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Now, I know that it's (relatively) easy to calculate, say, $r^{a+bi}$ (using the fact that, for $z_1, z_2\in \mathbb{C}, {z_1}^{z_2}=e^{z_2\ln(z_1)}$ and $\ln(z_1$) can just be found using: $\ln(a+bi)=\ln[\sqrt{a^2+b^2}]+i \cdot \arctan(\frac{b}{a})$ ).

Anyway, how would I go about calculating, say, $i^j, k^i$ etc., or, more generally, $(a_1+b_1i+c_1j+d_1k)^{a_2+b_2i+c_2j+d_2k}$ (I know that exponentiating a complex number (to another non-real complex number) produces a non-unique result, so I assume the same would apply further up the hypercomplex ladder; if that's the case, I'm only concerned with the 'principal' value)?

I obviously don't want a general formula or anything like that; just some intuition and a method by which I could calculate such a thing.

And, finally (because I really like to push my luck), can this method for quaternions be extended to higher number systems (i.e. $\mathbb{O, S},$ etc.) to give an analogous result?

Thanks

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There is $e^\xi$ for quaternion $\xi,$ simply because there is a nice representation in 4 by 4 real matrices, if for no better reason. However, since you lack commutativity, you can no longer make much sense out of $e^{\xi + \chi}.$ So bases other than $e$ itself are a bit optimistic. I guess positive reals... –  Will Jagy Mar 7 at 23:18

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There is no point in trying to generalise the base $x$ of exponentiation $x^y$ to be a quaternion, since already for $x$ a complex number (other than a positive real) and non-integral rational $y$ there is no unique natural meaning to give to $x^y$. (For instance $z^{2/3}$ could be interpreted as asking for the square of a cube root of$~z$, or for the cube root of $z^2$, and in both cases there are not one but (the same) three candidates; an attempt to force a single outcome for instance by fixing a preferred cube root for every complex number would make the two interpretations differ for certain$~z$.) Anyway, if anything $x^y$ is going to be equivalent to $\exp(\ln(x)y)$ or $\exp(y\ln(x))$ (giving you some choice in case of non-commutatvity) for some meaning of $\ln x$. So the whole effect of using a strange $x$ is to multiply the exponent by a constant; one is better off just writing that multiplication explicitly and sticking to the exponential function $\exp$.

There is no problem at all to extend $\exp$ to a function $\Bbb H\to\Bbb H$, by the usual power series. In fact every non-real quaternion spans a real subalgebra isomorphic to$~\Bbb C$, which will be $\exp$-stable, and restricted to it $\exp$ will behave just as the complex exponential function. Of course one can only expect $\exp(x+y)=\exp(x)\exp(y)$ to hold if $\exp(x)$ and $\exp(y)$ commute, which essentially is the case when $x$ and $y$ are le in the same subalgebra isomorphic to$~\Bbb C$ (and hence commute).

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