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.

I am note sure whether you can generally find such an exponential factorization.  However, by the Baker-Campbell-Hausdorff theorem, we have
$$   e^{A + B + \frac{1}{{2}} \left[{A},{B}\right] + \cdots} = e^A e^B,$$
where $ \left[{A},{B}\right] $, is the commutator, the antisymmetric sum of the objects $ A, B $.
Consider that commutator for your case
$$\begin{aligned}\left[{v},{iw}\right]&=v i w - i w v \\ &=i \left( { v w - w v} \right) \\ &= 2 i \left( { v \wedge w } \right).\end{aligned}$$
Note that this commutator expansion is specific to $\mathbb{R}^{3}$ since the pseudoscalar $ i $ happens to commute with all grades in that case.
This expansion shows that if $ v, w $ are colinear, one must have
$$e^{v + iw} = e^v e^{iw}.$$
If $ v, w $ were not colinear, but the higher order commutators happended to be zero, one could say
$$   e^{v + iw + i (v \wedge w)} = e^v e^{iw},$$
but that is not the type of decomposition that you are seeking.
