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 -
- What is $e^{qw}$ using Euler's Formula Extension?
- What is $e^{qw} * e^{qw} = e^{qw2}$, where $e^{qw2}$ is shown in the Euler's forumla extension?