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0

You probably need to distribute every product over sums in your expression in order for Conjugate to propagate properly: distributeProducts[expr_] := Replace[expr, t : Times[___] :> Distribute[t], {0, \[Infinity]}] distributeProducts@ Conjugate[ fc (2 c E^((2 I fc \[Pi] (R1 + R2))/\[ConstantC]) mu1 + d E^((4 I fc \[Pi] R1)/\[ConstantC]) mu2)] ...


3

For k/z = 1, and integrating by parts: ClearAll["Global`*"] Iv[t_] := Exp[-a*t]*t^2; u[t_] := ArcTanh[Sqrt[(t^2 - 1)/t^2]]; v = Integrate[Iv[t], t]; Du = Simplify@D[u[t], t]; Int == u[t]*v - Integrate[Du*v, t] HoldForm[Integrate[Exp[-a*t]*t^2*ArcTanh[Sqrt[(t^2 - 1)/t^2]], {t, 1, Infinity}] == Limit[-((E^(-a t) (2 + 2 a t + a^2 t^2) ArcTanh[Sqrt[(-1 + ...


4

Preliminary post For k/z = 1 there is a closed form result: I[1,1,a]=((4 + a^2) BesselK[0, a] + a (4 BesselK[1, a] + a BesselK[2, a]))/(2 a^3) The derivation and the extension to k/z != 1 requires some manual interaction to help Mathematica which we will show in the following. Solution Summary As the integral to be calculated is returned unevaluated ...


9

You can take Michael Trott's code and modify it a bit to easily plot these surfaces Import["http://www.mathematicaguidebooks.org/V6/downloads/\ RiemannSurfacePlot3D.m"] rsurf[func_] := Grid[{{RiemannSurfacePlot3D[w == func, Re[w], {z, w}, ImageSize -> 400, Coloring -> Hue[Rescale[ArcTan[1.4 Im[w]], {-Pi/2, Pi/2}]], PlotPoints ...


2

Here's a polynomial interpolation method, which can be be found in Chapter 5 of Boyd (2014). nn = 64; z0 = w1 + w2; rr = 1.1 w1; ff = N[WeierstrassP[z0 + rr #, inv] - L, Precision[#]] &; wprec = MachinePrecision; tj = 2 Pi*Range[0, nn - 1]/nn; wj = N[Exp[I tj], wprec]; fj = ff /@ wj; (* f[zj] *) aa = InverseFourier[fj]/Sqrt[nn]; (* Rough check of ...


1

You can use the TargetFunctions option of ComplexExpand[]: FullSimplify[ComplexExpand[(2 (ArcCot[E^(-((I ϕ)/2)) r] + ArcCot[E^((I ϕ)/2) r]))/π, TargetFunctions -> {Re, Im}], r > 1 && -π < ϕ < π] (2 (ArcCot[Sec[ϕ/2] (r + Sin[ϕ/2])] + ArcCot[r Sec[ϕ/2] - Tan[ϕ/2]]))/π One could probably argue for the ...


1

The answer is completely satisfactory. It returns the result in terms of Arg which is always a real number. If look for its imaginary part Im@ComplexExpand[(2 (ArcCot[E^(-((I \[Phi])/2)) r] + ArcCot[E^((I \[Phi])/2) r]))/\[Pi]] 0 So your answer is a Real quantity.


1

$Version (* "10.4.1 for Mac OS X x86 (64-bit) (April 11, 2016)" *) expr = Assuming[{r > 1, -π < ϕ < π, ϕ ∈ Reals}, Sum[ r^-n 4/π Sin[n π/2]/n Cos[n ϕ/2], {n, 1, ∞}]] // FullSimplify expr2 = Assuming[{r > 1, -π < ϕ < π, ϕ ∈ Reals}, expr // ComplexExpand[#, TargetFunctions -> {Re, Im}] & // ...


10

I can't find any confirmations in the documentation, but through numerical and visual checks I think when at least one input to Mod is not real, we have This definition doesn't equal the definition one would think to have over the reals, so my guess is a piecewise definition is used to modify the function over the real line. Testing mod[z_, n_, d_] := z ...


2

Try this: Sqrt[Im[z1]^2 - 2 Im[z1] Im[z2] + Im[z2]^2 + (Re[z1] - Re[z2])^2] // ComplexExpand // Simplify (* ((z1 - z2)^4)^(1/4) *) Have fun!


10

The following is not particularly fast and could be more accurate but does make progress toward the goals set in the Question. To begin, consider the analytical properties of f, as defined in the Question. By observation, it has branch points at {I Sqrt[6/5] ξ, I Sqrt[2/5] ξ} and their conjugates. Poles are obtained by p /. ...


3

Let, $$ f=\frac{z-e^{i \pi/2 n}}{z^{2 n}+1} $$ Then, in Mathematica code, the following seems to work, Limit[f, z -> Exp[(I Pi)/(2 n)], Assumptions -> n >= 1/2] which gives the result, $$ -\frac{e^{i \pi/2 n}}{2n} $$ The problem isn't that Mathematica is handling the $0/0$ limit incorrectly. It's that Mathematica, without additional information, ...



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