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3

This seems like a bug. Additional examples that would need to be explained if it is not: foo[2 I u Sin[x]] /. foo[Complex[0, _] u Sin[x]] :> bar foo[2 I u Sin[x]] /. foo[Complex[0, _] (p : u) Sin[x]] :> bar foo[2 I u Sin[x]] /. foo[Complex[0, _] HoldPattern[u] Sin[x]] :> bar foo[2 I u Sin[x]] bar bar So the problem is unrelated to ...


4

I don't know why Exp[2 I u Sin[x]] /. Exp[Complex[0, a_] u Sin[x]] :> a doesn't work as expected, except that MatchQ[Exp[2 I u Sin[x]], Exp[Complex[0, a_] u Sin[x]]] gives False However, since the apparently more general MatchQ[Exp[2 I u Sin[x]], Exp[Complex[0, a_] u_ v_]] gives True you could use Exp[2 I u Sin[x]] /. Exp[Complex[0, ...


1

It seems that a direct replacement (using ReplaceAll and RuleDelayed) may be adequate: Arg[1 + I a] /. Arg[1 + I*x_] :> ArcTan[x] ArcTan[a]


1

Starting with your code, corrected as in my comment, z[x_, y_] := x + y*I F[z_] := (25*Pi*z*I)/(1 + 10*Pi*z*I) you can plot the imaginary part of F as follow ContourPlot[Im[F[z[x, y]]], {x, -.2, .2}, {y, -.2, .2}, PlotRange -> All, Contours -> Range[-5, 5, .5], ContourLabels -> True] and similarly for other quantities. Many different ...


1

RegionPlot[ Norm[x + I y + .15] < 0.6 && Pi/4. <= Arg[x + I y] <= Pi, {x, -1, 1}, {y, -.1, 1}, GridLines -> {{-.15}, {0}}, PlotPoints -> 60, AspectRatio -> Automatic]


1

If it is just the displayed form you are after, you can also go with HoldForm like so: HoldForm@D[Conjugate[g[x]],x] This will carry over throughout the notebook without further ado, until you call ReleaseHold on it. I hope this might be of some help to you.


1

This can be done by directly defining the outcome of Derivative when applied to g in the two combinations that you seem to be interested in: Derivative[1][g][x_] := d[g[x]] Derivative[1][Conjugate][g[x_]] := d[Conjugate[g[x]]]/d[g[x]] On the second line, I used the fact that g is a generic function whose derivative under a Conjugate by default invokes ...


1

Someone else posted a working answer yesterday, which is gone today; maybe it was deleted by the author for some reason. The solution was to define a function arg[num_] := ArcTan[ComplexExpand[Im[num]]/ComplexExpand[Re[num]]]


3

If you mean the vertical line, this is created by only one point. p = Plot[N[Im[isinh]], {e, 0, 2}, PlotRange -> Full, Exclusions -> None, WorkingPrecision -> Infinity]; list = Cases[p, Line[x_] :> x, -1]; ListPlot[list, PlotRange -> All] If you delete this point then: point = Cases[p, {x_ /; Abs[x - 1] < 0.01, y_ /; y < 0.5}, ...


2

Reversing the order of integration produces a solution: ans= Integrate[HeavisideTheta[1 - x - y]/(x 100^2 - y (1 - y) 90^2), {y, 0, 1}, {x, 0, 1}, PrincipalValue -> True] (* Log[(100*10^(38/81))/(81*19^(19/81))]/10000 *) N[ans] (* 0.000060027526501455836 *) Solutions of this sort are what I would expect based on outlining a pencil-and-paper ...



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