Let
$$z = e^{-\frac{a}{b + ic}}$$
(where $i$ is the imaginary unit and $a,b,c \in \mathbb{R}$) be a complex number in exponential form.
Write $z$ in the following form:
$$z = e^{- A}e^{-B}$$
with $A \in \mathbb{R} \setminus \{ 0 \}$, $B \in \mathbb{C}$.
I tried
$$\frac{a}{b + ic} \frac{b - ic}{b - ic} = \frac{a}{b^2 + c^2} (b - ic)$$
So
$$A = \frac{ab}{b^2 + c^2}, \ B = -\frac{iac}{b^2 + c^2}$$
Is this the only way to split the above exponent $a / (b + ic)$, or are there other ways to get a real $A$ and a purely imaginary (or complex) $B$?
Purely imaginary $B$
The real $A$ is unique. This is because when $z=\rho e^{i\theta}\in\mathbb{C}\backslash\{0\}$ where $\rho\in\mathbb{R}^+$ and $\theta\in\mathbb{R}$, $\rho$ is unique as $\rho=|z|$. To have the desired form, the only possibility for $A$ is such that $e^{-A}=\rho$, and that is $A=-\ln{\rho}=\ln{\frac{1}{\rho}}$.
$B$, on the other hand, is not unique. Recall that $$\forall\theta\in\mathbb{R},\,e^{i\theta}=\cos\theta+i\sin\theta$$ and since $\cos$ and $\sin$ are $2\pi$-periodic, we have $$\forall\theta\in\mathbb{R},\forall n\in\mathbb{Z},e^{i\theta}=e^{i(\theta+2\pi n)}$$
Thus, if $B$ is such that $z=e^{-A}e^{-B}$ then for any $n\in\mathbb{Z}$, $B_n=B+2\pi n$ satisfies $z=e^{-A}e^{-B_n}$ and $B_n$ is purely imaginary.
Complex $B$
When you don't ask for $B$ to be purely imaginary, even $A$ isn't unique:$$z=e^{-A}e^{-B}=e^{-\frac{A}{2}}e^{-\left(\frac{A}{2}+B\right)}=e^{-(A+B)}$$