Say I have a multi vector of real geometric algebra of dimension 2. I define the Clifford conjugate as follows:
$$ (A+Xe_0+Ye_1+Bi)^\ddagger=A-Xe_0-Ye_1-Bi $$
where $A$ is a scalar, $e_0$ and $e_1$ are the generators, and $Bi=Be_0e_1$ is a pseudo-scalar.
In the complex conjugate case, we know that $(e^{it})^\dagger=e^{-it}$, such that $(e^{it})^\dagger e^{-it}=1$.
Is this the case with a multi-vector as well? Can we claim that :
$$ (e^{A+Xe_0+Ye_1+Bi})^\ddagger=e^{A-Xe_0-Ye_1-Bi} \tag{A} $$
Or is the fact that the commutator $[Xe_0+Ye_1,Bi]\neq 0$ frustrates the identity?
Let:
$$T_{1}= a_{s} + e_{0}a_{0} + e_{1} a_{1} + (e_{0}e_{1}) a_{01}\\ T_{2}= b_{s} + e_{0}b_{0} + e_{1} b_{1} + (e_{0}e_{1}) b_{01}$$
and $\ddagger$ defined by: $$(A+Xe_0+Ye_1+B(e_{0}e_{1}))^\ddagger=A-Xe_0-Ye_1-B(e_{0}e_{1})$$
where $\{a_{s}, a_{0}, a_{1}, a_{01}, b_{s}, b_{0}, b_{1},b_{01}, A, B,X,Y \} \subset \mathbb{R}\;$.
We have: $$T_{1}T_{2} = (a_{s}b_{s}+a_{0}b_{0}+a_{1}b_{1}- a_{12}b_{12}) + \\+e_{0}(a_{0}b_{s} + b_{0}a_{s} + a_{01}b_{1} - b_{01}a_{1}) + \\ +e_{1}(a_{1}b_{s} + b_{1}a_{s} + a_{0}b_{01} - b_{0}a_{01}) + \\ + (e_{0}e_{1})(a_{01}b_{s}+ b_{01}a_{s} + a_{0}b_{1} - b_{0} a_{1})$$
We can check that the following relationships hold:
$$(T_{1} + T_{2})^{\ddagger} = T_{1}^{\ddagger} + T_{2}^{\ddagger}\\ \forall r\in\mathbb{R},\; (rT_{1})^{\ddagger} = r(T_{1}^{\ddagger}) \\ (T_{1}T_{2})^{\ddagger} = T_{2}^{\ddagger} T_{1}^{\ddagger}$$
From this we see that for a given term $T$ and $n\in \mathbb{N}$ we have $$(T^n)^{\ddagger} = (T^{\ddagger})^n $$
We have: $$e^{T} = \sum_{k=0}^{\infty} \frac{T^{k}}{k!}$$ From this it follows that: $$(e^{T})^{\ddagger} = (\sum_{k=0}^{\infty} \frac{T^{k}}{k!})^{\ddagger} = \sum_{k=0}^{\infty} (\frac{T^{k}}{k!})^{\ddagger} =\\ =\sum_{k=0}^{\infty} \frac{(T^{k})^{\ddagger}}{k!} = \sum_{k=0}^{\infty} \frac{(T^{\ddagger})^{k}}{k!} = e^{(T^{\ddagger})} $$
Thus the equation $(A)$ holds.
Edit: I've computed an explicit formula for $e^{A+Xe_0+Ye_1+B(e_{0}e_{1})}$. Assume $X^2 + Y^2 > B^2$. Let $n = \sqrt{X^2 + Y^2 - B^2}$. Then:
$$e^{A+Xe_0+Ye_1+B(e_{0}e_{1})} = \\=e^A (\cosh(n) + \frac{\sinh(n)}{n} (X e_{0} + Y e_{1} + B(e_{0}e_{1}) ))$$ Conversely, assume $X^2 + Y^2 < B^2$. Let $n = \sqrt{B^2 - (X^2 + Y^2)}$. Then: $$e^{A+Xe_0+Ye_1+B(e_{0}e_{1})} = \\=e^A (\cos(n) + \frac{\sin(n)}{n} (X e_{0} + Y e_{1} + B(e_{0}e_{1}) ))$$
Finally, assume that $X^2+Y^2 = B^2$. Then: $$e^{A+Xe_0+Ye_1+B(e_{0}e_{1})}= e^A(1 + Xe_0+Ye_1+B(e_{0}e_{1})) $$
We can recover the ordinary exponential by setting $X,Y,B=0$, the complex exponential (Euler formula) by setting $X,Y=0$, and the split-complex exponential by setting $Y,B=0$.