Is it possible to specify a domain for Z Math Surface or in GN that also automatically excludes imaginary results?

Question

In this question I want to specify a domain of $0<=x<=1$ and $0<=y<=1$ in the Z Math Surface (object menu Add > Math Function > Z Math Surface) for the equation. I also need it to exclude the imaginary result.

(y-x+1-((y-x+1)**2 -4*y if 4*y < (y-x+1)**2 else 0 )**(1/2) ) / 2

I tried ((y-x+1-((y-x+1)**2 -4*y if 4*y < (y-x+1)**2 else 0 )**(1/2) ) / (2)) if x >= 0 and x <= 1 and y >= 0 and y <= 1 else 0

And also tried (((y if y >= 0 and y <= 1 else 0)-(x if x >= 0 and x <= 1 else 0)+1-(((y if y >= 0 and y <= 1 else 0)-(x if x >= 0 and x <= 1 else 0)+1)**2 -4*(y if y >= 0 and y <= 1 else 0) if 4*(y if y >= 0 and y <= 1 else 0) < ((y if y >= 0 and y <= 1 else 0)-(x if x >= 0 and x <= 1 else 0)+1)**2 else 0 )**(1/2) ) / (2))

But nothing can get me the correct result as when I plot it with python which should look like this:

How do I tell Z Math Surface to use a range of values within $0<=x<=1$ and $0<=y<=1$ that automatically exclude the imaginary results? Is that even possible? It probably is not possible with the math surface functions? Perhaps possible with geometry nodes?

EDIT: Also tried XYZ Math Surface which will still necessitate an if/else statement to exclude the imaginary part and will also give a squarish result:

(v-u+1-((v-u+1)**2 -4*v if 4*v < (v-u+1)**2 else 0 )**(1/2)) / 2

If there is a Geometry Nodes solution please feel free to post it.

Answer 1

Unfortunately, Z Math Surface uses the X Size and Y Size parameters to set a symmetric range of $-X Size \le X \le X Size$ and $-Y Size \le Y \le Y Size$. See below for an ugly way to make this work, but here's a simpler way using XYZ Function surfaces:

Simple solution

You can accomplish what you want using an X, Y, Z Surface, by setting $X = U$, $Y = V$, and using the U Min, U Max, V Min and V Max parameters. You also need to write the $Z$ equation using $U$ and $V$. Here's a simple example:

Here's your equation in parametric form: $$(v-u+1-((v-u+1)**2 -4*v if 4*v < (v-u+1)**2 else 0 )**(1/2) ) / 2$$

swapping domain solution

To do this with Z Math Surface you need to change your equation. Consider using a math surface with the domain $-.5 \le X \le .5$ by setting X Size to .5. In the original equation replace $X$ with $X+.5$ everywhere. This will give you a resulting domain of $0 \le X \le 1$. The problem is that the surface will be displayed offset by .5. You can remedy this, of course, by moving the surface once it is created to compensate, but now you have fudged both the surface and the domain. (Obviously this will work just as well for $Y$.)