What is $i^j$ for quaternions?

Question

Given complex numbers, we can calculate e.g. $i^i$.

Given quaternions, how can we calculate something like $i^j$? Wolfram Mathematica choked on that and googling did not produce any useful results. My guess is that this could be something ill defined, similar to quaternion derivative or, perhaps, even worse.

Answer 1

$$ i^j = (e^{i\pi/2})^j = e^{ij\pi/2} = e^{k\pi/2} = k $$

Answer 2

How do we meaningfully calculate $i^i$? One way which comes to mind is to write

$i = e^{\pi i \ 2}, \tag 1$

and thus

$i^i = (e^{\pi i / 2})^i; \tag 2$

if we now assume the rule

$(a^b)^c = a^{bc}, \; a, b, c \in \Bbb C, \tag 3$

is valid, then (2) becomes

$i^i = (e^{\pi i / 2})^i = e^{\pi i^2 / 2} = e^{-\pi / 2}. \tag 4$

So far, so good; however, the equation (1) does not uniquely define $i$ as an exponential; indeed, we have

$i = e^{(2n + 1/ 2)\pi i}, \; n \in \Bbb Z, \tag 5$

whence

$i^i = (e^{(2n + 1/2)\pi i})^i = e^{(2n + 1/2)\pi i^2} = e^{-(2n + 1/2)\pi}, n \in \Bbb Z. \tag 6$

If we accept these results, we may move on to attempting to calculate $i^j$; from (5), assuming (3) holds for quaternions,

$i^j = (e^{(2n + 1/2)\pi i})^j = e^{(2n + 1/2)\pi ij} = e^{(2n + 1/2)\pi k}$ $= e^{2n\pi k} e^{\pi k / 2} = e^{\pi k / 2} = k, \; n \in \Bbb Z. \tag 7$

It is perhaps worth noting that

$i^j = k = ij, \tag 8$

and that via cyclic permutation $i \to j \to k$ we find

$j^k = i = jk, \tag 9$

and

$k^i = j = ki. \tag{10}$

Note Added in Edit, Friday 19 June 2020 12:51 PM PST: A few words in response to the comment of Konstantin Konstantinov on this answer. The defining relations of the quaternion algebra,

$i^2 = j^2 = k^2 = -1, \tag{11}$

$ij = k, jk = i, ki = j, \tag{12}$

found in this wikipedia entry, lead naturally to the formulas

$e^{2n\pi i} = e^{2n \pi j} = e^{2n\pi k} = 1, n \in \Bbb Z, \tag{13}$

from which we see that for

$t \in [0, 2\pi) \tag{14}$

$e^{(2n\pi + t)i} = e^{2n\pi i}e^{t\pi i} = e^{t\pi i}, n \in \Bbb Z, \tag{15}$

and corresponding formulas with $j$ and $k$ replacing $i$. From this we see that there are a countably infinite number of distinct solutions to (15), viz.

$2n\pi + t, \; n \in \Bbb Z, \tag{16}$

with corresponding solution sets for the $j$ and $k$ variants of (15). We further observe that

$e^{i \theta} = \cos \theta + i \sin \theta, \tag{17}$

again with the corresponding relations for $j$ and $k$; of course (17) gives rise to (5) in the usual manner:

$e^{(2n + 1/ 2)\pi i} = \cos ((2n + 1/2)\pi) + i \sin ((2n + 1/2)\pi)$ $= i\sin((2n + 1/2)\pi) = i\sin((1/2) \pi) = i, \; n \in \Bbb Z, \tag{18}$

and again, the corresponding statements hold for $j$ and $k$. End of Note.

Answer 3

I know I'm late to this, but there is one thing I'm not seeing in the other responses. For some reason $$ i=e^{\frac{\pi}{2}i+2\pi ni }, n\in\mathbb{Z} $$ but since we're dealing with quaternions, it shouldn't be unreasonable to say that $$ i=e^{\frac{\pi}{2} i + 2 \pi x i + 2 \pi y j + 2 \pi z k }, \text{ for } x,y,z\in\mathbb{Z}\\ \text{since } e^{2 \pi x i} = e^{2 \pi y j} = e^{2 \pi z k} = 1 $$ In this case $$ i^{j} = (e^{\frac{\pi}{2} i + 2 \pi x i + 2 \pi y j + 2 \pi z k })^{j}\\ = e^{\frac{\pi}{2} ij + 2 \pi x ij + 2 \pi y jj + 2 \pi z kj }\\ = e^{\frac{\pi}{2} k + 2 \pi x k - 2 \pi y - 2 \pi z i }\\ = e^{\frac{\pi}{2} k} e^{ 2 \pi x k} e^{ - 2 \pi y} e^{ - 2 \pi z i }\\ = k \cdot 1 \cdot e^{ - 2 \pi y} \cdot 1\\ = e^{ - 2 \pi y} k$$ which is not well-defined since the choice for $y$ (and less importantly $x$ and $z$) is arbitrary.

Note that even with all of this, it shouldn't work out so nicely in general due to quaternions not being commutative as $(q_{1}q_{2})^{n}\neq q_{1}^{n} q_{2}^{n}$ unless $q_{1}q_{2} = q_{2}q_{1}$, but a lot of the cases here are ultimately multiplying things by 1 (or some other real number), which might commute well, but I'm unsure.

Similarly, the order of addition of exponents isn't changed so that it can be used more as shorthand since $e^{q_{1}+q_{2}}$ isn't exactly well-defined as $e^{q_{1}}e^{q_{2}}\neq e^{q_{2}}e^{q_{1}}$ in general.

A little bit of a tangent, but another property that may not necessarily hold, especially if the previous line did without additive commutativity in the exponent (except where $e^{q}=1$) is that $e^{q_{1}}=e^{q_{2}}\Rightarrow q_{1} = q_{2}$. Assuming it did hold, then using $i$ as above and a similar $j$ and $k$, you could show that $ij=k \Rightarrow k = ni + mj \text{ for some } n,m\in\mathbb{Q}$, which is clearly false.