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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
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$i^{i^i} = \cos \left( {\frac{1}{2}{e^{ - \pi /2}}\pi } \right) + i\sin \left( {\frac{1}{2}{e^{ - \pi /2}}\pi } \right)$ –  Gamma Function Jul 9 '13 at 17:12
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@JacobMayle math.stackexchange.com/a/439833/19379 –  M Turgeon Jul 9 '13 at 17:13
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Very nice Mahdi +1 @MahdiKhosravi –  Babak S. Jul 9 '13 at 18:01
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we have interesting answer, but what is the interpretation of $i^i$ ? –  lmorin Jul 10 '13 at 1:01

5 Answers 5

up vote 17 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

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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43  
Interesting proof. –  Lord Soth Jul 9 '13 at 17:05
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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
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+1 I am amazed how people are still afraid of computers. –  ABC Jul 9 '13 at 17:08
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smbc-comics.com/comics/20130120.gif x) –  Schlomo Jul 9 '13 at 17:09
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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

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

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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