FindRoot with two equations

*To*: mathgroup at smc.vnet.net*Subject*: [mg24885] FindRoot with two equations*From*: AES <siegman at stanford.edu>*Date*: Sat, 19 Aug 2000 04:45:57 -0400 (EDT)*Organization*: Stanford University*Sender*: owner-wri-mathgroup at wolfram.com

I'm working with a lengthy function fC that contains many variables x1, x2, x3, . . ., all purely real, but with some explicit I's in the formula (the "Sqrt[-1]" kind of I). I want to set the variables x3, x4, and upward to fixed values, and then solve for values of x1, x2 that make f=0. I break fC into its real and imaginary parts that are functions of x1 and x2 only by creating a rule myValues = {x3->x30, x4->x40, etc} and then breaking fC into real and imaginary parts using [1] fR[x1_, x2_] = ComplexExpand[Re[fC]] /. myValues fI[x1_, x2_] = ComplexExpand[Im[fC]] /. myValues At this point I should have two equations in two unknowns, so I try FindRoot[{ fR[x1, x2] == 0, fI[x1, x2] == 0}, {x1, x10}, {x2, x20}] Several hours of messing with different approaches to this produce different variations on the error message FindRoot :: "frnum": "Function is not a length 2 list of numbers at (X1, x2) = (x10, x20)" I've done all the checks on my formulas and expressions I can think of -- is there some secret I'm missing here? Thanks siegman at stanford.edu [1] As a cranky side note, there must a special slot somewhere for whoever thought up the ComplexExpand[Re[fC]] notation -- instead of the much more intuitive syntax Re[ComplexExpand[fC]] -- and then (so far as I can find) documented it nowhere in the Mathematica manual or Help messages.