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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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.
Try seeing where the boundary curves gets mapped to. Remember 1/z maps lines and circles into lines and circles. That should be enough infromation.Quote:
Originally Posted by mndi1105
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?Quote:
Originally Posted by mndi1105
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}]
Oh yea, you need to try and do these mapping by hand first before you start relying on Mathematica right? Mathematica is like a 50 caliber rifle: Fun to use but you need to be careful using 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, -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}}]];
