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1

NSolve with adequate precision works well eqns = Sin[z + Sin[z + Sin[z]]] == Cos[z + Cos[z + Cos[z]]] && -3 < Re[z] < 3 && -3 < Im[z] < 3; roots = NSolve[eqns, z, WorkingPrecision -> 20]; And @@ (eqns /. roots) (* True *) Note that there are a large number of roots roots // Length (* 883 *)


0

Let's see what are the regions of interest: n = 4; Partition[ RegionPlot[#[[2]][#[[1]][Sqrt[e + x^2 + Sqrt[(x^2) (2*e + x^2)]]], 0], {x, -n, n}, {e, -n, n}, PlotLabel -> #, AxesLabel -> Automatic, LabelStyle -> Medium] & /@ Tuples@{{Re, Im}, {Less, Greater}}, 2] ...


4

Edited to simplify the derivation and reduce the length of the final result. If R21 and R32 are pure imaginary, then the equation is solved easily. R21 /. Solve[eqns /. {Im[R21] -> -I R21, Im[R32] -> -I R32}, term][[1]] (* (I p (2 b q^2 x + 2 b q^2 y - 4 b c x y - p^2 x y - 4 b c x z - p^2 x z))/(6 b p^2 q^2 - 4 a b q^2 x - p^2 q^2 x - q^4 x - ...


2

Here's how I handle roots of integer power: myroot[x_, n_, k_] := Root[#^n - x &, k] allroots[multivalexpr_] := FixedPoint[ Function[{expr}, Flatten[MapAt[ Function[{ex}, Replace[ex, myroot[sym_, i_] :> (myroot[sym, i, #] & /@ Range[i])]], expr, FirstPosition[expr, myroot[_, _], {0}]]]], multivalexpr] Then ...


2

I think what you want is something like this: Conjugate[f[x_, y_, z_]] ^:= cf[x, y, z] Derivative[d__][cf][x__] := Conjugate[Derivative[d][f][x]] D[Conjugate[f[x, y, z]], x] Conjugate[Derivative[1, 0, 0][f][x, y, z]] All I did here is to define the derivative of the function f to be another function cf which then can be given the property you want. ...


0

Limit[(1 - z)/(1 - Conjugate[z]), z -> 1, Direction -> #, Assumptions -> Element[z, Complexes]] & /@ {-1, 1, I, -I} (* {1, 1, -1, -1} *) Output is identical to that of Limit[z/Conjugate[z], z -> 0, Direction -> #, Assumptions -> Element[z, Complexes]] & /@ {-1, 1, I, -I}


0

It seems that Residue (and Series on which it is based) is unable to analyze all the cases in which the formula for the residue might vary. Reduce can do it, within its limitations, if we reduce the system 1/function == 0 && nu == pole. Here's the OP's example. fn = (w^(n/2 + I nu) ws^(-(n/2) + I nu))/(n/2 + I nu)^2; We need to add some ...


1

Like the OP, I find that Assuming[x_ ∈ Reals, FullSimplify[Conjugate[Exp[I x1] x2]]] (* E^(-I x1) x2 *) works, but Assuming[x_ ∈ Reals, FullSimplify[Conjugate[Exp[I x1] x2 + x3]]] (* E^(I x1) x2 + x3 *) Assuming[x_ ∈ Reals, FullSimplify[Conjugate[Exp[I x1] x2 + Exp[I x3] x4]]] (* E^(I x1) x2 + E^(I x3) x4 *) return the wrong answer without ...


1

a = 2; y = 0.0000000001; d = y; z = I a + I y + d - I t; Plot[Im[Log[z^2 + a^2]], {t, -1, 5}, PlotRange -> All, GridLines -> {None, {-Pi, Pi}}] Everything looks ok to me, considering that the imaginary part is practically negligible. Plot[Re[z^2 + a^2], {t, -1, 5}, PlotRange -> All] Plot[Im[z^2 + a^2], {t, -1, 5}, PlotRange -> All]


5

The choice of branch cuts is made by re-defining the argument of the complex number under the square root. This can be done by using the approach of this answer: arg[z_, σ_: - Pi] := Arg[z Exp[-I (σ + Pi)]] + σ + Pi; sqrt[x_, σ_: - Pi] := Sqrt[Abs[x]] Exp[I arg[x, σ]/2] Here, the parameter $\sigma$ is the location of the branch cut of the square root. ...


5

Not all parts of your file are recognized as numeric data during the Import. u = Import["file.dat"] Head /@ Flatten[u] $\ $ {String, Integer, Integer, String} You can convert it for example with SetAttributes[stringToNumber, Listable]; stringToNumber[s_String] := ToExpression[StringReplace[s, {"e" -> " 10^", "i" -> " I"}]]; stringToNumber[s_] := ...


6

One possibility: try this (you can use Import on your file instead): lines = ImportString[ "1+1e-18i 24 42 23.43e-23i", "Lines"] Then: result = Map[StringSplit, lines] And: Map[Interpreter["ComplexNumber"], result, {2}] This gives: {{1.` + 1.`*^-18 I, 24}, {42, 0.` + 2.343`*^-22 I}}



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