# Conditional Plot of a function if another function satisfies something and in the case of complex variable

Let's say we have function $$g(\zeta)$$ and another map $$f(\zeta)$$, where $$\zeta$$ is a complex variable. I am trying to plot a graph of $$f(\zeta)$$ but given the condition of $$g(\zeta)$$.

For instance, I am trying to have a command like: $$\text{plot}\ f(\zeta)\ \text{blue if}\ g(\zeta)\leq 1$$

A similar question has been asked here: Conditional options in Plot

I followed the answer in that question but I cannot get what I want. Here is what I have tried:

1. Firstly, I define the conditional plot command, as we are dealing with two variables $$(\zeta=x+iy)$$, I changed "plot" in the original command to contour plot.

ConditionalPlot[fun_,condition_, varrange_, trueopts_, falseopts_] :=
Module[{plottrue, plotfalse},
plottrue = ContourPlot[If[condition, func], varrange, trueopts];
plotfalse = ContourPlot[If[Not[condition], func], varrange, falseopts];
Show[plottrue, plotfalse, PlotRange -> All]]

2. Then, I used this command to plot:

ConditionalPlot[f (x + Iy), g (x + Iy) <= 1, {x, -3, 3}, {y, -3, 3},
{Filling -> Axis, FillingStyle -> LightGreen, PlotStyle -> Dashed},
{Filling -> Axis,FillingStyle -> LightRed, PlotStyle -> Thick}]


However, the output of Mathematica is just the code I have entered, without producing the graph. I don't really know where the problem is.

I also tried to use explicit formula of f(x+Iy) and g(x+Iy) and the problem is still there.

I also tried this command:

ConditionalPlot[f (x + Iy),g (x + Iy) <= 1, {x, -3, 3}, {y, -3, 3},
{PlotStyle -> Dashed},{PlotStyle -> Thick}]


since I think maybe Filling--Axis is not right for contour plot, however it still cannot work.

Any idea is welcome! Thank you!

I also tried another way to do this question

The idea is from this post: Plot 3D with conditions and variables dependent on each other

Here is my command:

ContourPlot[
ConditionalExpression[f (x + Iy), g (x + Iy) <= 3 && g (x + Iy) >= 1],
{x, -3, 3}, {y, -3, 3},
PlotPoints -> 100, MaxRecursion -> 4]


However, Mathematica then produced me a graph with nothing. Just empty. So I think there is something it cannot understand...

### Edit

There are quite a few minor errors in your code and, I think, two major ones:

• f and g have to be real-valued functions of a complex variable.
• you should be plotting f with Plot3D if you want to use options like Filling.

I also think that for Plot3D, the filling option Filling -> Bottom will give a better plot than Filling -> Axis.

Taking the above into account and fixing the minor errors as well, produces the following:

Clear[ConditionalPlot]
ConditionalPlot[func_, condition_, xrng_, yrng_, trueopts_, falseopts_] :=
Module[{plottrue, plotfalse},
plottrue = Plot3D[If[condition, func], xrng, yrng, trueopts];
plotfalse = Plot3D[If[Not[condition], func], xrng, yrng, falseopts];
Show[plottrue, plotfalse, PlotRange -> All]]

f[z_] := Re[z^2]
g[z_] := Im[z]

ConditionalPlot[f[x + I y], g[x + I y] <= 1, {x, -3, 3}, {y, -3, 3},
{Filling -> Bottom, FillingStyle -> LightGreen,
PlotStyle -> Lighter[Green, .2], Lighting -> "Neutral"},
{Filling -> Bottom, FillingStyle -> LightRed,
PlotStyle -> Lighter[Red, .2], Lighting -> "Neutral"}] As to the matter you bring up in your comment this answer, what you suggest will work.

Clear[ConditionalPlot]
ConditionalPlot[func_, condition_, xrng_, yrng_, opts_] :=
Module[{plottrue},
plottrue = Plot3D[If[condition, func], xrng, yrng, opts];
Show[plottrue]]

ConditionalPlot[f[x + I y], g[x + I y] <= 1, {x, -3, 3}, {y, -3, 3},
{Filling -> Bottom, FillingStyle -> LightGreen,
PlotStyle -> Lighter[Green, .2], Lighting -> "Neutral"}] But don't think that now the use of Module and Show are overkill? Wouldn't the following, which gives the same plot, be better?

Clear[ConditionalPlot]
ConditionalPlot[func_, condition_, xrng_, yrng_, opts_] :=
Plot3D[If[condition, func], xrng, yrng, opts]

• Amazing!!! I have an another question that, what should I do if I want Mathematica do nothing if the condition is false? Feb 13, 2019 at 1:57
• Is that okay if I just remove plotfalse in the show command? Feb 13, 2019 at 1:59