations $\begin{gathered} {(x + yi)} + (x’ + y’i) = (x’ + y’i) + (x + yi),\\ \{(x + yi) + (x’ + y’i)\} + (x” + y”i) = (x + yi) + \{(x’ + y’i) + (x” + y”i)\},\\ (x + yi) (x’ + y’i) = (x’ + y’i) (x + yi),\\ (x + yi)\{(x’ + y’i) + (x” + y”i)\} = (x + yi)(x’ + y’i) + (x + yi)(x” + y”i),\\ \{(x + yi) + (x’ + y’i)\}(x” + y”i) = (x + yi)(x” + y”i) + (x’ + y’i)(x” + y”i),\\ (x + yi) \{(x’ + y’i) (x” + y”i)\} = \{(x + yi) (x’ + y’i)\} (x” + y”i),\end{gathered}$ 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 $$(x + yi) – (x’ + y’i)$$ as $(x + yi) + \{- (x’ + y’i)\} = x + yi + (-x’ – y’i) = (x – x’) + (y – y’)i;$ or again, as the number $$\xi + \eta i$$ such that $(x’ + y’i) + (\xi + \eta i) = x + yi,$ which leads to the same result. And $$(x + yi)/(x’ + y’i)$$ is defined as being the complex number $$\xi + \eta i$$ such that $(x’ + y’i) (\xi + \eta i) = x + yi,$ or $x’ \xi – y’ \eta + (x’ \eta + y’ \xi)i = x + yi,$ or $\begin{equation*}x’ \xi – y’ \eta = x,\quad x’ \eta + y’ \xi = y. \tag{4}\end{equation*}$

Solving these equations for $$\xi$$ and $$\eta$$, we obtain $\xi = \frac{xx’ + yy’}{x’^{2} + y’^{2}},\quad \eta = \frac{yx’ – xy’}{x’^{2} + y’^{2}}.$ This solution fails if $$x’$$ and $$y’$$ are both zero, i.e. if $$x’ + y’i = 0$$. 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 $$\overline{OB}$$ by $$\overline{OC}$$ is that of finding $$D$$ so that the triangles $$COB$$$$AOD$$ are similar, and this is evidently possible (and the solution unique) unless $$C$$ coincides with $$0$$, or $$\overline{OC} = 0$$. (2) The numbers $$x + yi$$, $$x – yi$$ are said to be conjugate. Verify that $(x + yi)(x – yi) = x^{2} + y^{2},$ so that the product of two conjugate numbers is real, and that $\frac{x + yi}{x’ + y’i} = \frac{(x + yi)(x’ – y’i)}{(x’ + y’i)(x’ – y’i)}\\ = \frac{xx’ + yy’ + (x’y – xy’)i}{x’^{2} + y’^{2}}.$

## 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 $x’\xi – y’\eta = 0,\quad x’\eta + y’\xi = 0.$ These equations give $$\xi = 0$$, $$\eta = 0$$, $\xi + \eta i = 0,$ unless $$x’ = 0$$ and $$y’ = 0$$, or $$x’ + y’i = 0$$. Thus $$x + yi$$ cannot vanish unless either $$x’ + y’i$$ or $$\xi + \eta i$$ vanishes.

## 41. The equation$$i^{2} = -1$$.

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 $i^{2} = ii = (0 + 1i) (0 + 1i) = (0 \cdot 0 – 1 \cdot 1) + (0 \cdot 1 + 1 \cdot 0)i = -1.$ Similarly $$(-i)^{2} = -1$$. Thus the complex numbers $$i$$ and $$-i$$ satisfy the equation $$x^{2} = -1$$.

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 $$ii = i^{2}$$ by $$-1$$ whenever it occurs. Thus, for example, \begin{aligned} (x + yi) (x’ + y’i) &= xx’ + xy’i + yx’i + yy’i^{2}\\ &= (xx’ – yy’) + (xy’+ yx’)i.\end{aligned}

## 42. The geometrical interpretation of multiplication by $$i$$.

Since $(x + yi)i = -y + xi,$ it follows that if $$x + yi$$ corresponds to $$\overline{OP}$$, and $$OQ$$ is drawn equal to $$OP$$ and so that $$POQ$$ is a positive right angle, then $$(x + yi)i$$ corresponds to $$\overline{OQ}$$. 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 $$xx’ + yx’i$$, $$(x + yi)y’i$$ as $$-yy’ + xy’i$$, and $$(x + yi) (x’ + y’i)$$ as the sum of these displacements, i.e. as $xx’ – yy’ + (xy’ + yx’)i.$