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Knowing that for quaternions Euler's identity is:

$q = a + bi + cj + dk$ with a,b,c,d real numbers

$\sqrt{b^2+ c^2 + d^2} = r > 0$

$e^q = e^{a + r\sqrt{-1}} = e^ae^{r\sqrt{-1}} = e^a(\cos(r) + \sqrt{-1} \sin(r)) = e^a(\cos(r) + \frac{\sin(r)}{r}(bi + cj + dk))$

Using above method, where $e^q = e^ae^{r\sqrt{-1}}$, what is $d^ne^{nx\sqrt{-1}}*e^q$ ?

Edit: The original question above utilized a flawed assumption that a quaternion $e^q$ could be multiplied directly by a simple complex number set $d^n(a+bi)^n$=$d^ne^{nx\sqrt{-1}}$. After discovering this mistake the revision of the question is as follows.

By converting the complex number set into a quaternion using Cayley-Dickson construction:

$e^{x\sqrt{-1}} = e^{ix} = a + bi$

$q = a + bi + cj + dk$ with a,b,c,d real numbers from single complex number set

Complex numbers are defined in terms of real numbers, such that:

$z = a + bi, i^2=−1$

Same process with two complex numbers [produces quaternion] :

$qw = z1 + z2j$, $j^2=−1, ij=−ji=k$,

$qw = a + bi + aj + bij$

$\equiv$ $qw = a + bi + aj + bk$

Given the above method -

Complex number: $d^n(a+bi)^n = d^ne^{nxi}$

as quaternion would be: [correct me if I'm wrong here]

$qw = d^na^n + d^nb^ni + d^na^nj + d^nb^nk$


Updated questions -

  1. What is $e^{qw}$ using Euler's Formula Extension?
  2. What is $e^{qw} * e^{qw} = e^{qw2}$, where $e^{qw2}$ is shown in the Euler's forumla extension?
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  • $\begingroup$ What is $x$? Cheers! $\endgroup$ – Robert Lewis Jan 15 at 7:09
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    $\begingroup$ Same as it is in Euler's formula with complex numbers, $e^{xi}$= $e^{x\sqrt{-1}}$ where $x$ is a real number. $\endgroup$ – Du'uzu Mes Jan 15 at 7:15
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    $\begingroup$ More to the point, what is $\sqrt{-1}$? (In the quaternions, there are lots of solutions to $q^2=-1$). $\endgroup$ – Lord Shark the Unknown Jan 15 at 7:16
  • $\begingroup$ For $d^ne^{nx\sqrt{-1}}$ it's just a simple complex number in Euler's Formula, which is then multiplied by the quaternion extended into Euler's Formula. Not sure if that's the way to write it out though. $\endgroup$ – Du'uzu Mes Jan 15 at 7:22
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    $\begingroup$ You are correct that for the quaternions extension $\frac{bi+cj+dk}{r}$ is a square root of −1. $\endgroup$ – Du'uzu Mes Jan 15 at 7:45

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