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How can we go about finding a Moebius map that fixes the set $\{z_1=x+iy,\,\,\, z_2={1\over iy-x}\}$ for some $x,y\in \mathbb R$ that does not correspond to rotation about any arbitrary axis of the Riemann sphere?

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    Glad to help. I'll go ahead and put it as an actual answer, since it wor$k$ed out for you.2012-04-27

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Choose a third point $z_3$ distinct from $z_1,z_2$. The map has to satisfy $(w,z_1,z_2,w_3)=(z,z_1,z_2,z_3)$ for some $w_3$ distinct from $z_1,z_2$, so we need only determine how to appropriately choose $w_3$.

Let $Z_k$ be the point on the sphere corresponding to $z_k$ ($k=1,2,3$). The line through $Z_1,Z_2$ is an axis of the sphere, and so there is a unique plane $H$ in $\mathbb{R}^3$ normal to this axis and containing $Z_3$. If $C$ is the intersection of the sphere $H$, it isn't difficult to see that $C$ would be mapped to itself by any rotation about the axis, so $w_3$ must be chosen not to lie on the line or circle in the complex plane corresponding to $C$.

Having done so, it remains only to show that the given transformation isn't a rotation about any other axis of the sphere, either, but this is not difficult, as $z_1,z_2$ should be the only points fixed by the transformation.