Infinite tetration of $z$ where $z=i^i$

Question

1) Proof that $z=i^i$ is a real number

Euler's identity;

$$e^{i\pi} + 1 = 0$$

can be manipulated in order to obtain the result:

$$e^{i\pi} = -1$$

Raising both sides of the equality to the power of $i$ gives, after simplification:

$$e^{-\pi} = i^{2i}$$

Square rooting both sides gives:

$$e^{-\frac{\pi}{2}} = i^{i} \approx 0.2078 $$

2) The formula for the infinite tetration of some $z$

For the general case of some base $z$ , using the Lambert W function, one can find that, if:

Let $c = z^{z^{z^{...}}} = z^{1/z} $, '$...$' denoting the iterated exponentiation (tetration) of $z$.

Then:

$$c = -\frac{W(-\ln(z))}{\ln(z)}$$

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3) Evaluation for $z=i^i$ as shown in 1)

Now let in 2) the value $z = i^{i} = e^{-\frac{\pi}{2}} \approx 0.2078 $ as found in 1), then it can be computed that the infinite tetration of $z=i^{i}$ looking like $$ c = (z)^{(z)^{(z)^\ldots}} = (i^i)^{(i^i)^{(i^i)^\ldots}} $$ is equal to:

$$c=\frac{2W(\frac{\pi}{2})}{\pi} \approx 0.4745409995$$

Is this correct?!

Answer 1

You made a small mistake:

$$c=z^{z^{z^{\cdots}}}=i^{i^{i^{\dots}}}$$

Hence:

$$z=i\ne i^i$$

$$i=e^{i\pi/2}$$

So, reevaluating, manipulating your formula with a Lambert W function identity:

$$c=e^{-W(-\ln(e^{i\pi/2}))}=e^{-W(-i\pi/2)}=\frac{W(-i\pi/2)}{-i\pi/2}$$

This should then evaluate to the actual answer.