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This bug has been fixed in V11.2.0 or earlier. Regression in V13.0.0 or earlier.
Consider
Plot[Im[(1 + (-1)^(2/3)*x)/((-1)^(2/3) + x)], {x, 0, 1}]
At least in Mma 11 it gives the following picture:
Meanwhile,
Plot[Im[ComplexExpand[(1 + (-1)^(2/3)*x)/((-1)^(2/3) + x)]], {x, 0,
1}]
gives
What is going on here?
It seems that it was fixed in V11.2.0 or earlier. It was fine until V12.0.0 or later. And there's a regression in V13.0.0 or earlier.
V11.2.0
V12.0.0
V13.0.0
Plotis doing some transformation that it shouldn't. $\endgroup$Plot[Im[1/((-1)^(2/3) + x)], {x, 0, 1}]works as expected. $\endgroup$Plot[Im[(1 + (-1)^(2/3)*x)/((-1)^(2/3) + x)], {x, 0, 1}, WorkingPrecision -> 10]. It works with a black-box machine-precision function:f[x_?NumericQ] := Im[(1 + (-1)^(2/3)*x)/((-1)^(2/3) + x)]; Plot[f[x], {x, 0, 1}]. And a third way:Plot[Im[(1 + (-1.)^(2/3)*x)/((-1)^(2/3) + x)], {x, 0, 1}]$\endgroup$_Complexpattern. Maybe Mma erroneously assumes the argument ofImis real as there are noComplexheads? But then whyPlot[Im[1/((-1)^(2/3) + x)], {x, 0, 1}]work fine? $\endgroup$Plot[N@Im[(1 + (-1)^(2/3)*x)/((-1)^(2/3) + x)], {x, 0, 1}]. AddEvaluate, and YEP:Plot[Evaluate@N@Im[(1 + (-1)^(2/3)*x)/((-1)^(2/3) + x)], {x, 0, 1}]$\endgroup$