a = 0; b = 10; ContourPlot[{y + 3 ArcTan[y/(x - a)]}/20, {x, -b,
  b}, {y, -b, b}, Exclusions -> {x == a}, PlotPoints -> 50,
ContourStyle -> {Red}, Frame -> True, Axes -> True,
ContourShading -> None, PlotLegends -> {"\[Psi](x,y)"}]

2楼: Originally posted by xzczd at 2015-01-07 22:28:07
很遗憾，如果你坚持要用这个方程，那么这就是个不可能的任务，因为ContourPlot忠实地画出了你输入的东西：

a = 0;
Limit & /@ {-1, 1}
(* {y + (3 \ Sqrt)/(2 y), y - (3 \ Sqrt)/(2 y)} *)


如你所见 ...

a = 0;
Limit[y + 3 ArcTan[y/(x - a)], x -> 0, Direction -> #] & /@ {-1, 1}
(* {y + (3 \[Pi] Sqrt[y^2])/(2 y), y - (3 \[Pi] Sqrt[y^2])/(2 y)} *)

a = 0; b = 10;
ContourPlot[{y + 3 Arg[(x - a) + I y]}/20, {x, -b, b}, {y, -b, b},
  ContourStyle -> Red, Frame -> True, Axes -> True,
  ContourShading -> None] /.
Graphics[a__] :>
  Legended[Graphics[a],
   Placed[LineLegend[{Red}, {"\[Psi](x,y)"}], After]]

2楼: Originally posted by xzczd at 2015-01-07 22:28:07
很遗憾，如果你坚持要用这个方程，那么这就是个不可能的任务，因为ContourPlot忠实地画出了你输入的东西：

a = 0;
Limit & /@ {-1, 1}
(* {y + (3 \ Sqrt)/(2 y), y - (3 \ Sqrt)/(2 y)} *)


如你所见 ...

z1 = -(0.5*Log[x^2 + y^2] - Arg[x + y I])/(2 Pi);
ContourPlot[{z1}/300, {x, -10, 10}, {y, -10, 10},
ContourStyle -> {Red}, Frame -> True, Axes -> False,
ContourShading -> None,
Exclusions -> {x == 0, y == 0}, PlotPoints -> 50]

3楼: Originally posted by kabaisun at 2015-01-07 23:00:15
错位纠正了，还有一个疑问：ArcTan求角度和Arg求的不一样吗？...