# 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:

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

# 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?!

• Exponentiation is not commutative, so we should not expect the infinite tetrations of $i$ and $i^i$ to agree, i.e., that $$i^{i^{i^{i^{\cdots}}}} = (i^i)^{(i^i)^{\cdots}}.$$ Indeed, the number $\alpha := \frac{2}{\pi} W \left(\frac{\pi}{2}\right)$ does not satisfy $i^\alpha = \alpha$ for any choice of branch. One does find a solution to this equation (for a particular choice of branch), however, at $\approx 0.438 + 0.361 i$. – Travis Jan 22 '16 at 14:48
• Is your question about finding value of $i^i$ – Archis Welankar Jan 22 '16 at 15:26
• you are missing the "i"; $W(-\frac{\pi i}{2})$ should work, but it is difficult to work with the lambert W function in the complex plane. I get $-W(-\frac{\pi i}{2})/\frac{\pi i}{2}\approx 0.4383+0.3606i$ – Sheldon L Jan 24 '16 at 18:18
• @ArchisWelankar His question is on $$i^{i^{i^{\dots}}}$$ – Simply Beautiful Art Jan 25 '16 at 23:56
• The subject-line is about tetration with base $z=i$, and the last question is about tetration with base $z = i^i \approx 0.2078...$. It's difficult to decide how to help ... – Gottfried Helms Mar 5 '17 at 14:22

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

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.