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This question raised from this one. Solving $i^i=x$ we get $x=e^{(i \frac{\pi}{2} + i2k\pi)i}=e^{-\frac{\pi}{2} - 2k\pi}$ $(k \in \mathbb{Z})$, than what about the values of $x = i^{i^{i^{i^{.^{.^{.{^ \infty}}}}}}}$? Is there an argument to show that $i^{i^{i^{i^{.^{.^{.{^ \infty}}}}}}} \neq \mathbb{C} \setminus\{0\}$?

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    The tower is countable, so it can't equal $\mathbb{C}$.2011-09-30

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I'm not completely sure how to interpret $x = i^{i^{i^{i^{.^{.^{.{^ \infty}}}}}}}$, but I suspect it should be equivalent to $x = i^{x}$, which is equivalent to $x=e^{(i \frac{\pi}{2} + i2k\pi)x}$. Since the right-hand side of this is of the form $e^z$, it never takes the value $0$, so $0 \not\in i^{i^{i^{i^{.^{.^{.{^ \infty}}}}}}}$ and $i^{i^{i^{i^{.^{.^{.{^ \infty}}}}}}} \neq \mathbb{C}$.

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    your interpretation is exactly what I mean, and I forget that trivial detail. I edit the question, thanks2011-03-29
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Suppose in $y = i^x = \exp(x * i \pi /2) $ the variable $x$ were purely real, then $y$ could only be purely real, if $x=2*k, k \in \mathbb{N} $ such that $ y=\exp( k i \pi)$ and thus $y \in (-1,1) $.

But if $y \in (-1,1) $ it cannot be that $x=y$ since $x=2 k$ so in $x=i^x$ there is no purely real solution fox $x$ and it is needed that $x \in \mathbb{C} $