# $i$ to the power of $i$ to the power of $i$ infinitely? [duplicate]

How could you evaluate $$i$$ to the power of $$i$$ which is the to the power of $$i$$ with infinitely many $$i$$ exponents? My initial idea was to set it all equal to x which allowed me to write:

$$i^x=x$$

I tried a couple of different ways to solve this but struggled to get anywhere. When I put this equation in Wolfram alpha they use $$W_n$$ term that I am unfamiliar with. Could someone please explain to me what this means or if there is any alternative ways to solve this problem?

• It is $i^{i \cdot i \cdot i \cdots }$ which does not converge. Do you mean a power tower with $i$ instead? See also here.
– Gary
Oct 14, 2022 at 5:42
• Hi :) I don't believe, that the sequence $(a_n)_n$ with $a_{n+1}=a_n^i,\ a_0=i$ converges. It's periodic, since $i^4=1$. Oct 14, 2022 at 5:43
• It is not hard to see that the sequence is $i, e^{-{\pi \over 2}}, -i, e^{{\pi \over 2}},...$ Oct 14, 2022 at 5:45
• Sorry I meant each i is to the power of the last i not to the power of the whole thing, I'm not to sure how to format that correctly it gives me an error message when I try to type it like that.
– Tom
Oct 14, 2022 at 5:50
• Generally, equations like $a^x=x$ involve something like "Lambert's W function" and won't have nice solutions that are easily written down. Maybe there is some complex analysis that at least proves one or more solutions exist though. Oct 14, 2022 at 5:52

From the euler formula we know that $$i=e^{\frac{i\pi}{2}}$$ Which means that $$i^i= \left(e^{\frac{i\pi}{2}}\right)^i=e^{\frac{i^2\pi}{2}}$$ and for n powers of $$i$$ $$(((i^i)^i)^i)^{\dots}=e^{\frac{i^n\pi}{2}}$$ As you can see the limit does not exist since $$i^4$$ oscillates between $$1$$,$$-1$$, $$i$$, and $$-i$$.
• $i^i = e^{-{\pi \over 2}}$. Oct 14, 2022 at 5:47
• @copper.hat true, i was just writing $-1$ as $i^2$ to show the trend
• I think the question is about the power tower. That is, the limit of the sequence defined recursively by $z_1=i$, $z_{n+1}=i^{z_n}$. You are (in my opinion mistakenly) looking at the recursion $z_{n+1}=z_n^i$ instead. Oct 14, 2022 at 9:50