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Math Help - sketch

  1. #1
    Junior Member
    Joined
    Nov 2009
    Posts
    51

    sketch

    sketch region where f(z)=z^3

    0<r<2/3
    0<theta<pi/4
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  2. #2
    Super Member
    Joined
    Aug 2008
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    We have f(z)=z^3,\quad 0<r\leq 2/3,\quad 0\leq t\leq \pi/4 then consider f(z)=r^3 e^{3 it} and just for now, let r=1, and t go from 0 to \pi/4 then wouldn't that be an arc of radius one but going from 0 to 3\pi/4? Same dif for all the other arcs as the radius goes from zero to 2/3 then we have the piece with radius 0 to (2/3)^3 going from 0 to 3\pi/4 as shown in the plot below. Here's the Mathematica code to draw f(z) in some region defined by the region parameters. Try and find a machine with Mathematica and experiment with it.

    Code:
    mapregion[f_, region_, xrange_, yrange_] := 
       Module[{real, imag, rplot, transform, newplot}, 
        real = ComplexExpand[Re[f /. z -> x + I*y]]; 
         imag = ComplexExpand[Im[f /. z -> x + I*y]]; 
         rplot = RegionPlot[region, {x, xrange[[1]], xrange[[2]]}, 
           {y, yrange[[1]], yrange[[2]]}, PlotPoints -> 75, 
           AxesLabel -> {Style["x", 20], Style["y", 20]}, Frame -> None, 
           Axes -> True, AspectRatio -> 1, PlotRange -> {xrange, yrange}]; 
         transform = rplot /. GraphicsComplex[pnts_, data__] :> 
            GraphicsComplex[({real, imag} /. {x -> #1[[1]], y -> #1[[2]]} & ) /@ 
              pnts, data]; transform = transform /. RGBColor[x_, y_, z_] -> 
            RGBColor[1, 0, 0]; newplot = Show[{transform}, 
           PlotRange -> {xrange, yrange}, AxesLabel -> {Style["u", 20], 
             Style["v", 20]}, Frame -> None, Axes -> True]; 
         GraphicsArray[{{rplot, newplot}}]]; 
    
    mapregion[z^3, Inequality[0, Less, Abs[x + I*y], LessEqual, 2/3] && 
       0 <= Arg[x + I*y] <= Pi/4, {-1, 1}, {-1, 1}]
    Attached Thumbnails Attached Thumbnails sketch-region2.jpg  
    Last edited by shawsend; February 2nd 2010 at 05:21 AM.
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