Complex exponential arrangement > 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:$$z=e^{-A}e^{-B}=e^{-\frac{A}{2}}e^{-\left(\frac{A}{2}+B\right)}=e^{-(A+B)}$$