# What's the power of quaternion $[e^{a} * (cos(r) + \frac{sin(r)}{r}(bi+cj+dk)) ]^ 2$?

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.

http://www.euclideanspace.com/maths/algebra/realNormedAlgebra/quaternions/functions/power/index.htm

• 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. – Marc van Leeuwen Jan 22 at 13:16
• 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. – Du'uzu Mes Jan 22 at 13:32