Verify that the mapping z --> z^2, (2 € C) will map the line y = -3 to a parabola. Find its Cartesian equation and sketch it. Determine the direction of the movement of the image point as the pre-image point moves from left to right on the line y = -3.

Answer:1. We have the complex map [tex]w=z^2[/tex] which transform the plane [tex](x,y)[/tex] to the plane [tex](u,v)[/tex]. This is[tex]u+iv = (x+iy)^2 = x^2-y^2 +2xyi[/tex].Now, the complex numbers on the line [tex]y=-3[/tex] are of the form [tex]z=x-3i[/tex]. if we substitute those values into the map we get[tex]u+iv = (x-i3)^2 = x^2-9^2 -6xi[/tex].So,[tex]\begin{cases}u&=x^2-9\\v&=-6x\end{cases}[/tex].This means that we can consider the above equation as the parametric equation of the image of the line [tex]y=-3[/tex] by the map [tex]w=z^2[/tex]. It is not difficult to notice that this is the parametric equation of a parabola. In order to obtain its Cartesian equation in the plane [tex](u,v)[/tex] we only need to "eliminate" the parameter [tex]t[/tex]. This can be done by the following relation:[tex]u-\frac{v^2}{36} = -9[/tex]that gives the Cartesian equation[tex]u=\frac{v^2}{36}-9[/tex]. In the usual [tex](x,y)[/tex] coordinates we have the equation[tex]x=\frac{y^2}{36}-9[/tex].In the attached figure we have its graph.2. From the parametric equation we can see that if [tex]x[/tex] moves from [tex]-\infty[/tex] to [tex]\infty[/tex] the image point moves from the branch of the parabola situated on the first quadrant crossing the horizontal axis when [tex]x=0[/tex] and then going through the branch on the fourth quadrant to infinity.