# ContourPlot of implicit function doesn't work when I square the equation [duplicate]

When I try this

ContourPlot[e (-208 Sqrt[1 - e^2] - 205 Sqrt[4 - e^2] +
2 e^2 (Sqrt[1 - e^2] + Sqrt[4 - e^2])) + 400 Sin[\[Phi]]== 0, {\[Phi],
0, 2*\[Pi]}, {e, -1, 1}]


It works and give nice plot but when I square the equation

ContourPlot[(e (-208 Sqrt[1 - e^2] - 205 Sqrt[4 - e^2] +
2 e^2 (Sqrt[1 - e^2] + Sqrt[4 - e^2])) + 400 Sin[\[Phi]])^2== 0, {\[Phi],
0, 2*\[Pi]}, {e, -1, 1}]


Nothing comes out; same thing happens if I take Abs[]. I don't understand why these two are not equivalent to each other. Since usually I deal with complex equations, I need let the norm equal to zero, thus either square or Abs comes out. Is there a way to let the latter one works? Thanks

## marked as duplicate by ssch, Sjoerd C. de Vries, halirutan♦, Mr.Wizard♦Oct 22 '13 at 7:25

• If you're lucky you can do it symbolically too, with f being your function: sols = Solve[f[\[Phi], e]^2 == 0, {e}]; ParametricPlot[Evaluate[{\[Phi], e} /. sols], {\[Phi], 0, 2 Pi}] – ssch Oct 22 '13 at 4:49