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

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-(v^2)/(36) = -9`

that gives the Cartesian equation

`u=(v^2)/(36)-9`. In the usual
`(x,y)` coordinates we have the equation

`x=(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.