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Last night I learned the amazing fact that $i^i=e^{-\pi/2}$
So I started computing other powers and would like to get confirmation that:
$i^{i^i}=e^{-i\pi/2}$
$i^{i^{i^{i}}}=e^{\pi/2}$
$i^{i^{i^{i^{i}}}}=e^{i\pi/2}$
Then the pattern continues
(I tried an online calculator to confirm ${i^i}^{i^i}$ but got an imaginary part, and a real part that didn't match)
Now I'm reminded of those infinite power towers you see on Twitter and wondering what would happen if you took $i^{i^{i^{i^{i^{.{^{.^{.}}}}}}}}$
I tried writing $z = i^{i^{i^{i^{i^{.{^{.^{.}}}}}}}}$ then rewriting this as $z=i^z$ thus $z^{1/z}=i$ but not sure where to go from here, and I'm not even sure if what I'm doing makes sense. Any thoughts?

Sydney Carton
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    I think you're calculating $(i^i)^i$ and $((i^i)^i)^i$ and so on; but the appropriate calculation for that notation would be $i^{(i^i)}$ and $i^{(i^{(i^i)})}$ and so on. – Greg Martin Sep 27 '22 at 16:07
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    Does this answer your question? What is the value of i^i?. Here it is explained how to do this exponentiation (and to iterate it). – Dietrich Burde Sep 27 '22 at 16:10
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    The power tower seems to converge to $$0.4382829367 + 0.3605924719\cdot i$$ If I remember right, we can calculate this fix-point with the Lambert-W-function , if it converges at all. – Peter Sep 27 '22 at 16:13
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    Wow, thanks Greg. I assumed these were the same but not at all. So how on earth would you go about evaluating something like $i^{e^{-\pi/2}$? Also thanks for the information Peter. I will look into that function. – Sydney Carton Sep 27 '22 at 16:18
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    continuing @Peter: In PariGP use t=exp(-LambertW(-log(I))) That reproduces the given value – Gottfried Helms Oct 02 '22 at 23:18

2 Answers2

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Actually there are infinit solutions: $$ \begin{align*} z &= \mathrm{i}^{\mathrm{i}}\\ z &= (\mathrm{i})^{\mathrm{i}}\\ z &= (\mathrm{e}^{\ln(\mathrm{i})})^{\mathrm{i}}\\ z &= (\mathrm{e}^{(\frac{\pi}{2} + 2 \cdot k_{k \in \mathbb{Z} \cdot \pi}) \cdot \mathrm{i}})^{\mathrm{i}}\\ z &= \mathrm{e}^{(\frac{\pi}{2} + 2 \cdot k_{k \in \mathbb{Z} \cdot \pi}) \cdot \mathrm{i} \cdot \mathrm{i}}\\ z &= \mathrm{e}^{(\frac{\pi}{2} + 2 \cdot k_{k \in \mathbb{Z} \cdot \pi}) \cdot \mathrm{i}^{2}}\\ z &= \mathrm{e}^{(\frac{\pi}{2} + 2 \cdot k_{k \in \mathbb{Z} \cdot \pi}) \cdot (-1)}\\ z &= \mathrm{e}^{-(\frac{\pi}{2} + 2 \cdot k_{k \in \mathbb{Z} \cdot \pi})}\\ z &= \mathrm{e}^{-\frac{\pi}{2} - 2 \cdot k_{k \in \mathbb{Z} \cdot \pi}}\\ \\ \text{only if } k = 0 \text{:}\\ z &= \mathrm{e}^{-\frac{\pi}{2} - 2 \cdot k_{k \in \mathbb{Z} \cdot \pi}}\\ z &= \mathrm{e}^{-\frac{\pi}{2} - 2 \cdot 0 \cdot \pi}\\ z &= \mathrm{e}^{-\frac{\pi}{2}}\\ \end{align*} $$

It makes a diffrens when you calculate $$ i^{(i^{i})} \text{ or } (i^{i})^{i} $$ 'cause

$$ \begin{align*} \mathrm{i}^{(\mathrm{i}^{\mathrm{i}})} &\text{ or } (\mathrm{i}^{\mathrm{i}})^{\mathrm{i}}\\ \mathrm{i}^{\mathrm{e}^{-\frac{\pi}{2} - 2 \cdot k_{k \in \mathbb{Z} \cdot \pi}}} &\text{ or } \mathrm{i}^{\mathrm{i} \cdot \mathrm{i}}\\ \mathrm{i}^{\mathrm{e}^{-\frac{\pi}{2} - 2 \cdot k_{k \in \mathbb{Z} \cdot \pi}}} &\text{ or } \mathrm{i}^{\mathrm{i} \cdot \mathrm{i}}\\ \mathrm{i}^{\mathrm{e}^{-\frac{\pi}{2} - 2 \cdot k_{k \in \mathbb{Z} \cdot \pi}}} &\text{ or } -\mathrm{i}\\ \end{align*} $$

$$ \begin{align*} z &= \mathrm{i}^{\mathrm{i}^{\mathrm{i}^{.{^{.^{.}}}}}}\\ z &= \mathrm{i} \uparrow\uparrow \infty\\ z &= \mathrm{i}^{z} \quad\mid\quad (\text{ })^{\frac{1}{z}}\\ z^{\frac{1}{z}} &= \mathrm{i} \quad\mid\quad (\text{ })^{-1}\\ (z^{\frac{1}{z}})^{-1} &= \mathrm{i}^{-1}\\ z^{-1 \cdot \frac{1}{z}} &= \frac{1}{\mathrm{i}}\\ (z^{-1})^{\frac{1}{z}} &= \frac{\mathrm{i}}{\mathrm{i}^{2}}\\ (\frac{1}{z})^{\frac{1}{z}} &= \frac{\mathrm{i}}{-1}\\ (\frac{1}{z})^{\frac{1}{z}} &= -\mathrm{i}\\ &\Rightarrow z = \frac{\operatorname{W}(\ln(\frac{1}{i}))}{\ln(\frac{1}{\mathrm{i}})}\\ z &= \frac{\operatorname{W}(\ln(\frac{1}{\mathrm{i}}))}{\ln(\frac{1}{\mathrm{i}})}\\ z &= \frac{\operatorname{W}(\ln((-\frac{\pi}{2} + 2 \cdot k_{k \in \mathbb{Z} \cdot \pi}) \cdot \mathrm{i})))}{\ln((-\frac{\pi}{2} + 2 \cdot k_{k \in \mathbb{Z}} \cdot \pi) \cdot \mathrm{i})} \quad\mid\quad \text{let us use } k = 0\\ z &= \frac{\operatorname{W}(\ln((-\frac{\pi}{2}) \cdot \mathrm{i})))}{\ln((-\frac{\pi}{2}) \cdot \mathrm{i})} \quad\mid\quad \text{let us use the main branch of the lambert W-Function}\\ z &= 0.438282936727032111626975163551264824267897351646394603609221240... + 0.360592471871385485952940526906000653826577030786027004741451298... \cdot \mathrm{i} \end{align*} $$

PS A nice calculator for stuff like this is Wolfram|Alpha...

Kevin Dietrich
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Complex exponents aren't well-defined. Some exponents with a non-real base and a non-real exponent yield infinitely many different values. In fact, over the complex numbers, even $1^\pi$ can take on infinitely many different values depending on how many whole turns $1$ is considered to be.

mathlander
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