# Are complex numbers the set of values of a tower of imaginary units?

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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How is the infinite tower defined? The last inequality makes only sense if $x$ is a set. Your notation may allow that barely, but I is this your intention? –  shuhalo Mar 27 '11 at 15:11
$i^i$ is just $e^{-\frac{\pi}{2}}$ which is a proven transcendental. How do you include the $k$? –  Chulumba Mar 27 '11 at 15:14
@Martin As in the cited question, the tower is defined as the solution of the equation $i^x - x = 0$. –  Emanuele Natale Mar 28 '11 at 23:56
@Chulumba $e^{-\frac{\pi}{2}}$ is just one possible value of $i^i$, as @Gottfried Helms answers in your comment in the original cited question –  Emanuele Natale Mar 29 '11 at 0:05
The tower is countable, so it can't equal $\mathbb{C}$. –  Dan Brumleve Sep 30 '11 at 9:48

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 $i^{i^{i^{i^{.^{.^{.{^ \infty}}}}}}} \neq \mathbb{C}$.
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 $x \in \mathbb{C}$