Finding where $\frac{\operatorname{Log} z}{e^z - e}$ is differentiable and calculating it In order to take the derivative of
$$\frac{\operatorname{Log} z}{e^z - e}$$
I did it symbolically using the product rule to arrive at
$$f'(z) = \frac{1}{z}(e^z-e)^{-1}+(\operatorname{Log} z)( -(e^z-e)^{-2}e^z)$$
I know that $\operatorname{Log} z$ is differentiable everywhere except at the negative real line. Also, $\frac{1}{e^z-e}$ can't be differentiable at $z=1$, but are there more points? I don't think so, because $e^z-e=0\iff z=1$, therefore the set where the function is differentiable is the intersection of the 2 sets I described?

Noting $\lim_{z\to1}\frac{\log z}{e^z-e}=\frac1e$, define $f(1)=\frac1e$. Now
\begin{eqnarray}
 \lim_{\Delta z\to0}\frac{f(1+\Delta z)-f(1)}{\Delta z}&=&\lim_{\Delta z\to0}\frac{\log(1+\Delta z)-e^{\Delta z}+1}{e\Delta z(e^{\Delta z}-1)}\\
&=&\lim_{\Delta z\to0}\frac{\log(1+\Delta z)-e^{\Delta z}+1}{\Delta z^2}\frac{\Delta z}{e^{\Delta z}-1}\\
&=&-1
\end{eqnarray}
and one has that $f'(z)$ is analytic (diferentiable) at $z=1$.
