Can one make a piecewise function on the manifold of complex numbers $\mathbb{C}^1$ that works on two metric spaces? A metric space is a set together with a metric, a means to tell the distance between two points. One is familiar with piecewise functions being discontinuous between the pieces. One could decide to pick different metric/distance functions for each piece function. Consider this simple two-part function:
$$ f(z) = \left\{ \begin{array}{ll} f_1 = z/|z| & \rm{if \;Re}(z) = 0\; and\; Im(z) \ne 0 \\ f_2 = (\cosh(Re(z)), \sinh(Re(z)) & \end{array} \right. $$
The first function $f_1$ has a metric function of a norm, $z=\sqrt{a^2+b^2}$. The norm is zero or a positive number. Because the square of the second function $f_2$ is equal to positive one, a square could serve as a metric function (the way one measures distance), $z=a^2-b^2$. A square is a signed measure since it can be negative.
After extensive clarification, it sounds like the question can be broken down as follows:
Part 1: Can the codomain of a function be the union of two different metric spaces?
Sure. It is unclear how this would come up naturally, but if there are two metric spaces $A$ and $B$, then it is possible to define some function $f : \mathbb{C} \rightarrow (A \cup B)$.
Part 2: How could one clearly mark this in a piecewise function, if the underlying sets of $A$ and $B$ are the same?
I'd be curious to see other people's suggestions, but here is a possible solution:
$$f(z) = \left\{ \begin{array}{ll} z/|z| \in A & \text{if Re}(z) = 0 \text{ and Im}(z) \ne 0 \\ \cosh(\text{Re}(z)) + i \sinh(\text{Re}(z)) \in B & \text{if Re}(z) \ne 0 \text{ or } z = 0 \\ \end{array} \right. $$
Alternatively, to make this more explicit one could use a disjoint union, sometimes called a "tagged union". For example if we have multiple metric spaces $M_i$, we could define a disjoint union $$D = \bigcup_i \{(x,i) : x \in M_i \}.$$ Then we could define a function $g : \mathbb{C} \rightarrow D$.
In your case, the function could be defined as:
$$ \begin{aligned} g &: \mathbb{C} \rightarrow \left( \{(z,0) : z \in A\} \cup \{(z,1) : z \in B\} \right) \\ g(z) &= \left\{ \begin{array}{ll} (z/|z|,\ 0) & \text{if Re}(z) = 0 \text{ and Im}(z) \ne 0 \\ (\cosh(\text{Re}(z)) + i \sinh(\text{Re}(z)),\ 1) & \text{if Re}(z) \ne 0 \text{ or } z = 0 \\ \end{array} \right. \end{aligned} $$
You could in principle just tag with the set itself, so you might prefer this (I am not sure how common this is, but the meaning should still be clear):
$$ \begin{aligned} h &: \mathbb{C} \rightarrow \left( \{(z,A) : z \in A\} \cup \{(z,B) : z \in B\} \right) \\ h(z) &= \left\{ \begin{array}{ll} (z/|z|,\ A) & \text{if Re}(z) = 0 \text{ and Im}(z) \ne 0 \\ (\cosh(\text{Re}(z)) + i \sinh(\text{Re}(z)),\ B) & \text{if Re}(z) \ne 0 \text{ or } z = 0 \\ \end{array} \right. \end{aligned} $$