Euler's identity extended into quaternions is:

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

for the below: $\sqrt{b^2+ c^2 + d^2} = r > 0$, and $\frac{bi+cj+dk}{r}$ = $\sqrt{-1}$,

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

Therefore, what is the higher power of $[e^{a} * (cos(r) + \frac{sin(r)}{r} (bi+cj+dk)) ]^ 2$ ?

I found this site on the powers of quaternions, but it doesn't address quaternions in Euler's form.


  • $\begingroup$ Don't say a quaternion equals $\sqrt{-1}$ if you just mean to say that its square is $-1$. If there is any standard choice for what $\sqrt{-1}$ means (which I would say there is not), it is defintiely not the value in the question. $\endgroup$ – Marc van Leeuwen Jan 22 at 13:16
  • $\begingroup$ I have not stated the quaternion equals $\sqrt{-1}$ but that for the following quaternion equations the written $\sqrt{-1}$ is $\frac{bi+ cj + dk}{r}$. This is the common reference point, as it's lack of clarification can lead to confusion due to it's many possibilities. $\endgroup$ – Du'uzu Mes Jan 22 at 13:32

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