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 \(z\).

$\leftarrow$ 231. The connection between the logarithmic and inverse trigonometrical functions Main Page 233. The series for \(\cos z\) and \(\sin z\) $\rightarrow$