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2

I seems clear to me that Nasser in his comment has the right answer. To make this clear to everyone, let's look at a simple case of the OP's problem. f[x_, y_] = {x + I x^2, y + I y^2}; df = D[f[x, y], x] {1 + 2 I x, 0} The OP argues that somehow ComplexExpand[df] should remove any trace of I from {1 + 2 I x, 0}; i.e., that 1 + 2 I x and x should ...

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ReplaceAll works on "the structure", not the "pretty printed" form. FullForm[I] gives Complex[0, 1] and I /. Complex[0, 1] -> Complex[0, -1] gives -I

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If you have a list $b$ of complex numbers, the quickest way I have found to plot them is: ListPlot[Table[{Re[b[[k]]], Im[b[[k]]]}, {k, 1, 4}]] Here $b$ has 4 complex numbers in it.

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For basic function construction see: http://reference.wolfram.com/language/tutorial/MakingDefinitionsForFunctions.html You did not explain how you want to plot a complex valued function but here's a start: c = -0.5 + 0.2 I; z[0] := c z[n_Integer?Positive] := z[n - 1]^2 + c Array[z, 20] {-0.29 + 0. I, -0.4159 + 0.2 I, -0.367027 + 0.03364 I, -0.366423 + ...

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p[z_] = z^2 + z; c[t_] = Cos[t] + Sin[t] I; ParametricPlot[{Re[p[c[t]]], Im[p[c[t]]]}, {t, 0, 2 Pi}]

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You might find useful the following rule: rule={Complex[0, 1] -> -I, Complex[0, -1] -> I}; For example, let us look at the following mildly complex expression: expr = (a + b*I + Sqrt[c - d*I])/(a - b*I + Sqrt[c + d*I])^(3/2); One finds the instance expr /. {a -> 1, b -> 1, c -> 1, d -> 1} // N (* 0.546701 + 0.402882 I *) ...

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The differential equation, its initial condition, and its boundary conditions are translationally invariant in space. Consequently, the solution must be independent of x and y. Indeed, solving the equations as given in the Question does give a spatially constant solution that oscillates in time. For instance, DensityPlot[Evaluate[Re[A[x, y, 10000]] /. ...

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