Can we take the exponential function of a vector? Main question:
If $x=(x_1,x_2)\in \mathbb R^2$, can we have the exponential function of the vector, i.e.
$$
e^{i \pi x}=e^{i \pi (x_1,x_2)} \quad? \tag 1
$$
Follow-up question:
Can we in some way "split" $e^{i \pi (x_1,x_2)}$?
I'm thinking of the rule the $e^{ia}e^{ib}=e^{i(a+b)}$, where $a,b\in \mathbb R$.
A test, if we use the standard basis and write the vector as
$$
x=(x_1,x_2)=\hat e_1 x_1+\hat e_2x_2 \tag 2
$$
Can we now write $(1)$ as
\begin{align}
e^{i \pi (x_1,x_2)} &=
e^{i \pi (\hat e_1 x_1+\hat e_2x_2)}  \tag 3\\
&=
e^{i \pi (\hat e_1 x_1)}
e^{i \pi (\hat e_2 x_2)}
\quad? \tag 4
\end{align}
Is this legitimate?
 A: That is equivalent to saying that $\exp(e_i)$ is defined for each $i \in \{1,2\}$ where $e_i's$ are basis vectors. When it comes to analytic functions, I often think of its Taylor expansion.
$$
\exp(x): = 1 + x + x^2/2 + \cdots
$$
If $x$ is a real number, the definition can be used to find the corresponding function value. It can also be applied to matrices for which exponentiation is well defined (in this case the $1$ in the right-hand side of the equation should be replaced by the identity I). However, for a vector $x$, $x^2$ is not defined in general. If you define the exponentiation of vectors, it would make sense, but you can't in reality as you know.
A: As pointed out in the comments and the answer by Hermis14, the exponential function is usually generalized via its infinite series. And this is a problem since the square of a vector is not usually defined. However, in Geometric Algebra/a Clifford algebra, we usually have $\mathbf v^2=\Vert\mathbf v\Vert^2$ (the square of the length/magnitude/norm), and may add scalars and vectors independently (kind of like real and imaginary parts of a complex number).
In that sort of nonstandard setting, we have:
\begin{align}\exp(\mathbf v)&=1+\mathbf v+\frac{\Vert\mathbf v\Vert^2}{2!}+\frac{\Vert\mathbf v\Vert^2}{3!}\mathbf v+\cdots
\\&=\begin{cases}1 & \text{if }\mathbf{v}=\boldsymbol{0}\\1+\Vert\mathbf v\Vert\frac{\mathbf v}{\Vert\mathbf v\Vert}+\frac{\Vert\mathbf v\Vert^2}{2!}+\frac{\Vert\mathbf v\Vert^3}{3!}\frac{\mathbf v}{\Vert\mathbf v\Vert}+\cdots & \text{otherwise}\end{cases}
\\&=\begin{cases}1 & \text{if }\mathbf{v}=\boldsymbol{0}\\\cosh\left(\Vert\mathbf v\Vert\right)+\frac{\mathbf v}{\Vert\mathbf v\Vert}\sinh\left(\Vert\mathbf v\Vert\right) & \text{otherwise}\end{cases}
\end{align}
where $\cosh$ and $\sinh$ are the hyperbolic trig functions. You can then insert $i\pi$ if you wish.
