Sketching the complex function f(z)=z^3

Start by declaring the function :

f[z_] = z^3 u[x_, y_] = ComplexExpand[Re[f[x + I * y]]] v[x_, y_] = ComplexExpand[Im[f[x + I * y]]]

In the previous cell compare what happens if you put u[x_,y_]=Re[f[x+I*y]]

I . Plot the level curves for the 2 - variable function  Abs[f[z]] = (u (x, y)^2 + v (x, y)^2)^(1/2) :

ContourPlot[Abs[f[x + I * y]], {x, -1, 1}, {y, -1, 1}] ;

ContourPlot[Abs[f[x + I * y]], {x, -1, 1}, {y, -1, 1}, Contours30, ContourShadingFalse] ;

II . Simultaneously plot the level curves for the 2 - variable functions  u and v :

p1 = ContourPlot[u[x, y], {x, -1, 1}, {y, -1, 1}, Contours50, ContourShadingFa ... , y], {x, -1, 1}, {y, -1, 1}, PlotPoints50, ContourShadingFalse] ; Show[p1, p2] ;

III . Plot the pre - image curve C ' in the z - plane of a curve C in the w - plane . If the c ... s equation  h (u, v) = 0 RowBox[{then C ' has equation h (u (x, y), v (x, y)), =, 0.}]

One common choice of curve C in the w-plane is (a) a horizontal line v=k or (b) a vertical line
u=k,where k is a constant. In these cases the pre-image curves C' have equations (a) v(x,y)=k
or (b) u(x,y)=k, which are level curves for v(x,y) and u(x,y) respectively. Thus the preimages of horizontal and vertical lines in the w-plane have already been represented in the pictures in part II.

Let's try another line say v=u+1.This leads to the xy-equation v(x,y)-u(x,1)-1=0
for the pre-image curve C'.

v[x, y] - u[x, y] - 1

ContourPlot[v[x, y] - u[x, y] - 1, {x, -5, 5}, {y, -5, 5}, Contours {0}, ContourShadingFalse, PlotPoints40] ;

Or if we consider the parallel family of lines v=u+k, k=constant.This leads to the xy-equation v(x,y)-u(x,1)=k
which is level curve for v(x,y)-u(x,y). For the choices of k=-10,-9,...-1,0,1,...10 try:

ContourPlot[v[x, y] - u[x, y], {x, -3, 3}, {y, -3, 3}, ContoursRange[-10, 10], ContourShadingFalse, PlotPoints40] ;

IV . Plot the image curve C ' in the w - plane of a curve C in the z - plane .   For ... = x (t), y = y (t) then C ' has parametric equations u = u (x (t), y (t)), v = v (x (t), y (t)) .

Let's consider C as the horizontal line y=k, which we can parametrize by x=t,y=k. Then C' has parametrization u=u(t,k), v=v(t,k):

kk = 4 ; ParametricPlot[{u[t, kk], v[t, kk]}, {t, -3, 3}] ;

Table[ParametricPlot[{u[t, kk], v[t, kk]}, {t, -3, 3}], {kk, -3, 7}] ; Show[%] ;

Created by Mathematica  (October 1, 2004)