The exponential limit. Let $$z$$ be any complex number, and $$h$$ a real number small enough to ensure that $$|hz| < 1$$. Then $\log(1 + hz) = hz – \tfrac{1}{2}(hz)^{2} + \tfrac{1}{3}(hz)^{3} – \dots,$ and so $\frac{\log(1 + hz)}{h} = z + \phi(h, z),$ where $\begin{gathered} \phi(h, z) = -\tfrac{1}{2}hz^{2} + \tfrac{1}{3}h^{2}z^{3} – \tfrac{1}{4}h^{3}z^{4} + \dots,\\ |\phi(h, z)| < |hz^{2}| (1 + |hz| + |h^{2}z^{2}| + \dots) = \frac{|hz^{2}|}{1 – |hz|},\end{gathered}$ so that $$\phi(h, z) \to 0$$ as $$h \to 0$$. It follows that $\begin{equation*} \lim_{h\to 0} \frac{\log(1 + hz)}{h} = z. \tag{1} \end{equation*}$

If in particular we suppose $$h = 1/n$$, where $$n$$ is a positive integer, we obtain $\lim_{n\to \infty} n\log \left(1 + \frac{z}{n}\right) = z,$ and so $\begin{equation*} \lim_{n\to \infty} \left(1 + \frac{z}{n}\right)^{n} = \lim_{n\to \infty} \exp\left\{n\log\left(1 + \frac{z}{n}\right)\right\} = \exp z. \tag{2} \end{equation*}$ 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 $\frac{\log(1 + tz + hz) – \log(1 + tz)}{h} = \frac{1}{h}\log\left(1 + \frac{hz}{1 + tz}\right)$ which tends to the limit $$z/(1 + tz)$$ as $$h \to 0$$. Hence $\begin{equation*} \frac{d}{dt} \{\log(1 + tz)\} = \frac{z}{1 + tz}. \tag{3} \end{equation*}$

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 $$\phi(t) = \psi(t) + i\chi(t)$$ is a complex function of $$t$$, whose real and imaginary parts $$\phi(t)$$ and $$\chi(t)$$ possess derivatives, then \begin{aligned} \frac{d}{dt}(\exp\phi) &= \frac{d}{dt}\{(\cos\chi + i\sin\chi) \exp\psi\}\\ &= \{(\cos\chi + i\sin\chi) \psi’ + (-\sin\chi + i\cos\chi)\chi’\} \exp\psi\\ &= (\psi’ + i\chi’)(\cos\chi + i\sin\chi) \exp\psi\\ &= (\psi’ + i\chi’) \exp(\psi + i\chi) = \phi’ \exp\phi,\end{aligned} so that the rule for differentiating $$\exp\phi$$ is the same as when $$\phi$$ is real. This being so we have \begin{aligned} \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{equation*} \tag{4} \end{equation*}\end{aligned} Here both $$(1 + tz)^{m}$$ and $$(1 + tz)^{m-1}$$ have their principal values.