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 \[\log(1 + hz) = hz – \tfrac{1}{2}(hz)^{2} + \tfrac{1}{3}(hz)^{3} – \dots,\] 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.