I'm having trouble finding the power of a dual quaternion ($Q=q_r+q_dε$) raised to some real number $n$: $Q^n=(q_r+q_dε)^n=?$ It is known that for a quaternion $q=e^{\vec{u}}=\cos(|\vec{u}|)+\frac{\vec{u}}{|\vec{u}|}\sin(|\vec{u}|)$, its power to a number $n$ would come to be $q^n=e^{n\vec{u}}=\cos(n|\vec{u}|)+\frac{\vec{u}}{|\vec{u}|}\sin(n|\vec{u}|)$. Something similar happens for dual numbers $(a+bε)^n=a^n+nba^{n-1}ε$. However, this is true for numbers whose multiplication is commutative, since for the case $n=2$ we have $(a+bε)^2=a^2+(ab+ba)ε=a^2+2baε$ given that $ab=ba$. Nevertheless, this is not true for quaternions.
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2Use the matrix representation, then you can diagonalize it will be clearer and you won't make commutation mistakes – julio_es_sui_glace Jun 26 '23 at 15:18
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Arbitrary exponentiation of arbitrary quaternions is not uniquely defined (and hence not uniquely defined for dual quaternions). For instance, what is $(-1)^{1/2}$? There are infinitely many quaternion square roots of $-1$. – Nicholas Todoroff Jun 26 '23 at 18:29
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As a reference for the question you could use Matrix Representation of Dual Quaternions from Mehdi Jafari, Mücahit Meral and Yusuf Yayli (Section $5$.: $n$th Roots of Matrices of Dual Quaternions) with $n^{-1} = x$. There you will also find a step-by-step guide for everything. But it is largely based on mtrices and with little analytical derivation. – The Art Of Repetition Jun 27 '23 at 00:15