236. The exponential limit

The exponential limit. Let $z$ be any complex number, and $h$ a real number small enough to ensure that $|hz|<1$. Then $$\log (1+hz)=hz–\frac{1}{2}(hz{)}^2+\frac{1}{3}(hz{)}^3–\dots ,$$ and so $$\frac{\log (1+hz)}{h}=z+\varphi (h,z),$$ where $$\begin{array}{c}\varphi (h,z)=-\frac{1}{2}h{z}^2+\frac{1}{3}{h}^2{z}^3–\frac{1}{4}{h}^3{z}^4+\dots ,\\ |\varphi (h,z)|<|h{z}^2|(1+|hz|+|h^2{z}^2|+\dots )=\frac{|h{z}^2|}{1–|hz|},\end{array}$$ so that $\varphi (h,z)\to 0$ as $h\to 0$. It follows that $$\begin{array}{}\text{(1)}& \underset{h\to 0}{lim}\frac{\log (1+hz)}{h}=z.\end{array}$$

If in particular we suppose $h=1/n$, where $n$ is a positive integer, we obtain $$\underset{n\to \mathrm{\infty }}{lim}n\log (1+\frac{z}{n})=z,$$ and so $$\begin{array}{}\text{(2)}& \underset{n\to \mathrm{\infty }}{lim}{(1+\frac{z}{n})}^n=\underset{n\to \mathrm{\infty }}{lim}\exp \{n\log (1+\frac{z}{n})\}=\exp z.\end{array}$$ This is a generalisation of the result proved in § 208 for real values of $z$.

From (1) we can deduce some other results which we shall require in the next section. If $t$ and $h$ are real, and $h$ is sufficiently small, we have $$\frac{\log (1+tz+hz)–\log (1+tz)}{h}=\frac{1}{h}\log (1+\frac{hz}{1+tz})$$ which tends to the limit $z/(1+tz)$ as $h\to 0$. Hence $$\begin{array}{}\text{(3)}& \frac{d}{dt}\{\log (1+tz)\}=\frac{z}{1+tz}.\end{array}$$

We shall also require a formula for the differentiation of $(1+tz{)}^m$, where $m$ is any number real or complex, with respect to $t$. We observe first that, if $\varphi (t)=\psi (t)+i\chi (t)$ is a complex function of $t$, whose real and imaginary parts $\varphi (t)$ and $\chi (t)$ possess derivatives, then $$\begin{array}{rl}\frac{d}{dt}(\exp \varphi )& =\frac{d}{dt}\{(\cos \chi +i\sin \chi )\exp \psi \}\\ & =\{(\cos \chi +i\sin \chi ){\psi }^{\prime }+(-\sin \chi +i\cos \chi ){\chi }^{\prime }\}\exp \psi \\ & =({\psi }^{\prime }+i{\chi }^{\prime })(\cos \chi +i\sin \chi )\exp \psi \\ & =({\psi }^{\prime }+i{\chi }^{\prime })\exp (\psi +i\chi )={\varphi }^{\prime }\exp \varphi ,\end{array}$$ so that the rule for differentiating $\exp \varphi$ is the same as when $\varphi$ is real. This being so we have $$\begin{array}{rl}\frac{d}{dt}(1+tz{)}^m& =\frac{d}{dt}\exp \{m\log (1+tz)\}\\ & =\frac{mz}{1+tz}\exp \{m\log (1+tz)\}\\ & =mz(1+tz{)}^{m-1}.\begin{array}{}\text{(4)}& \end{array}\end{array}$$ Here both $(1+tz{)}^m$ and $(1+tz{)}^{m-1}$ have their principal values.