I was trying to make a Möbius transformation $M:\Bbb C \to \Bbb C$ with $2$ fixed points. What I did was:
Make a degree 2 formula $\frac{az+b}{cz+d}=cz^2+dz-az-b=0 $ so $c\ne 0$ and
choose $a,b,c,d$ arbitrarily.
So I want $(a,b,c,d)=(0,2,3,0)$ giving me $M(z)=\frac{2}{3z}$
I can solve for the fixed points like this:
$3z^2 -2 =0\implies z=\pm \sqrt{\frac{2}{3}}$ are two fixed points.
I graph it and I get the following 
It is not a circle or a straight line? But I thought Möbius transforms took straight lines to circles or more straight lines?
Is $ M(z)=z= \frac{2}{3z}$ still a Möbius transform, even though it turns a line into a hyperbola?
Tip: For typing ö,(German Umlaut) hold Alt and type 148.
Möbius transformations map the set of lines and circles to the set of lines and circles.
If you restrict your $M$ to the real axis, what you see in the picture is the graph of the function, not the image.
Since $M$ is real valued on the real axis, the image is actually the projection of the graph on the $y$-axis. I.e. the images is the real axis itself ($M(\infty) = 0$ to fill in the gap). No contradiction here.