I am not sure how to solve the following exercise on Möbius transformations:
Let $D=\{z:|z-1|\le\sqrt{2}\wedge|z+1|\le\sqrt{2}\}$ and $f(z)=\frac{-2}{z+i}$. Find the image of the set D through the function f.
I drew the set D. Then I computed some points in it and they went to something like parabola. I would like to draw that new set and I am not sure how should I do it. Any help?
On
Hint: Define the maps $$g:z\mapsto z+i$$$$h:z\mapsto \frac{1}{z}$$$$k:z\mapsto -2z$$ and note that $f=k\circ h\circ g$. The image under these three maps are somewhat easy to find. I.e. find $g(D)$, then $h(g(D))$ and finally $f(D)=k(h(g(D)))$.
$g$ translates each point by $+i$
$h$ sends $z$ to a point with modulus $1/|z|$ and $\arg(1/z)=-\arg(z)$ (reflection about the real axis)
$k$ sends $z$ to a complex number with modulus $2|z|$ and $\arg(-2z)=\arg(z)+\pi$ (rotation $\pi$ radians clockwise).
Hint: Möbius transformations take lines and circles to lines or circles, right? And they're continuous. This means that you can see what happens to the boundary components $|z-1| = \sqrt{2}$ and $|z+1| = \sqrt{2}$, and the image of $D$ will be one of the regions bounded by the images of those two arcs. Desmos (try it) gives that those images will be two lines with slopes approximately $\pm 1$ so the image of $D$ will look like what's above the graph of $|x|$ after some vertical shift. To do it analytically, note that $$f(z) = \frac{-2z-2i}{|z+i|^2}$$and set $z = 1+\sqrt{2}e^{it}$, $z = -1+\sqrt{2}e^{it}$ to see what happens.