Let $x$ and $y$ be two complex numbers, so $x=\Re(x)+i\Im(x)$ and $y=\Re(y)+i\Im(y)$ such that $\Re(x)$, $\Im(x)$, $\Re(y)$ and $\Im(y)$ are real numbers.
Assume that $$x=e^{({-y^{-1}}+c)},$$ where $c \in \mathbb{C}$ is a constant.
How to analyse the behaviour of $y$ when $x \rightarrow 0$?
Thanks in Advance.
The following is very detailed so that you won't miss any point.
Let's write $y=a+ib$ where $a,b$ are real number. Let's write also $\rho=|y|=\sqrt{a^2+b^2}$ (just for easing the reading).
It's well known that $y^{-1}=\dfrac{a-ib}{\rho^2}=\rho^{-2}(a-ib)=a\rho^{-2}-ib\rho^{-2}$
Then $$x=e^{-y^{-1}+c}=e^ce^{-y^{-1}}=e^c\cdot e^{-a\rho^{-2}+ib\rho^{-2}}=e^c\cdot e^{ib\rho^{-2}}\cdot e^{-a\rho^{-2}}$$
Thus $$|x|=\lambda\cdot e^{-a\rho^{-2}}$$ where $\lambda=|e^c|>0$ (remember also, $e^{-a\rho^{-2}}$ is a positive number and $|e^{ib\rho^{-2}}|=1$).
Choose $\varepsilon>0$ then $$|x|<\varepsilon\Leftrightarrow e^{-a\rho^{-2}}<\varepsilon/\lambda$$
When $\varepsilon$ is small enough (that is, $\varepsilon/\lambda<1$) then $\ln{(\varepsilon/\lambda)}<0$ and we finally have $$\dfrac{-a}{\rho^2}<\ln{(\varepsilon/\lambda)}\Leftrightarrow\rho^2<\dfrac{-a}{\ln{(\varepsilon/\lambda)}}$$
First, if $a<0$ as the logarithm is negative: you've got a negative square and this cannot be. If $a=0$ then $\rho=0$ but $y$ cannot be null. This forces $a>0$.
Second, let's $\varepsilon\to 0$ then the right member tends toward $0$ and this means $\rho\to 0$.
Finally : when $x\to 0$ then $|x|\to 0$ and $y\to 0$ with a real part strictly positive.
In other words, if $|x|$ is very small, then $y$ is in a right side of a small disc centered on $0$.