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Do you know how to compute $i^i$? (P.S. I assume you want the principal value?) – Hurkyl Jul 9 '13 at 17:02
$i^{i^i} = \cos \left( {\frac{1}{2}{e^{ - \pi /2}}\pi } \right) + i\sin \left( {\frac{1}{2}{e^{ - \pi /2}}\pi } \right)$ – Gamma Function Jul 9 '13 at 17:12
@JacobMayle – M Turgeon Jul 9 '13 at 17:13
Very nice Mahdi +1 @MahdiKhosravi – Babak S. Jul 9 '13 at 18:01
we have interesting answer, but what is the interpretation of $i^i$ ? – Were_cat Jul 10 '13 at 1:01
up vote 18 down vote accepted

$i^i=e^{i\log i}$

Now on principal branch,using $i=e^{i\pi/2}\implies \log i=i\pi/2$ gives $i^i=e^{-\pi/2}$

Therefore, $i^{i^i}=i^{e^{-\pi/2}}=e^{e^{-\pi/2}\log i}=e^{i(\pi e^{-\pi/2})/2}=\cos\left(\pi \frac{e^{-\pi/2}}{2}\right)+i\sin\left(\pi \frac{e^{-\pi/2}}{2}\right)$

and hence its imaginary part is $\neq 0$ as $ \frac{e^{-\pi/2}}{2}$ is not an integer.

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Don't you mean $\frac{\pi e^{-\pi/2}}{2}$ instead of $\pi e^{\frac{-\pi/2}{2}}$? – Gustav Larsson Jul 9 '13 at 19:40

Complex powers may have more than one value. In our case $$ i^i=e^{i \log i}=\exp \left(i \left(\ln(1)+i \frac{\pi}{2}+2\pi k i\right)\right)=e^{-\frac{\pi}{2}+2\pi k} $$ where $k$ is an integer. Thus $$i^{i^i}=e^{i^i \log(i)}=\exp\left(e^{-\frac{\pi}{2}+2\pi k}\cdot\left(i \frac{\pi}{2}+2\pi l i\right)\right), $$

which is $e$ to an imaginary power. It is therefore a point on the unit circle, but it can never be chosen real.

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Techinically, you'd need to show that the imaginary power can't be a multiple of $\pi$. – Thomas Andrews Jul 9 '13 at 17:14
@ThomasAndrews you are right, but for the principal branch at least it is clear. – user1337 Jul 9 '13 at 17:21
I'm wondering how to prove $e^{-\pi/2+2\pi k}$ is irrational? – Yuchen Liu Jul 19 '13 at 7:24

Wolfram Alpha gives the answer to be:

$0.94715899... + 0.320764449... i$.

Therefore it is not a real number, as the answer has an imaginary component to it.

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Interesting proof. – Lord Soth Jul 9 '13 at 17:05
I never said it was a proof. I was just answering the first question that was asked: is it real? – Sujaan Kunalan Jul 9 '13 at 17:06
+1 I am amazed how people are still afraid of computers. – Mlazhinka Shung Gronzalez LeWy Jul 9 '13 at 17:08
30 x) – Schlomo Jul 9 '13 at 17:09
Wolfram Alpha is a fine tool if what you really want is an (approximate) answer. A question like this, nobody needs the answer, so the only goal is implicitly to understand the answer. The answer itself is about as useful as writing out the digits $e^\pi$ in base $7$. – Thomas Andrews Jul 9 '13 at 17:11

We have $i^i=(e^{i\pi /2})^i=e^{-\pi /2}$. Then, $$i^{i^i}=i^{e^{-\pi /2}}=(e^{-i\pi /2})^{e^{-\pi /2}}=e^{-i\pi e^{-\pi /2}/2}=\cos (\pi e^{-\pi /2}/2)-i\sin(\pi e^{-\pi /2}/2) $$ Now $\sin(\pi e^{-\pi /2}/2)$ is non-zero, since $e^{-\pi /2}/2$ is not an integer.

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First find all the values of $i^i$. Then, if $\beta$ is one of those values, a possible value of $i^{i^i}$ is $(e^{i\pi/2})^{\beta}$. Or $(e^{-3i\pi/2})^{\beta}$.

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