The equation $2 \cosh(3.1786803659501505 z) = z$? Let $a$ be a positive real number and $z$ a complex number.
I was wondering about the equation $2 \cosh(a z) = z$ where we solve for $z$.
Clearly if $z$ is a solution than so is its conjugate.
It appears that if $0<a<c$ (for some real $c$) then the equation $2\cosh(a z) = z$ has 2 solutions.
And also if $a>c$ then we have more than 2 solutions to $2 \cosh(a z) = z$.
(if $a$ is slightly larger than $c$ the equation has 4 solutions)
So I wonder about $c$ and I have $4$ questions:
$1)$ $c$ is about  $3.1786803659501505$ but can $c$ be expressed by an integral or a limit ?
$2)$ How many solutions does $2 \cosh(c z) = z$ have ? Is it $2$ or $4$ ?
$3)$ I also wonder if I can call $2 \cosh((c+\epsilon) z) = z$ a bifurcation near $c$ ? I think not because the 4 zero's are always quite far away from eachother whereas for the bifurcations of e.g. the logistic map we get zero's that are arbitrary close. 
$4)$ Similar to question $1)$ : Consider the 4 zero's of $2 \cosh((c+\epsilon) z) = z$.
Can those zero's be expressed by an integral or a limit ? How about the sum of some of the zero's ?
Btw is this considered chaos theory , algebra or complex analysis ? Or all of that ? Im a bit insecure about the title and tags I should use. Is it ok to use many decimals in a title ?
 A: Too long for a comment.
Note that $f(z) = 2\cosh(az)-z$ is an entire function with an essential singularity at $\infty$ (for $a\neq 0$). Because of Picard's great theorem we would expect that there are infinitely many zeros of $f$ (Picard implies that the equation $f(z) = b$ has infinitely many solutions for every complex $b$ with at most two exceptions).
Of course, it's conceivable that $0$ is one of the exceptional values (I'll think about that some more), but even for $a$ close to $0$, I can find many solutions to your equation. For example, if $a = 0.1$, here are just a few solutions:
\begin{align}
z &= 2.041835988 \\
z &= 35.7606533293 \\
z &= 44.49177815\pm73.07181538i \\
z &= -40.28136974\pm39.12782185i \\
z &= 68.66860456\pm957.4697985i \\
z &= -68.99027078\pm988.9051729i
\end{align}
A: Generally speaking, zeroes of a complex function don't appear in isolation as a parameter is continuously varied.  They can appear and disappear in sets (generically, pairs) via bifurcations, and can move around, but can't just come and go as they please.
In your case, the appearance of "additional" roots has more to do with the basins of attraction of your root finder than it does with your function.  If you ask WolframAlpha to solve $2\cosh(az)=z$ with $a=3.17$, it finds two roots (at $z\approx0.08\pm 0.48 i$); if you ask it with $a=3.18$, it finds the same two roots, moved slightly, plus two more roots (at $z\approx -0.21 \pm 1.45 i$).  But in fact these "new" roots were already present at $a=3.17$, and have simply moved into the range where the root finder detects them.  To find them for smaller $a$, you can give WolframAlpha a hint, by first transforming $z\mapsto (-0.21+1.45i) + z;$ this moves one of the "new" roots near the origin, where it may be easier to find.  And indeed, solving $2\cosh(a(z-0.21+1.45i))=z-0.21+1.45i$ with $a=3.17$ does yield a root at $z\approx -0.004 + 0.009 i$.
