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I am using the function ImplicitRegion[].

RegionPlot[ImplicitRegion[x^2 + y^2 == -1, {x, y}]]

enter image description here

which gives me a circle, though in the documentation variables are said to be real.

Adding additional assumption does not change anything:

RegionPlot[ Assuming[{x, y} \[Element] Reals, 
  ImplicitRegion[   x^2 + y^2 == -1, {{x, -\[Pi], \[Pi]}, {y, -\[Pi], \[Pi]}}]]]

How to fix this issue?

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If ImplicitRegion was working properly you would not get a plot. If you are trying to plot the equation:

funcs = y /. Solve[x^2 + y^2 == -1, y];

ReImPlot[funcs, {x, -Pi, Pi},
 PlotStyle -> {AbsoluteThickness[3], Automatic},
 PlotLegends -> Automatic]

enter image description here

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  • $\begingroup$ But is there a way to tell ImplicitRegion that (x,y) are real and thus there are no solutions? I am using this function in NIntegrate so it's important not to get fake contours. $\endgroup$ – Alexander Nikolaenko Apr 8 at 6:07
  • $\begingroup$ ImplicitRegion does not take an Assumptions option so you have to explicitly include the domain constraint. RegionPlot[ImplicitRegion[x^2 + y^2 == -1 && Element[{x, y}, Reals], {x, y}], {x, -Pi, Pi}, {y, -Pi, Pi}] will return an empty plot. $\endgroup$ – Bob Hanlon Apr 8 at 6:15
  • $\begingroup$ I tried changing -1 to 1 in your example and it gave an empty plot again. $\endgroup$ – Alexander Nikolaenko Apr 8 at 7:57
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    $\begingroup$ Yes; however, you indicated that your actual problem is integration. Both Integrate[1, {x, y} \[Element] ImplicitRegion[x^2 + y^2 == 1, {x, y}]] and Integrate[1, {x, y} \[Element] ImplicitRegion[x^2 + y^2 == -1, {x, y}]] return the expected results. As do Integrate[1, {x, y} \[Element] ImplicitRegion[x^2 + y^2 == 1 && Element[{x, y}, Reals], {x, y}]] and Integrate[1, {x, y} \[Element] ImplicitRegion[x^2 + y^2 == -1 && Element[{x, y}, Reals], {x, y}]] $\endgroup$ – Bob Hanlon Apr 8 at 13:00
  • $\begingroup$ oh, I see, thanks! Perhaps the problem is related to RegionPlot. $\endgroup$ – Alexander Nikolaenko Apr 8 at 15:43

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