Question

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

Answer:

1. We have the complex map $w=z^2$ which transform the plane $(x,y)$ to the plane $(u,v)$. This is

$u+iv = (x+iy)^2 = x^2-y^2 +2xyi$.

Now, the complex numbers on the line $y=-3$ are of the form $z=x-3i$. if we substitute those values into the map we get

$u+iv = (x-i3)^2 = x^2-9^2 -6xi$.

So,

$\begin{cases}u&=x^2-9\\v&=-6x\end{cases}$.

This means that we can consider the above equation as the parametric equation of the image of the line $y=-3$ by the map $w=z^2$.

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 $(u,v)$ we only need to "eliminate" the parameter $t$. This can be done by the following relation:

$u-\frac{v^2}{36} = -9$

that gives the Cartesian equation

$u=\frac{v^2}{36}-9$. In the usual $(x,y)$ coordinates we have the equation

$x=\frac{y^2}{36}-9$.

In the attached figure we have its graph.

2. From the parametric equation we can see that if $x$ moves from $-\infty$ to $\infty$ the image point moves from the branch of the parabola situated on the first quadrant crossing the horizontal axis when $x=0$ and then going through the branch on the fourth quadrant to infinity.