232. The power series for $\exp z$

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 $$\begin{array}{}\text{(1)}& \exp z=1+z+\frac{z^2}{2!}+\dots .\end{array}$$ 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 $$\begin{array}{}\text{(2)}& F(z)F(h)=F(z+h).\end{array}$$ 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 $$\frac{f(k)–1}{k}=i\{1+\frac{ik}{2!}+\frac{(ik{)}^2}{3!}+\dots \};$$ and so, if $|k|<1$, $$\left|\frac{f(k)–1}{k}–i\right|<(\frac{1}{2!}+\frac{1}{3!}+\dots )|k|<(e–2)|k|.$$ Hence $\{f(k)–1\}/k\to i$ as $k\to 0$, and so $$\begin{array}{}\text{(3)}& f^{\prime }(y)=\underset{k\to 0}{lim}\frac{f(y+k)–f(y)}{k}=if(y).\end{array}$$

Now $$f(y)=F(iy)=1+(iy)+\frac{(iy{)}^2}{2!}+\cdots =\varphi (y)+i\psi (y),$$ where $\varphi (y)$ is an even and $\psi (y)$ an odd function of $y$, and so $$\begin{array}{rl}|f(y)|& =\sqrt{\{\varphi (y){\}}^2+\{\psi (y){\}}^2}\\ & =\sqrt{\{\varphi (y)+i\psi (y)\}\{\varphi (y)–i\psi (y)\}}\\ & =\sqrt{F(iy)F(-iy)}=\sqrt{F(0)}=1;\end{array}$$ and therefore $$f(y)=\cos Y+i\sin Y,$$ where $Y$ is a function of $y$ such that $-\pi <Y\le \pi$. 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 $$\frac{K}{k}=\{\frac{\cos (Y+K)–\cos Y}{k}\}/\{\frac{\cos (Y+K)–\cos Y}{K}\}.$$ 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 $$f^{\prime }(y)=(-\sin Y+i\cos Y)\frac{dY}{dy}.$$ But we have seen already that $$f^{\prime }(y)=if(y)=-\sin Y+i\cos Y.$$ Hence $$\frac{dY}{dy}=1,Y=y+C,$$ 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 $z$.