Calculate $i ^ {i+1}$ and also $i^{i^{i^{\dots}}}$

Question

I just wanted to ask the following questions please. The first I have is calculate $i^{(i+1)}$ and also $i^i$ I was just wondering if anyone can nudge me in the right direction to solve these questions.

many thanks!

I'm also interested in seeing how this is done. I guess all comes to the definition but I have been searching for it for so long. Can someone or at least give some reference to the theory which defines complex exponentiation? Pls :) — sve, Nov 11 '13 at 13:26
Similar question: http://math.stackexchange.com/questions/191572/prove-that-ii-is-a-real-number — N.U., Nov 11 '13 at 13:27
It is a nice idea to compute I^I^I^I...For I^I, we get the know result Exp[- Pi /2]. For I^I^I, the result is quite nice and simple. Increasing the power tower, the expressions are very esthetical (at least for me). — Claude Leibovici, Nov 11 '13 at 13:40
@DonAntonio Sure, but it's also not so difficult; write $b = re^{i\theta}$, use the product law for $\log$, and notice there is a periodic set of solutions. — user7530, Nov 11 '13 at 14:00
The product law for exponents/logarithms does not work with complex numbers: one must choose a branch cut for the complex logarithm, an infinitely-valued function. — DonAntonio, Nov 11 '13 at 14:18

score 4 · Answer 1 · answered Nov 11 '13 at 13:34

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Choosing the usual branch for the complex logarithm, we get

$$i^i=e^{i\,\text{Log}\,i}=e^{i\left(\log 1+i\arg i\right)}=e^{i\cdot\frac{\pi i}2}=e^{-\frac\pi2}$$

Thus

$$i^{i+1}=ii^i=e^{-\frac\pi2}\cdot i$$

answered Nov 11 '13 at 13:34

DonAntonio

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Do all these things come from complex analysis? – sve Nov 11 '13 at 13:36
Yup, @Inwvr ... – DonAntonio Nov 11 '13 at 13:37
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Thank you! This is great! Forgot about that pesky property! :P – Mathsider Nov 16 '13 at 10:06

score 0 · Answer 2 · edited Nov 11 '13 at 13:43

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$i^{i+1}=\{e^{i\pi/2}\}^{i+1}=e^{(i+1)i\pi/2}=e^{i\pi/2-\pi/2}=ie^{-\pi/2}$

Here, we made use of the identity, $e^{i\theta}=cos(\theta)+isin(\theta)$.

For the second quantity (post-edit, different from question in title?), just divide the previous by $i$.

edited Nov 11 '13 at 13:43

DonAntonio

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answered Nov 11 '13 at 13:37

dan

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You should check your message before you post it, @dan – DonAntonio Nov 11 '13 at 13:37

score 0 · Answer 3 · answered Nov 11 '13 at 14:01

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One may also view $i^i$ as the set $\{e^{-\pi(2k+1)}\mid k\in\mathbb Z\}$.

answered Nov 11 '13 at 14:01

Michael Hoppe

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Calculate $i ^ {i+1}$ and also $i^{i^{i^{\dots}}}$

3 Answers3