# How to compute the infinite tower of the complex number $i$, that is$^{\infty}i$

Let $x = i^{i^{i^{i^{.^{.^{.{^ \infty}}}}}}}$. This is the solution of the equation $i^x - x = 0$ . I used Euler's identity to find a solution. But I haven't yet found the real and imaginary parts of the solution. Are there more solutions? If so why did I miss them?

-
Shouldn't this be $i^x-x=0$ ? – Gottfried Helms Jan 15 '11 at 14:06
@Loy That is, $i$ raised to $i$ raised to $i$...infinitely many times. This is similar to say the problem of finding the exact value of $sqrt{2-sqrt{2-sqrt{2-sqrt{2-\ldots}}}}$ – Chulumba Jan 15 '11 at 14:09
@Helms, yes, thanks. – Chulumba Jan 15 '11 at 14:10
since ii=exp(−π/2), you can have at most 4 options in which one could be a solution. exp(−π/2), exp(−iπ/2) exp(π/2), and exp(iπ/2) depending upon how many terms you use. But it is going up to inf. I can not figure out limit of this series. But none of these options satisfy your equation. You got a typo in your question at best! – Dilawar Jan 15 '11 at 14:13
Ok, now its seem to be fine. – Dilawar Jan 15 '11 at 14:14

In case you really mean $I^x - x = 0$ you can find this by the iteration

 x = <some initial value>
repeat
x = I^x   // where I is the imaginary unit
until convergence


You'll get approximately $x = 0.438282936727 + 0.360592471871*I$ (using Pari/GP, for instance)

However, you can also find the value using the lambert-w-function.
[update] using the lambert-w:

let $\lambda=\ln(i)$
then
$i^x = x$
$1 = x* \exp( -x \lambda)$
$-\lambda = -x \lambda * \exp( -x \lambda)$
$W(-\lambda) = - x \lambda$
so
$x = \frac{W(-\ln(i))}{-\ln(i)}$

(Whether this is more "exact" is rather a question "exact in terms-of-what?")

[end update]

-
Thanks. I am interested in the exact solution. However, I have found it interesting to know that iteration can also be used for complex numbers. – Chulumba Jan 15 '11 at 14:24

Solving $i^i=x$ we get $e^{(i \frac{\pi}{2} + i2k\pi)i}=e^{-\frac{\pi}{2} - 2k\pi}$, than the values of that tower are dence in $\mathbb{C}$, or maybe equal (that's a good question I think). In mathematical softwares, one among the values is choosen, usually setting $k=0$ in the expression above.

-
$i^i$ is just $e^{-\frac{\pi}{2}}$ which is a proven transcendental. How do you include the $k$? – Chulumba Mar 27 '11 at 15:15
if $i^i=e^{- {\pi \over 2}}$ then it is also $i^i=e^{- {\pi \over 2}+ k*2 \pi i}$ . For integer $k$ the exp-function is periodic by $2 \pi i$ in its argument. – Gottfried Helms Mar 27 '11 at 15:53
@Chulumba You should review branch cuts and multivalued functions. – Alex B. Mar 27 '11 at 16:44