# Getting an Accurate Transformed Region

I would like to get an accurate plot of the image of concentric circles under the transformation $$f(z) = \log(1+z).$$

I've defined $$\cal R$$ as the union of a few circles:

p[x_, y_][\[Alpha]_] := x^2 + y^2 - \[Alpha]^2;
m = Table[ImplicitRegion[p[x, y][\[Alpha]] == 0, {x, y}],
{\[Alpha], Range/7}];
\[ScriptCapitalR] = RegionUnion[m];
a = Region[\[ScriptCapitalR], BaseStyle -> RGBColor[0, 0, .8, .7],
Frame -> True];


Now the function $$f(z)$$ is defined in terms of its real and imaginary parts:

f = Evaluate[{1/2 Log[(1 + x)^2 + y^2], ArcTan[y/(1 + x)]}] &;

\[ScriptCapitalE] = TransformedRegion[\[ScriptCapitalR], f];
b = Region[\[ScriptCapitalE], BaseStyle -> RGBColor[1, 0, 0, .7],
Frame -> True];


$$\cal E$$ is the transformed region. We then plot $$a$$ and $$b$$ the regions defined by $$\cal R$$ and $$\cal E$$, respectively.

GraphicsRow[{a, b}] My question is this: All of the red curves look nice with the exception of the outermost one. This curve should go off to infinity (to the left) as $$z \rightarrow -1$$ but Mathematica wants to connect it. Any suggestions?

UPDATE

Although the answers in the comments work and are expedient, there still remains a question. Obviously, we cannot get the solution all the way out to the point at infinity. Still what if we wanted to plot a solution valid in the region $$x \ge -10$$, for example? How can we improve the accuracy by, for example, specifying more sample points as Mathematica does its computations?

• Truncating the plot range on it should work. Aug 5 '19 at 20:20
• GraphicsRow[{a, Show[b,PlotRange->{{-5,1},{-2,2}}]}]
– Bill
Aug 5 '19 at 20:30
• Yes, this is a concise elegant solution! Works very well for this case. Thank you!
– mjw
Aug 5 '19 at 20:44

GraphicsRow[{a, Show[b, PlotRange -> {{-5, 1}, {-2, 2}}]}] 