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6

I suspect that ContourPlot doesn't really have a notion of the orientation of curves in a way that will let you preserve them after applying arbitrary transformations. ParametricPlot is probably the way to go. Start with parametric representations of the hyperbolas with the desired orientation: hyperbola1[t_] := {#, -#} &@{Csc[t], -Cot[t]}; ...


6

Still guessing... is the expected output something like the following? ContourPlot[{Re[(u + I v)^2] == 1, Im[(u + I v)^2] == 2, u == 1, v == 2}, {u, -4, 4}, {v, -4, 4}, Axes -> True, Frame -> False, BaseStyle -> Arrowheads[{{.05, .5}}], PlotLegends -> "Expressions", ContourStyle -> ...


2

If you accept the position of the arrowheads ... r1 /. Line -> Arrow


22

A major point behind the video is that Mobius transformations are simplest when viewed on the sphere. Thus, we'll never actually define a Mobius transformation - we'll do that part on the sphere. Of course, we will need to project back and forth. Here are the stereo graphic projection and it's inverse implemented as compiled functions for speed. This is ...


3

The linear system is easily solved generally by first calulating the MatrixExp. Then we can extract the real and imaginary parts. Here we go The complex matrix A = {{0, 1}, {-2, -I}}; The matrix exp At = MatrixExp[t A] (* {{1/3 (2 Cos[t] + Cos[2 t] + I (2 Sin[t] - Sin[2 t])), 1/3 (I (-Cos[t] + Cos[2 t]) + Sin[t] + Sin[2 t])}, {1/ 3 (2 I (Cos[t] ...


5

Almost a dozen alternatives with timings: ClearAll[f1, f2, f3, f4, f5, f6, f7, f8, f9, f10, f11]; f1[w_] := w /. {a_, Complex[_, b_]} :> {a, b}; f2[w_] := {#, Im@#2} & @@@ w; (* my favorite ... credit: ubpdqn *) f3[w_] := w /. Complex[_, b_] :> b; f4[w_] := MapAt[Im, w, {{All, -1}}]; f5[w_] := Module[{s = Im /@ w}, s[[All, 1]] = w[[All, 1]]; s]; ...


3

w = { {0.01, 99 + 0.00001414 I}, {0.15, 6.6370108 + 0.003144129 I}, {0.25, 3.9515722 + 0.00854493297 I}, {6, 0.10041 + 0.28132187 I} }; res = Replace[w, {x_, complex_} :> {x, Im @ complex}, {1}] {{0.01, 0.00001414}, {0.15, 0.00314413}, {0.25, 0.00854493}, {6, 0.281322}} ListLogPlot[res, PlotTheme -> "Detailed"]


1

Sorry, this is too long for a comment, but I thought it might help! Update I also tried, interestingly, c = Table[Exp[-t (1 + 256 I Pi )], {t, 0, 1, 1/30000}]; // AbsoluteTiming (* 0.109373 seconds *) d = Table[Exp[t (-1 - 256 I Pi )], {t, 0, 1, 1/30000}]; // AbsoluteTiming (* 0.140624 seconds *) Though this method in fact produces a slightly different ...


1

How about using NSolve like this. As you know z roots are 2-sets. but I have tried the one of things. eqn[x_, y_, z_] := z^2 + 3 z + x^2 + y^2 == 0 opts := Sequence[ PlotLegends -> Automatic, Exclusions -> {x^2 + y^2 == 9/4}, ExclusionsStyle -> Red, ColorFunction -> "SunsetColors"]; rec = ContourPlot[ Re[z] /. NSolve[eqn[x, y, ...


0

Solve[{(1-I)/Sqrt[2]==Exp[I alpha] Tan[beta],0<beta<Pi,0<alpha<2Pi}//ComplexExpand, {alpha,beta},Method->Reduce]


1

Just to add a couple of more observations to Nasser's. Case 6 As Daniel Lichtblau hints at in a comment, if we use an exact 37/10 in place of the approximate 3.7, we get an exact result with a zero imaginary component: Integrate[PDF[NormalDistribution[14, 37/10], x], {x, 15, Infinity}] N@% (* 1/2 Erfc[(5 Sqrt[2])/37] 0.393476 *) Case 7 Such a small ...


4

You may use the following: Reduce[(1 - I)/Sqrt[2] == ExpToTrig[Exp[I alpha]]*Tan[beta] && 0 < beta < Pi && 0 < alpha < 2 Pi]; {ToRules[%]} // FullSimplify {{alpha -> (7 π)/4, beta -> π/4}, {alpha -> (3 π)/4, beta -> (3 π)/4}} Although I'm not sure why it doesn't work without the ExpToTrig[] thing



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