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This question already has an answer here:

I was looking at Quaternions at Wikipedia - I was trying to find the value of $i^j$ etc...

Wikipedia lists $q^\alpha$ where $\alpha$ is real, but I can't find the value of $i^j$. Any clues?

The answer given here doesn't give an explicit solution.

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marked as duplicate by beep-boop, abiessu, Jonas Meyer, Potato, Lord_Farin Mar 25 '15 at 17:48

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  • $\begingroup$ From the result of my naive attempt at an answer, at least one possible approach results in $i^j=k$ and simultaneously $i^j=-k$ based on what assumptions are in place. I take this as a sign that this particular value might not be defined without taking some extra assumptions in hand. $\endgroup$ – abiessu Mar 24 '15 at 19:12
  • $\begingroup$ if $i^j=k=ij$, then $i^j-ij=0$, so $i(i^{j-1}-j)=0$, so $i^{j-1}-j=0$, but $i$ and $j$ are orthoganol. @abiessu $\endgroup$ – JonMark Perry Mar 25 '15 at 2:29
  • $\begingroup$ I'm not sure I follow that particular line... Why is there a problem with that particular result? $\endgroup$ – abiessu Mar 25 '15 at 4:44
  • $\begingroup$ it doesn't sound right, but I got to thinking about it, and since $i^i$ is real, I don't suppose it makes sense. if $ij=-k=ji$, then $i^j−ji=0$, so $(i^{j−1}−j)i=0$, so $i^{j−1}−j=0$. $\endgroup$ – JonMark Perry Mar 25 '15 at 4:53
  • $\begingroup$ And this all points to more reasons why the value might not be defined or worse, might not be definable. $\endgroup$ – abiessu Mar 25 '15 at 5:12
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Wecan define the exponential of a quaternion $e^q$ , $q \in \mathbb{H}$, see: Exponential Function of Quaternion - Derivation. Than we can use such definition for define $p^q$ , $q,p \in \mathbb{H}$ as $p^q=e^{(\log p)q}$, with some attention to the definition of $\log p$ that is a multivalued function. See the logarithm of quaternion.


Added after the comments.

For a quaternion $q$ theexponential $e^q$ is a well defined quaternion, so also $(e^q)^p$ is well defined, but, in general $(e^q)^p \ne (e^p)^q$ . this is not a strange result since $\mathbb{H}$ is a non commutative ring and also other properies of the exponential function for a field are not valid in a ring.

We know that every quaternion $z=a+ib+jc+kd=a+\mathbf{v}$ can be write in polar form as:$ z=|z| e^ {\mathbf{x} \theta} = |z|\left(\cos \theta +\mathbf{n}\sin \theta \right)= e^{\log |z| + \mathbf{x} \theta}$, where : $$ \begin{split} & |z|= \sqrt{a^2+b^2+c^2+d^2}\\ & \cos \theta=\dfrac{a}{|z|} \qquad \sin \theta=\dfrac{|\mathbf{v}|}{|z|}\\ & \mathbf{x}=\dfrac{\mathbf{v}}{|z|\sin \theta} \end{split} $$

so we can define a ''principal'' logarithm as:

$$ \log z= \log |z| + \theta \mathbf{x} $$ and we have $z=e^{\log z}$

Now it seems to me that we can well define: $q^p = \left(e^{\log q}\right)^p=e^{p\log |z|}e^{\theta\, \mathbf{x}\,p}$

The ''problem'' is that $e^{\theta\, \mathbf{x}\,p} \ne e^{\theta\,p \mathbf{x}}$. But is this really a problem or simply a property of quaternon exponential?

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  • $\begingroup$ Is there any particular reason to define it this way instead of $p^q=e^{(\log p)q}$? $\endgroup$ – abiessu Mar 25 '15 at 14:28
  • $\begingroup$ Good point! I forgot non-commutativity. I edit. $\endgroup$ – Emilio Novati Mar 25 '15 at 14:39
  • $\begingroup$ I attempted to use the same definitions you're looking at (except pulling from the question-linked Wikipedia page), but when I did, I got to this question of "why would we define $(e^a)^b=e^{ab}$ rather than $e^{ba}$, or something else entirely" and decided that I wasn't smart enough to decide that. $\endgroup$ – abiessu Mar 25 '15 at 15:20
  • $\begingroup$ But since your answer covers basically the same ground as mine, I'll bring mine back with a little more clarity of definition... $\endgroup$ – abiessu Mar 25 '15 at 15:21
  • $\begingroup$ @EmilioNovati The point - here as on abiessu's answer - is that non-commutativity means that there isn't a canonical definition available; the exponential map $e^q$ is a different thing from being able to define arbitrary powers, and frankly it's an amazing piece of good fortune that we can use the one to define the other for real (and to a lesser extent complex) numbers. $\endgroup$ – Steven Stadnicki Mar 25 '15 at 16:02
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Expressing $i$ as $e^{i\frac\pi 2+2ni\pi}$ and defining $(e^a)^b=e^{ab}$, we can then write

$$i^j=e^{(i\frac\pi 2+2ni\pi)j}=e^{ij\frac\pi 2+2nij\pi}\\ =e^{k\frac\pi 2+2nk\pi}=k$$

This secondary translation follows directly from the linked definition of exponentiation as a sum.

Following further commented discussion, we can then do some manipulations:

$$i^j=k\implies i^j-ij=0\\ i(i^{j-1}-j)=0\implies i^{j-1}=j\\ i^{2j-1}=i^{j-1}i^j=jk=i$$

This clearly falls flat, as $i^{2j-1}=i^ji^{j-1}=kj=-i$ as well. Therefore, the simple derivation used above together with assuming that we can define $(e^a)^b=e^{ab}$ fails to produce a consistent model, and cannot be relied upon.

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    $\begingroup$ Actually, I suppose this also depends on the equivalence $(e^a)^b=e^{ab}$... $\endgroup$ – abiessu Mar 24 '15 at 18:54
  • $\begingroup$ This assumes that $(a^b)^c=a^{(bc)}$ (and not, for instance, $a^{(cb)}$ - is there any particular reason to believe that assumption? $\endgroup$ – Steven Stadnicki Mar 24 '15 at 18:54
  • $\begingroup$ @abiessu: By definition of exponential we have : $e^{i\pi/2}=\dfrac{2i}{\pi}$ $\endgroup$ – Emilio Novati Mar 25 '15 at 15:35
  • $\begingroup$ sorry... ignore my comment it's wrong. $\endgroup$ – Emilio Novati Mar 25 '15 at 15:43
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    $\begingroup$ we have: $i \cdot i^{(j-1)}=e^{\frac{\pi}{2}i}e^{\frac{\pi}{2}i(j-1)}\ne e^{\frac{\pi}{2}i(1+j-1)}=i^j$ $\endgroup$ – Emilio Novati Mar 25 '15 at 17:55

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