Exponential Function of Quaternion - Derivation The equation for the exponential function of a quaternion $q = a + b i + c j + dk$ is supposed to be $$e^{q} = e^a \left(\cos(\sqrt{b^2+c^2+d^2})+\frac{(b i + c j + dk)}{\sqrt{b^2+c^2+d^2}} \sin(\sqrt{b^2+c^2+d^2})\right)$$
I'm having a difficult time finding a derivation of this formula.  I keep trying to derive it, but I end up getting different results.  Would someone be able to point me to a proof of this formula or do the derivation here?
Note: I also don't understand why some people say $e^q = e^a e^{b i + c j + d k}$. Can you please explain this, too?
 A: The definition of quaternionic exponential is given by the absolutely convergent series
$$
e^z=\sum_{k=0}^\infty\dfrac{z^k}{k!}
$$
It is well known that, from this definition, if $x, y$ commute we have $e^xe^y=e^ye^x=e^{x+y}$.
Since real quaternions commute with all other quaternions, for $a \in \mathbb{R}$  we have $e^{a+z}=e^ae^z \; \forall z\in \mathbb{H}$ so, if $z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k} = a+\mathbf{v}$, we have $e^z=e^ae^\mathbf{v}$, where $\mathbf{v}$ is an imaginary (or vector) quaternion.
Now we have:
claim
If $ \mathbf{v}  \in \mathbb{H}_P$ is an imaginary quaternion, putting $\theta=|\mathbf{v}|$  we have:
$$
e^\mathbf{v}= \cos\theta + \mathbf{v}\;\dfrac{\sin \theta}{\theta}
$$
proof
We note that:
$$
\mathbf{v}^2= (b \mathbf{i}+c  \mathbf{j} +d  \mathbf{k})(b \mathbf{i}+c  \mathbf{j} +d  \mathbf{k})=
-b^2-c^2-d^2=-|\mathbf{v}|^2
$$
so:
$$
 \mathbf{v}^2= -\theta^2 
\quad,\quad  \mathbf{v}^3= -\theta^2\mathbf{v} 
\quad,\quad \mathbf{v}^4= \theta^4 
\quad,\quad \mathbf{v}^5= \theta^4 \mathbf{v}
\quad,\quad \mathbf{v}^6= -\theta^6 
\quad,\quad \cdots 
$$
and the series become.
$$
\begin{split}
e^\mathbf{v}&=\sum_{k=0}^\infty\dfrac{\mathbf{v}^k}{k!}=\\
%
&=1+\dfrac{\mathbf{v}}{1!}-\dfrac{\theta^2}{2!}-\dfrac{\theta^2\mathbf{v}}{3!}+\dfrac{\theta^4}{4!}+\dfrac{\theta^4\mathbf{v}}{5!}-\dfrac{\theta^6}{6!}+\cdots=\\
%
&=1+\dfrac{\theta\mathbf{v}}{1!\,\theta}-\dfrac{\theta^2}{2!}-\dfrac{\theta^3\mathbf{v}}{3!\,\theta}+\dfrac{\theta^4}{4!}+\dfrac{\theta^5\mathbf{v}}{5!\,\theta}-\dfrac{\theta^6}{6!}+\cdots=\\
%
&=\left(1-\dfrac{\theta^2}{2!}+\dfrac{\theta^4}{4!}-\dfrac{\theta^6}{6!}\cdots\right)+\dfrac{\mathbf{v}}{\theta}\left( \dfrac{\theta}{1!}-\dfrac{\theta^3}{3!}+\dfrac{\theta^5}{5!}\cdots\right)=\\
%
&=\cos\theta +\dfrac{\mathbf{v}}{\theta}\sin\theta
\end{split}
$$
So the exponential of a quaternion is:
$$
e^z = e^{a+\mathbf{v}}=e^a \left(  \cos  |\mathbf{v}| +\dfrac{\mathbf{v}}{|\mathbf{v}|} \,\sin |\mathbf{v}| \right)
$$
A: If $q=\alpha+i\beta+j\gamma+k\delta$, it is always possible to represent $q$ as $q=a+Ib$, where $a=\alpha$, $b=\sqrt{\beta^2+\gamma^2+\delta^2}$ and $I=\dfrac{i\beta+j\gamma+k\delta}{b}$. Note that $I^2=-1$. Since for complex numbers it holds $e^{a+ib}=e^a(\cos(b)+i\sin(b))$, the same holds for
$$
e^q=e^{a+Ib}=e^a(\cos(b)+I\sin(b))=e^a\left(cos\left(\sqrt{\beta^2+\gamma^2+\delta^2}\right)+\dfrac{i\beta+j\gamma+k\delta}{\sqrt{\beta^2+\gamma^2+\delta^2}}\sin\left(\sqrt{\beta^2+\gamma^2+\delta^2}\right)\right).
$$
