1. ## sketch

sketch region where f(z)=z^3

0<r<2/3
0<theta<pi/4

2. We have $f(z)=z^3,\quad 0 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}]