Re: Complex and Solve

*To*: mathgroup at smc.vnet.net*Subject*: [mg125162] Re: Complex and Solve*From*: Howard Lovatt <howard.lovatt at gmail.com>*Date*: Sat, 25 Feb 2012 01:56:24 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <201202211115.GAA08330@smc.vnet.net>

Thanks Bob, That is even better. I am somewhat surprised that you need to manually specify $Assumptions and that since this is global that Solve doesn't take this into account anyway. But this solution is much better than my original, so thanks again, -- Howard. On 23 February 2012 23:07, Bob Hanlon <hanlonr357 at gmail.com> wrote: > Actually I should have checked, with version 8 Solve can handle > inequalities so either Reduce or Solve works here. > > $Assumptions = rfe > 0 && l > 0; > > r = 1; > \[Omega] = 300; > vrms = 2; > irms = 1/2; > \[Phi] = 30 Degree; > > impedance = r + 1/(1/rfe + 1/(I \[Omega] l)) // FullSimplify; > > voltage = vrms (Cos[\[Phi]] + I Sin[\[Phi]]) // Simplify; > > Solve[{irms == voltage/impedance, $Assumptions}, {rfe, > l}][[1]] // FullSimplify > > {rfe -> (1/11)*(-7 + 30*Sqrt[3]), l -> (1/600)*(17 - 4*Sqrt[3])} > > Alternatively, > > Solve[{irms == voltage/impedance, Im[l] == 0, Im[rfe] == 0}, {rfe, > l}][[1]] // FullSimplify > > {rfe -> (1/11)*(-7 + 30*Sqrt[3]), l -> (1/600)*(17 - 4*Sqrt[3])} > > > Bob Hanlon > > On Thu, Feb 23, 2012 at 5:46 AM, <howard.lovatt at gmail.com> wrote: > > Thanks Bob, > > > > Your solution is much better than mine. I had no idea that Reduce could > be used like you showed. > > > > Bizarre that Solve and Reduce behave so differently! > > > > Thanks again, > > > > -- Howard. > > > > > > -- > Bob Hanlon > -- -- Howard.

**References**:**Complex and Solve***From:*howard.lovatt@gmail.com