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By definition $\ln = \log_e$ on complex numbers is given by $$ \ln(re^{i\theta}) = \ln(r) + i\theta $$ $(-\pi < \theta\leq \pi, r >0)$. Then $\ln(-1) = \pi i$. And $\ln(\pi i) = \ln(\pi) + i\pi/2$.

If $\ln^{\circ n}(z) = \ln\circ\ln\circ \dots \circ\ln$ ($n$ times), is it possible to find what the exact value of $$ \lim_{n \to \infty} \ln^{\circ n}(-1)\quad $$ is?

From just using a calculator it seems like this actually converges. And from starting with for example $-2$ it looks like it converges to the same number.

If it is not possible to find an exact value, how might one prove that this actually converges?

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Do you mean $0 < \pi \le 2 \pi$? – Robert Israel Dec 10 '12 at 19:34
@RobertIsrael: I made a mistake on that. – Thomas Dec 10 '12 at 19:35
(One of) the complex fixpoint(s) of the exp-function... $ \approx 0.318131505205 + 1.33723570143*I$ – Gottfried Helms Dec 10 '12 at 19:37
@GottfriedHelms: Yes, that is what I got from just using a calculator. Is it possible to find the exact value? – Thomas Dec 10 '12 at 19:38
In terms of the Lambert W function, the fixed points of the natural logarithm are $z_k = \exp(-W_{k}(-1))$. – mjqxxxx Dec 10 '12 at 19:38
up vote 2 down vote accepted

The limit, if it exists, is a fixed point of $\ln$, and therefore of $\exp$. Those fixed points are the branches of $-W(-1)$, where $W$ is the Lambert W function. The values with imaginary part in $(-\pi, \pi]$ are the $0$ branch at $.3181315052-1.337235701 i$ and the $-1$ branch which is the complex conjugate of that. The derivative of $\ln(z)$ is $1/z$. Since $|1/z| < 1$ for each of these fixed points, they are both attracting, i.e. if some $\ln^{\circ n}(-1)$ is close enough to one of them, it will approach it in the limit. In fact, if $p$ is one of these fixed points, $$|p - \ln(z) | = \left|\int_\Gamma \frac{d\zeta}{\zeta}\right| \le \frac{|z - p|}{|p| - |p-z|}$$ where $\Gamma$ is the straight line from $z$ to $p$. So the basin of attraction includes the open disk of radius $r$ around $p$ where $r/(|p|-r) = 1$, i.e. $r = |p|/2 = .6872785050$. In this case $\ln^{\circ 3}(-1) = .6645719224+.9410294873 i$ has distance approximately $.5263082050$ from $.3181315052+1.337235701 i$, and therefore future iterates will converge to that fixed point.

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