For what values does this method converge on the Lambert W function? Someone from another question had noted that the following statement
$$W(x)=\ln\left(\vcenter{\frac x{\ln\left(\vcenter{\frac{x}{\ln\left(\vcenter{\frac x{\ln(\dots)}}\right)}}\right)}}\right)$$
Can be found from the identity $W(x)=\ln(x)-\ln(W(x))$ where $W(z)$ is the Lambert W function and point convergence.
I was wondering for what values of $x$ does this converge, including $x\in\mathbb{C}$.  My assumption is that it will only work for $W(x)\in\mathbb{R}$, but I'm unsure.
 A: If $p$ is a fixed point of the differentiable function $f$, the iteration scheme $y_{n+1} = f(y_n)$ with initial point close (but not equal) to $p$ will converge to $p$ if $|f'(p)| < 1$, but will not converge to $p$ if $|f'(p)| > 1$.  In this case your function is $f(y) = \ln(x/y)$ with fixed point $W(x)$ (for the appropriate branch of logarithm), and
$f'(p) = -1/p$.  Thus it will converge if $|p| > 1$, and diverge if 
$|p| < 1$.  The "0" branch of Lamber W satisfies $|W(z)| > 1$ outside a heart-shaped region of the complex plane:

The boundary of this heart is the curve $z = \exp(i\theta + \exp(i\theta))$ for $0 \le \theta \le 2 \pi$.  Other branches have absolute value $> 1$ everywhere.
A: This is an elaboration and illustration of Robert Israel's answer.
Let
$$
f_x(z)=\log\left(\frac xz\right)
$$
The question is asking when the repeated composition of $f_x$ converges.
Since
$$
\lim_{z\to\mathrm{W}(x)}\left|\frac{f_x(f_x(z))-f_x(z)}{f_x(z)-z}\right|=\left|\,f_x'(W(x))\,\right|=\frac1{\left|W(x)\right|}
$$
there is a region around $z=\mathrm{W}(x)$ where composition of $f_x$ converges when $\left|W(x)\right|\gt1$.
$$
\left\{ze^z:\left|z\right|\gt1\right\}
$$
Rather than consider the region where the compositions converge, which is complicated by branch overlaps, it is easier to consider the region where the compositions don't converge. That is,
$$
\left\{ze^z:\left|z\right|\le1\right\}
$$
To trace out the boundary where $\left|z\right|=r$, we have
$$
\begin{align}
ze^z
&=re^{r\cos(\theta)}e^{i(\theta+r\sin(\theta))}\\
&=re^{r\cos(\theta)}\cos(\theta+r\sin(\theta))+ire^{r\cos(\theta)}\sin(\theta+r\sin(\theta))
\end{align}
$$

