We saw in § 212 It will be convenient now to use $$z$$ instead of $$\zeta$$ as the argument of the exponential function. that when $$z$$ is real $\begin{equation*} \exp z = 1 + z +\frac{z^{2}}{2!} + \dots. \tag{1} \end{equation*}$ Moreover we saw in § 191 that the series on the right-hand side remains convergent (indeed absolutely convergent) when $$z$$ is complex. It is naturally suggested that the equation (1) also remains true, and we shall now prove that this is the case.

Let the sum of the series (1) be denoted by $$F(z)$$. The series being absolutely convergent, it follows by direct multiplication (as in Ex. LXXXI. 7) that $$F(z)$$ satisfies the functional equation $\begin{equation*} F(z) F(h) = F(z + h). \tag{2} \end{equation*}$ Now let $$z = iy$$, where $$y$$ is real, and $$F(z) = f(y)$$. Then $f(y) f(k) = f(y + k);$ and so $\frac{f(y + k) – f(y)}{k} = f(y) \left\{\frac{f(k) – 1}{k}\right\}.$

But $\frac{f(k) – 1}{k} = i\left\{1 + \frac{ik}{2!} + \frac{(ik)^{2}}{3!} + \dots\right\};$ and so, if $$|k| < 1$$, $\left|\frac{f(k) – 1}{k} – i\right| < \left(\frac{1}{2!} + \frac{1}{3!} + \dots\right)|k| < (e – 2)|k|.$ Hence $$\{f(k) – 1\}/k\to i$$ as $$k \to 0$$, and so $\begin{equation*}f'(y) = \lim_{k \to 0} \frac{f(y + k) – f(y)}{k} = if(y). \tag{3}\end{equation*}$

Now $f(y) = F(iy) = 1 + (iy) + \frac{(iy)^{2}}{{2!}} + \dots = \phi(y) + i\psi(y),$ where $$\phi(y)$$ is an even and $$\psi(y)$$ an odd function of $$y$$, and so \begin{aligned} |f(y)| &= \sqrt{\{\phi(y)\}^{2} + \{\psi(y)\}^{2}}\\ &= \sqrt{\{\phi(y) + i\psi(y)\}\{\phi(y) – i\psi(y)\}}\\ &= \sqrt{F(iy) F(-iy)} = \sqrt{F(0)} = 1;\end{aligned} and therefore $f(y) = \cos Y + i \sin Y,$ where $$Y$$ is a function of $$y$$ such that $$-\pi < Y \leq \pi$$. Since $$f(y)$$ has a differential coefficient, its real and imaginary parts $$\cos Y$$ and $$\sin Y$$ have differential coefficients, and are a fortiori continuous functions of $$y$$. Hence $$Y$$ is a continuous function of $$y$$. Suppose that $$Y$$ changes to $$Y + K$$ when $$y$$ changes to $$y + k$$. Then $$K$$ tends to zero with $$k$$, and $\frac{K}{k} = \biggl\{\frac{\cos(Y + K) – \cos Y}{k}\biggr\} \bigg/ \biggl\{\frac{\cos(Y + K) – \cos Y}{K}\biggr\}.$ Of the two quotients on the right-hand side the first tends to a limit when $$k \to 0$$, since $$\cos Y$$ has a differential coefficient with respect to $$y$$, and the second tends to the limit $$-\sin Y$$. Hence $$K/k$$ tends to a limit, so that $$Y$$ has a differential coefficient with respect to $$y$$.

Further $f'(y) = (-\sin Y + i\cos Y) \frac{dY}{dy}.$ But we have seen already that $f'(y) = if(y) = -\sin Y + i\cos Y.$ Hence $\frac{dY}{dy} = 1,\quad Y = y + C,$ where $$C$$ is a constant, and $f(y) = \cos(y + C) + i\sin(y + C).$

But $$f(0) = 1$$ when $$y = 0$$, so that $$C$$ is a multiple of $$2\pi$$, and $$f(y) = \cos y + i\sin y$$. Thus $$F(iy) = \cos y + i\sin y$$ for all real values of $$y$$. And, if $$x$$ also is real, we have $F(x + iy) = F(x) F(iy) = \exp x(\cos y + i\sin y) = \exp(x + iy),$ or $\exp z = 1 + z + \frac{z^{2}}{2!} + \dots,$ for all values of $$z$$.