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What is the value of $i^{i^{i^\ldots}}$?

My effort is the following: If $z, \alpha \in \mathbb{C}$ with $z \neq 0$ then we can write $z^{\alpha}=e^{\alpha \log z} = e^{\alpha [ \log |z|+i \text{ Arg z} + 2 \pi i m]}$ where $ m=0, \pm 1, \pm 2, \ldots$ and Arg z is the principle argument of the complex number $z$. But in above expression we don't have the finite no of exponents, so I am not sure how should I use this formula to the solve above problem. Thanks and Regards!

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closed as unclear what you're asking by Did, suomynonA, JonMark Perry, user91500, Parcly Taxel Oct 20 '16 at 11:38

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    $\begingroup$ One doesn't write $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+\infty$ for an infinite geometric series, so by the same token on probably shouldn't write $\infty$ in the exponent here. $\endgroup$ – arctic tern Oct 19 '16 at 17:41
  • $\begingroup$ By $\infty $ there I mean I am taking exponents infinite times not infinity in the exponents. $\endgroup$ – Shubham Namdeo Oct 19 '16 at 17:44
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    $\begingroup$ If the expression makes sense, it should be a number $z$ with $i^z=z$, so $z=e^{\frac \pi2iz}$ $\endgroup$ – Hagen von Eitzen Oct 19 '16 at 17:48
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    $\begingroup$ Each time you do an exponent you have to pick a branch cut. Given all the possible branch choices I'd think the answer is the entire unit circle (or at least dense in the unit circle). $\endgroup$ – Paul Oct 19 '16 at 17:48
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Let us denote your power tower by $a$. Then assuming it converges \begin{gather*} i^a=a \\ \frac 1i=\left( \frac 1a \right)^{1/a} \\[6pt] -\ln i =\left( \frac 1 a\right)\ln\left( \frac 1a\right) \\[6pt] -\ln i=e^{\ln\left( \frac 1a\right)}\ln\left( \frac 1a\right) \\[6pt] \ln \left( \frac 1a\right) = \mathop{W}(-\ln i) \\[6pt] \frac 1a=-\frac{\ln i}{\mathop W (-\ln i)} \\[6pt] \end{gather*} \begin{align*} a &= -\frac{\mathop{W}(-\ln i)}{\ln i} \\[6pt] &= \frac{2i}{\pi}\mathop{W}\left( -\frac 12\pi i\right)\\[6pt] &\approx 0.44+0.36i \end{align*}

where $\mathop{W}$ is the Lambert W function.

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    $\begingroup$ "assuming it converges" Does it? $\endgroup$ – Did Aug 12 '18 at 3:52

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