\begin{equation}
f(z) = e^z - 3e^z - 4 = 0
\end{equation}
Let $w=e^z$, then
\begin{equation}
w^2 - 3w - 4 = 0 \\
(w-4)(w+1) = 0 \Rightarrow w = 4 \space or \space w = 1 \\
e^z = 4 \space or \space e^z = -1 \\
z = \log4 \space or \space z = \log(-1) \\
\end{equation}
When $e^z = 4$, the order is 1
\begin{equation}
f'(z) =2e^{2z} - 3e^{z} = 2 \times 4^2 - 3 \times 4 \neq 0
\end{equation}
When $e^z = -1$, the order is 1
\begin{equation}
f'(z) =2e^{2z} - 3e^{z} = 2 \times (-1)^2 - 3 \times (-1) \neq 0
\end{equation}
