Let's define :
$e^{-c k}=\frac{k-1}{k+1}$
we can write :
\begin{align}
-ce^{-c k}=-\frac{k-1}{k+1}c
& \iff -c(k+1)e^{-c(k+1)}=-(\sqrt k - 1)(\sqrt k + 1)ce^{-c} \\
\\
& \iff -c(k+1)e^{-c(k+1)}= -(\sqrt k - 1)(\sqrt k + 1)ce^{-c} \\
\\
& \iff \ln(-c(k+1)) - c(k+1)=- (\sqrt k - 1)(\sqrt k + 1)ce^c \\
\\
& \iff W(-(\sqrt k - 1)(\sqrt k + 1)ce^c) - c(k+1)= \ln(- (\sqrt k - 1)(\sqrt k + 1)ce^c) \\
\\
& \iff k = -\frac{c + \ln(- (\sqrt k - 1)(\sqrt k + 1)c) - W(-(\sqrt k - 1)(\sqrt k + 1)ce^c) + c }{c} \\
\\
& \iff k = -(2 + \ln(- (\sqrt k - 1)(\sqrt k + 1)c) - W(-(\sqrt k - 1)(\sqrt k + 1)ce^c)) \\
\\
& \iff k = -2 - \ln(- (\sqrt k - 1)(\sqrt k + 1)c) + W(-(\sqrt k - 1)(\sqrt k + 1)ce^c))
\\
\end{align}
This express the multiple solutions using the $W(-(\sqrt k - 1)(\sqrt k + 1)ce^c)$ Lambert function with $c=2i \ell$.