Exponential of a 3D multivector in Clifford's Geometric Algebra An arbitrary 3D multivector can be written as the superposition: $M=Z+F$, where $Z=a+bi$ with $a,b\in\mathbb R$ and $i$ been the pseudo scalalar of $(Cl_3)$. The remaining term $F=v+iw$, where $v$ and $w$ are vectors of $Cl_3$, have vector and bivector parts, respectively. Since $Z\in$ Cen $(Cl_3)$, the exponential of $M$ becomes
$$\exp(M)=\exp(Z+F)=\exp(Z)\exp(F).$$
To obtain a closed form of $\exp(M)$, I would like to ask if there is a way to show that $$\exp(F)\stackrel{?}{=}\exp(v)\exp(iw)\ldots.$$
Thank you very much in advance.

As noted by Peeter Joot, $e^F$ cannot be split into $e^{v}e^{iw}$ since $[v,iw] \neq 0$. So the hunt is on for a decomposition of $F$ into two commuting parts. Fortunatelly, by staring at Cayley tables we can discover that $\mathbb{R}_3$ is isomorphic to $\mathbb{R}^+_{3,1}$, the even subalgebra of the spacetime algebra, and thus that your $F$ can be viewed as a bivector in spacetime. And for bivectors, we can use the invariant decomposition to decompose them into commuting terms.
So adapting the invariant decomposition formula slightly using the isomorphism, we can find an $F_1$ and $F_2$ satisfying $F = F_1 + F_2$, $\lambda_i := F_i^2 \in \mathbb{R}$, and $[F_1, F_2] = 0$:
$$F_i = \frac{\lambda_i + \tfrac{1}{2} \langle F^2 \rangle_3}{F},$$
where the $\lambda_i$ are given by
$$\lambda_i = \tfrac{1}{2}  \langle F^2 \rangle_0 \pm \tfrac{1}{2} \sqrt{ \langle F^2 \rangle_0^2 - \langle F^2 \rangle_3^2 }.$$
So we have now found two commuting elements $F_1$ and $F_2$, which both square to scalars so they follow Euler's formula. Specifically, since $\lambda_1 \geq 0$ and $\lambda_2 < 0$, we find
$$
\begin{aligned}
e^{F_1} &= \cosh(\sqrt{\lambda_1}) + \frac{F_1}{\sqrt{\lambda_1}} \sinh(\sqrt{\lambda_1}) \\
e^{F_2} &= \cos(\sqrt{-\lambda_2}) + \frac{F_2}{\sqrt{-\lambda_2}} \sin(\sqrt{-\lambda_2})
\end{aligned}
$$
To conclude, the exponential of any multivector $M = Z + F$ is
$$ e^{M} = e^{a}e^{bi}e^{F_1}e^{F_2}. $$
So the invariant decomposition allows us to bypass the Baker-Campbell-Hausdorff formula. For more details and to see how this generalises to higher dimensions I would highly recommend reading the invariant decomposition paper.
