Question about Möbiustransformation of y-axis $+\infty$

Question

I'm struggling abit with Möbiustransformations, and this time with b)

Let $f(z)=\frac{2z}{z+2}$ and calculate what the transformation will be using $0$, $\pm2$, $\infty$, $\pm(1+i)$.

b) the y-axis plus $\infty$

I know, by definition that circles maps to circles or lines and lines to circles or lines. I also know that if the transformation contains $\infty$ its a line and that Möbiustransformations contains it's angles.

So far i've done

$f(0)=0$

$f(\infty)=2$

$f(1+i)=\frac{2+2i}{3+i}$=$\frac{4+2i}{5}$

$f(-1-i)=2$

Which i make an assumption that its not a line considering the placement of points.

I know that the x-axis transforms onto itself, and that angles preserves, so it would have to be a circle to conserve angles. The problem is that the points given does not transform onto a circle, so i dont really know how to interpret the information.

Another question is also:

When checking what transformation x-axis will be, i use only points on x-axis (0,2,-2) which i assume is correct, but for y-axis im not allowed to use any points on y-axis, but only $\pm(1+i)$, does this give me any valid information about the transformation of y-axis? Or is the idea of this to only use the property that angles preserves?

Would someone like to help me out and give me hint?

Thanks!

Answer 1

Let $C$ denote the image of the extended imaginary axis under the transformation $f$.

• $C$ does not contain the point $\infty$, therefore it is a circle, not a line.
• $f(0) = 0$, $f(\infty) = 2$.
• The $y$-axis intersects the $x$-axis at a right angle, and $f$ maps the (extended) $x$-axis onto itself. It follows that $C$ intersects the $x$-axis at a right angle.

So the image $C$ is a circle through the point $0$ and $2$ which intersects the real axis at a right angle. That characterizes the image uniquely.