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我想使用类似的东西:
等高线图 [Abs[z-1] == 2]
并将 z 定义为 = x + iy
我在某个地方看到了一个带有 With 函数的示例,但我再也找不到它了,而且我的所有尝试都没有成功。
是的你可以。您只需要确保With将ContourPlot:
With
ContourPlot
With[{z = x + I y}, ContourPlot[Abs[z - 1] == 2, {x, -2, 2}, {y, -2, 2}]]
您还可以使用Evaluate:
Evaluate
ContourPlot[ With[{z = x + I y}, Abs[z - 1] == 2] // Evaluate, {x, -2, 2}, {y, -2, 2}]
这通常是绘图函数的情况,它几乎总是以非标准方式评估它们的参数。