RE: To verify Cauchy-Riemann relations in complex variable graphically

*To*: mathgroup at smc.vnet.net*Subject*: [mg39208] RE: [mg39176] To verify Cauchy-Riemann relations in complex variable graphically*From*: "David Park" <djmp at earthlink.net>*Date*: Mon, 3 Feb 2003 01:11:01 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

My friend Rip Pelletier pointed me to a better method to illustrate the Cauchy-Riemann relations for analytic functions. He pointed me to Tristan Needham's book "Visual Complex Analysis". In Chapter 5, Section I - Cauchy-Riemann Revealed, the CR conditions are related to the complex mapping. If a small patch of squares are mapped by an analytic function, then they go into another small patch in which all the squares have been amplified and rotated in exactly the same way. Fortunately, we have the ComplexMap package in Mathematica and can easily illustrate this for your functions. For example, for the Sin function... Needs["Graphics`ComplexMap`"] With[ {x = 2, y = 2, del = 0.01, f = Sin}, Show[GraphicsArray[{CartesianMap[ Identity, {x - del, x + del, del/5}, {y - del, y + del, del/5}, Axes -> False, DisplayFunction -> Identity], CartesianMap[ f, {x - del, x + del, del/5}, {y - del, y + del, del/5}, Axes -> False, DisplayFunction -> Identity]}], ImageSize -> 500]]; A square patch maps into a rotated square patch. Just change f and/or the mapping points for other cases. Use a pure function for z^2. For a case that is not analytic, and so the CR relations do not hold, use f = # + 2Abs[#] &. The squares go to parallelograms. David Park djmp at earthlink.net http://home.earthlink.net/~djmp/ From: Narasimham G.L. [mailto:google.news.invalid at web2news.net] To: mathgroup at smc.vnet.net Is it possible to have a semi transparent view of surfaces so that one may verify slopes by ParametricPlot3D for Cauchy-Riemann relations? The following is program for 3 functions Z^2, Z^3, Sin[Z].It was expected to check slopes at the line of intersection of Re and Im parts. R1=x^2-y^2 ; I1= 2 x y ; z2r=Plot3D[R1 , {x,-Pi,Pi},{y,-Pi,Pi} ]; z2i=Plot3D[I1 , {x,-Pi,Pi},{y,-Pi,Pi} ]; Show[z2r,z2i] ; 'Top view >> Re,Im Intxn'; Plot[{x ArcTan[-Sqrt[2]+1],x ArcTan[Sqrt[2]+1]}, {x,-Pi,Pi} ]; R3=x^3 - 3 x y^2 ; I3= 3 x^2 y - y ^3 ; z3r=Plot3D[R3 , {x,-Pi,Pi},{y,-Pi,Pi} ]; z3i=Plot3D[I3 , {x,-Pi,Pi},{y,-Pi,Pi} ]; Show[z3r,z3i] ; 'Top view >> Re,Im Intxn'; Plot[{x,x (-Sqrt[3]+2) , x (-Sqrt[3]-2) }, {x,-Pi,Pi} ]; R2=Cosh[y] Sin[x] ; I2=Sinh[y] Cos[x] ; scr=Plot3D[R2,{x,-Pi/2,Pi/2},{y,-Pi/2,Pi/2}]; sci=Plot3D[I2,{x,-Pi/2,Pi/2},{y,-Pi/2,Pi/2}]; Show[scr,sci]; 'Top view >> Re,Im Intxn'; Plot[{ArcTanh[Tan[x]]},{x,-Pi/2,Pi/2 }]; -- Posted via http://web2news.com To contact in private, remove