I was just playing around with complex numbers and I came up with something that I found really intriguing, but very bizarre and puzzling.

**Note:** I am just having fun, so, please do not take the following very seriously..But sure, please tell me in the comments whether it is correct or not.

My first goal was to find the value of \(\displaystyle i^i\). I tackled this in the following way:

Let,

\[x = i^i\]

Now,

\[i = e^{i\frac{\pi}{2}}\]

Substituting this value of \(\displaystyle i\),

\[x = \left(e^{i\frac{\pi}{2}}\right)^i\]

\[\Rightarrow x = e^{i^2\frac{\pi}{2}}\]

\[x = e^{-\frac{\pi}{2}}\]

\[\Rightarrow \boxed{i^i = 0.2078}\]

After having found this value, I took it one step further (towards insanity!)

What is the value of \(\displaystyle i^{i^{i\ldots 'x' times}}\), where there are \('x'\) iotas.

Let \(\displaystyle f(x) = i^{i^{i\ldots 'x' times}}\)

Then,

\[f(x) = i^{f(x-1)}\]

Taking \(\ln\) on both sides,

\[\ln f(x) = f(x-1)\cdot \ln i\]

**Note:** I know \(\ln i\) doesn't exist, but as I said : just having fun.

Differentiate on both sides wrt \(x\),

\[\frac{1}{f(x)}f'(x) = f'(x-1) \ln i\]

Assuming \(\displaystyle f(x)\) is differentiable over all \(x\),

Take \(\displaystyle\lim_{x\to\infty}\) on both sides.

\[\frac{1}{\lim_{x\to\infty}f(x)}\lim_{x\to\infty}f'(x) = \lim_{x\to\infty}f'(x-1)\ln i\]

If \(\displaystyle\lim_{x\to\infty}f'(x) = L\), then \(\displaystyle\lim_{x\to\infty}f'(x-1) = L\)

Thus,

\[\frac{1}{\lim_{x\to\infty}f(x)} = \ln i\]

\[\Rightarrow \lim_{x\to\infty}f(x) = \frac{1}{\ln i}\]

Therefore,

\[i^{i^{i\ldots}} = \frac{1}{\ln i} = \frac{2}{i\pi}\]

I know this may be terribly wrong, but I was just bored of studying, and decided to do all this crazy stuff just for fun.

Anyways, please be nice in the comments and tell me what you think.!

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## Comments

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TopNewestDiscovered the following later on :

In the process, we did the following,

\[f(x) = i^{f(x-1)}\]

\[\ln f(x) = f(x-1)\ln i\]

Take \(\displaystyle\lim_{x\to\infty}\) on both sides,

\[\ln\left(\lim_{x\to\infty}f(x)\right) = \ln i \times\lim_{x\to\infty}f(x-1)\]

We found out in the end that,

\[\lim_{x\to\infty}f(x) = \lim_{x\to\infty}f(x-1) = \frac{1}{\ln i}\]

Substituting this value,

\[\ln\left(\frac{1}{\ln i}\right) = \ln i \times\left(\frac{1}{\ln i}\right)\]

\[-\ln\left(\ln i\right) = 1\]

\[\boxed{\ln\left(\ln i\right) = -1}\]

Now, Im sure I am crazy!

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Well magnitude of 1/logi isn't 1 but i^i^i.... should have magnitude 1 right

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But i^i^i should be (e-^minus pi/2)^i which is -i. Magnitude is 1, if you take modulus.

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Comment deleted Mar 11, 2015

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It's \(incorrect \)

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\(lne^{ipi/2}\) = \(ipi/2\)

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I assume you took log on both sides of the equation \(\displaystyle i = e^{\frac{\pi}{2}}\)...But, you see, \(\displaystyle = e^{\frac{i\pi}{2}}\)..So, what you wrote is incorrect..

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Not offencive but I think that assuming the limit exists is wrong

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Maybe. I wasn't offended. I expected corrections..This was purely for fun.

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To put my view in a better way, \(f(x)\) is somewhat like floor function. It accepts different values of x and returns some constant values. Also the floor function is derivable only in specific domains and always returns zero as the value of the function in a way independent of x.

Similarly here, I'm quite certain that when you cancelled the limits, you made the error there by cancelling zero on either side of the equation.

Try it without differentiating, then mostly the anomaly will not rise.

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Consider \( f(x) = x^2 \). Differentiating, we get \( f'(x) = 2x \). Now, \( f(x) \) can be written as \( x+x+x+\ldots+\text{ x times } \). Differentiating term-wise, we have \( 1 + 1 + 1 + \ldots + \text{ x times } = x \).

The problem is, in writing \(f(x) = x+x+x+\ldots+\text{ x times} \), we assume \( x \) to be an integer!

Similar to what's going on here.

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But searching google for ln(ln(I)) gives 0.451582705 + 1.57079633 i And so does Wolfram Alpha...

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Yes, indeed. There must be some flaw in my crazy proof. I dont know.

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It may be a bit surprising that \(i^i\) has an infinite number of real values. Recall that \(e^{i\theta}=e^{i(\theta+2\pi n)}\) for any integer \(n\). Thus, we have that \[i^i=(e^{i\frac{\pi}{2}+2i\pi n})^i=e^{-\frac{\pi}{2}+2\pi n}\] for any integer \(n\).

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The problem with complex numbers. They don't exist. :D

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it's close, but the Lambert W function is defined as satisfying the equation

\(z=W(z){ e }^{ W(z) } \)

so that in solving the equation for this infinite tower function

\(z={ i }^{ z }\)

it works out that

\(z=(W(\frac { \pi }{ 2i } ))(\frac { 2i }{ \pi } )\)

and that first term isn't quite \(1\), but is some complex number. This problem was already answered in the mid 1800s.

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AWESOME! That's kind of how my brain works, go off on some crazy tangent, and usually create a problem. :D

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Great minds think alike. :D

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i think gud work out really gud for you

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You can't differentiate the function because it is discrete: x must be a positive integer.

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x was originally defined as an integer. How come x is differentiable over x

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can you tell me i^i^i^i^.................infinity = ? if every one is having fun, why not i?

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In breaking rules we discover newer better rules. This was fun. I have for too long been a non serious mathematician to come in meaningfully but I can tell you that I loved this post.

Such craziness was in fact the default way we used to prep for IIT JEE. And that was in a different era when coaching was optional ( and it was a suboptimal approach too if you chose not to take it!)

Anyways long story short , Dirac did mathematical craziness like this that led to the concept of antimatter.

And your jottings here reminded me of pretty similar jottings lying somewhere in my old notebooks!

Also a fellow fanboy of Euler and specifically of Euler's formula! :)

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Which is (-2i/pi)

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