In a program of mine, I am solving numerically in the real domain (with limited precision) the equation $$\cosh(x)=a\,x$$ which does not present any difficulty. The problem is that $a$ is a function of many other variables and I forgot to check, before solving, that $a >1$.
According to Murphy's law, in the middle of zillions of calculations, there was a crash. Recomputing $a$ with illimited precision, it was just a bit smaller than $1$.
So, out of pure curiosity, I started looking at the complex solutions of $$\cosh(x)=(1-\epsilon)\,x \qquad \text{where} \qquad 0 <\epsilon <1$$ the numerical solutions of which being easily computed using Newton method using $x_0=1\pm i$. Making $\epsilon=10^{-k}$ I can obtain the solution with very high accuracy up to $k \sim 323\times 10^6$ (have a look here for an example). $x_*$ noting the solution $$\Re(x_*)=0.9760545183390074945207946979151719003359441073084578875553199$$ $$\Im(x_*)=\pm0.8709514657614091170725524453488181084144105977590102585922735$$ and none of these numbers is recognized by inverse symbolic calculators.
So, my questions are
what are these numbers ?
is it possible to have the asymptotics of $x_*$ when $\epsilon \to 0^+$ ?
Let $x=\zeta+\delta$ where $\zeta=\cosh(\zeta)$ and expand to get
\begin{align}\cosh(\zeta+\delta)&=(1-\epsilon)(\zeta+\delta)\\&=\zeta+\delta-\zeta\epsilon-\epsilon\delta\end{align}
Moving all of the $\delta$ to one side leaves us with
$$\cosh(\zeta+\delta)-\zeta-(1-\epsilon)\delta=-\zeta\epsilon$$
from which the LHS may be expanded and $\delta$ solved for via series reversion.
The first few terms can be found to be
$$\delta=\frac\zeta{1-\zeta'}\epsilon+\frac{\zeta^2+2-2\zeta'}{(1-\zeta')^3}\zeta\epsilon^2+\mathcal O(\epsilon^3)$$
where $\zeta'=\sinh(\zeta)=\sqrt{1+\zeta^2}$.