# Using Quaternion Extension of Eulers Formula what is $e^{qw} * e^{qw} = e^{qw2}$?

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?
• What is $x$? Cheers! – Robert Lewis Jan 15 at 7:09
• Same as it is in Euler's formula with complex numbers, $e^{xi}$= $e^{x\sqrt{-1}}$ where $x$ is a real number. – Du'uzu Mes Jan 15 at 7:15
• More to the point, what is $\sqrt{-1}$? (In the quaternions, there are lots of solutions to $q^2=-1$). – Lord Shark the Unknown Jan 15 at 7:16
• 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. – Du'uzu Mes Jan 15 at 7:22
• You are correct that for the quaternions extension $\frac{bi+cj+dk}{r}$ is a square root of −1. – Du'uzu Mes Jan 15 at 7:45