a) Find the image of the infinite strip 0 < y < 1/(2c) under the transformation w=1/z. Sketch the strip and its image.

b) Find the image of the quadrant x>1, y>0 under the transformation w=1/z.

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- November 15th 2008, 06:23 PMmndi1105Mappings by 1/z
a) Find the image of the infinite strip 0 < y < 1/(2c) under the transformation w=1/z. Sketch the strip and its image.

b) Find the image of the quadrant x>1, y>0 under the transformation w=1/z. - November 15th 2008, 06:56 PMThePerfectHacker
- November 16th 2008, 11:22 AMshawsend
Here's a Mathematica routine to plot the image of complex functions over user-defined regions. It has some limitations. I plotted the two regions below. mapregion takes the function, the region (such as 0<=y<=.5 && -10<=x<=10), and the x and y plot ranges ({-8,8}, {-8,8}). What's that hole doing in the plots anyway? Is is my mistake? How would you call mapregion to plot the region for the second problem?

Code:`mapregion[1/z, 0 <= y <=.5 && -10 <= x <= 10,`

{-8, 8}, {-8, 8}]

mapregion[1/z, 0.1 <= y <= 1 && -10 <= x <= 10,

{-8, 8}, {-8, 8}]

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, -5, 5},

{y, -2, 2}, PlotPoints -> 75,

AxesLabel -> {Style["x", 20],

Style["y", 20]}, Frame -> None,

Axes -> True]; transform =

rplot /. GraphicsComplex[pnts_,

data__] :> GraphicsComplex[

({real, imag} /. {x -> #1[[1]],

y -> #1[[2]]} & ) /@ pnts,

data]; newplot = Show[transform,

PlotRange -> {xrange, yrange},

AxesLabel -> {Style["u", 20],

Style["v", 20]}, Frame -> None,

Axes -> True]; GraphicsArray[

{{rplot, newplot}}]];