sketch

• Feb 1st 2010, 10:33 PM
stumped765
sketch
sketch region where f(z)=z^3

0<r<2/3
0<theta<pi/4
• Feb 2nd 2010, 04:10 AM
shawsend
We have $\displaystyle f(z)=z^3,\quad 0<r\leq 2/3,\quad 0\leq t\leq \pi/4$ then consider $\displaystyle f(z)=r^3 e^{3 it}$ and just for now, let r=1, and t go from 0 to $\displaystyle \pi/4$ then wouldn't that be an arc of radius one but going from 0 to $\displaystyle 3\pi/4$? Same dif for all the other arcs as the radius goes from zero to $\displaystyle 2/3$ then we have the piece with radius 0 to $\displaystyle (2/3)^3$ going from 0 to $\displaystyle 3\pi/4$ 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}]