Re: Abs and derivative problems

*To*: mathgroup at smc.vnet.net*Subject*: [mg14657] Re: Abs and derivative problems*From*: Paul Abbott <paul at physics.uwa.edu.au>*Date*: Sat, 7 Nov 1998 02:10:05 -0500*Organization*: University of Western Australia*References*: <71q9qu$j25@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

sylvan wrote: > I could not calculate the modulus of a complex expression containing > imaginary parts in both denominator and numerator with Mathematica. An > Example: > > (a + I b) / (c + I d) > > a,b,c,d (real) symbolic variables. > > In pratice, this should be absolutely trivial. ComplexExpand is not > effective. > How do you "tell" mathematica that your variables are real ?? Actually, in the trivial example above, you can use ComplexExpand[(a + I b) / (c + I d), TargetFunctions->{Re,Im}] to get what you want. > I included an example below (cell format, you can cut and paste). In the Notebook fragment below, I show a simple trick for computing the complex conjugate of complex expressions with real variables. The basic idea is to compute the conjugate using the replacement Complex[a_, b_] :> Complex[a, -b] actually implemented using the built-in SuperStar function (so that you can compute conjugates by raising them to the power *): SuperStar[x_] := x /. Complex[a_, b_] :> Complex[a, -b] The point then is that Abs[z]^2 = SuperStar[z] z which enables you to quickly and efficiently compute the type of expression you want. Notebook[{ Cell[BoxData[ \(TraditionalForm \`\(exp = \(b0\ \[CapitalDelta]z\ \((Es - \[ImaginaryI]\ \[Eta]\ \[Omega])\)\)\/\(\(-m\)\ \[Omega]\^2 + m\ \[Omega] - \[ImaginaryI]\ m\ \[Gamma]\ \[Omega] + b0\ \((Es - \[ImaginaryI]\ \[Eta]\ \[Omega])\)\); \)\)], "Input"], Cell[BoxData[ \(TraditionalForm \`\(x_\^*\) := x /. Complex(a_, b_) \[RuleDelayed] Complex(a, \(-b\))\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`\(\(exp\^*\) exp // ExpandAll\) // Simplify\)], "Input"], Cell[BoxData[ \(TraditionalForm \`\(b0\^2\ \[CapitalDelta]z\^2\ \((Es\^2 + \[Eta]\^2\ \[Omega]\^2) \)\)\/\(\((Es\^2 + \[Eta]\^2\ \[Omega]\^2)\)\ b0\^2 + 2\ m\ \[Omega]\ \((\(-\[Omega]\)\ Es + Es + \[Gamma]\ \[Eta]\ \[Omega])\)\ b0 + m\^2\ \((\[Gamma]\^2 + \((\[Omega] - 1)\)\^2)\)\ \[Omega]\^2\)\)], "Output"] }, Open ]] } ] > Also, replacement rules like //. z[t_] -> t^2 do not work well on > expressions like z'[t] + b z[t]. the result is z'[t] + b t^2... I > could not force it to Evaluate z'[t] or D[z[t], t]. One simple, general, and reasonably elegant way is to use pure functions, e.g., z'[t] + b z[t] /. z -> Function[{t}, t^2] Another is to explicitly compute all derivatives: rule = z[t] -> t^2; z'[t] + b z[t] /. {rule, D[rule,t]} Cheers, Paul ____________________________________________________________________ Paul Abbott Phone: +61-8-9380-2734 Department of Physics Fax: +61-8-9380-1014 The University of Western Australia Nedlands WA 6907 mailto:paul at physics.uwa.edu.au AUSTRALIA http://www.physics.uwa.edu.au/~paul God IS a weakly left-handed dice player ____________________________________________________________________