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1

Try Refine Refine[Conjugate[a], a \[Element] Reals] or Simplify[Conjugate[a], Assumptions :> {a \[Element] Reals}]


1

From a mathematical point of view, this is a rather trivial question. But doing it with Mathematica is IMO not quite trivial. Here is one way. We write z = x + i y and compute the real and imaginary parts of z* and z^2 by using ComplexExpand; this function assumes that the variables x and y are real! z = x + I y; {ComplexExpand /@ (Re[Conjugate[z]] == ...


4

You mean like this? Reduce[x^4 == 1 - I, x] // N x==-1.06955+0.212748 I||x==-0.212748-1.06955 I||x==0.212748+1.06955 I||x==1.06955-0.212748 I


3

It's not the exact output you requested but in case you are not aware of the second parameter of Rationalize: Rationalize[N[4/3 + I Sqrt[2]/3], 1*^-6] 4/3 + (272 I)/577 If your hybrid output really is desired then perhaps building on m_goldberg's deleted answer: # + Defer[#2 I] & @@ Rationalize /@ {Re@#, Im@#} & @ N[4/3 + I Sqrt[2]/3] 4/3 ...


1

Here's an interesting way to do it using complex phase-amplitude plots: hue = Compile[{{z, _Complex}}, {(1.0 Arg[-z] + \[Pi])/(2 \[Pi]), Exp[1 - Max[Abs[z], 1]], Min[Abs[z], 1]}, CompilationTarget -> "C", RuntimeAttributes -> {Listable}]; ComplexPlot[f_, {x0_, x1_, \[Delta]x_}, {y0_, y1_, \[Delta]y_}] := Image[hue[ f[#[[All, All, 1]], ...


1

Expanding on @Nasser's answer f = (1.6 - 0.005 I) E^(-0.5 p^2 + (2. I) p q + (6.5 - 11 q) q); Partition[ Plot3D[#[f], {p, -2, 2}, {q, -2, 2}, PlotRange -> All, PlotPoints -> 51, AxesLabel -> (Style[#, 14, Bold] & /@ {p, q, #["f"]})] & /@ {Re, Im, Abs, Arg}, 2] // Grid


1

Since the function is complex valued, you have to find its real and imaginary and plot these. This type of question has been asked before many times, so I am sure this is duplicate. Clear[p, q] f = (1.6 - 0.005 I) E^(-0.5 p^2 + (2. I) p q + (6.5 - 11 q) q) p1 = Plot3D[Re@f, {p, -2, 2}, {q, -2, 2}, PlotRange -> All] p2 = Plot3D[Im@f, {p, -2, 2}, {q, -2, ...


3

The following gives what you intended: Refine[Expand[P[x, y]^2], (x|y|beta) \[Element] Reals] (* ==> Conjugate[z[y]]^2/E^((2*I)*beta*x) + 2*Conjugate[z[y]]*z[y] + E^((2*I)*beta*x)*z[y]^2 *) In cases where you can live with expansion of complex exponentials into Sin and Cos you can also use ComplexExpand[P[x, y]^2, z[y], ...



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