Complex function to its 2D equivalent Is there a general way to convert functions that have complex values to ones that use 2D reals? So $f:\mathbb{C}\rightarrow\mathbb{C}$ and $g:\mathbb{R^2}\rightarrow\mathbb{R^2}$ such that if
$$g(c_1,c_2) = \begin{bmatrix}v_1 \\ v_2\end{bmatrix}$$
then
$$f(c) = v_1 + v_2i$$
For example, I'd want to convert this function:
$$f(x)=(x+3-2i)\cdot(x-1+3i)^2$$
I was working on an algorithm to look for zeroes in a given region and figured it would be convenient to define test cases as complex polynomials, as they can easily be constructed to have known zeroes. But the algorithm itself is defined as $\mathbb{R^2}\rightarrow\mathbb{R^2}$ and easier that way.
Alternatively, is there a way to convert complex polynomials like that?
So far I've converted before and after each evaluation using
c = complex(v1, v2)
v1 = real(c)
v2 = imag(c)

but this gets a bit tedious and ugly. And after making an actual algorithm, a complex function needs to be evaluated twice for each point if one doesn't want to store the values and more clutter:
f = @(x) (x - (-3+2i)) .* (x - (1-3i)).^2;
% God forbid this ever sees the light of day
g = @(x) [real( f( complex(x(1,:), x(2,:)) ) );
          imag( f( complex(x(1,:), x(2,:)) ) )];

 A: (Too long for a comment.)
It's not entirely clear what the question means to ask. The following works out the given example:
$$f(z)=(z+3-2i)\cdot(z-1+3i)^2$$
Let $\,z=x+iy\,$ with $\,x,y \in \mathbb{R}\,$, then after expanding and regrouping:
$$
\begin{align}
(x+iy+3-2i)\cdot(x+iy-1+3i)^2 &= \big(x+3 + (y-2)i\big)\big(x-1+(y+3)i\big)^2 \\[5px]
 &= \underbrace{x^3 + x^2 - 3 x y^2 - 8 x y - 2 x - y^2 - 16 y - 36}_{\textstyle v_1(x,y)} \\[5px]
 &\quad+ i \underbrace{(3 x^2 y + 4 x^2 + 2 x y + 16 x - y^3 - 4 y^2 - 2 y - 2)}_{\textstyle v_2(x,y)}
\end{align}
$$
It follows that $\,f(z)=v_1\big(\operatorname{Re}(z),\operatorname{Im}(z)\big) + i \, v_2\big(\operatorname{Re}(z),\operatorname{Im}(z)\big)\,$, so $\,g(x,y) = \begin{bmatrix}v_1(x,y) \\ v_2(x,y)\end{bmatrix}\,$ would satisfy the requirements. It is not the case in general that $\,v_1(x,y)=v_1(x)\,$ and $\,v_2(x,y)=v_2(y)\,$.
For an arbitrary non-polynomial $\,f\,$, one would have to define $\,v_1(x,y)= \operatorname{Re}\big(f(x+iy)\big)\,$ and $\,v_2(x,y)= \operatorname{Im}\big(f(x+iy)\big)\,$, but getting closed forms may not necessarily be pretty.
