What is the symmetry group of the circle as a complex curve?

Question

In a complex (projective) plane CP2 using homogeneous coordinates $(x, y, z)$ what is the group of (projective) transformations that leave the complex circle $$x^2 + y^2 = R^2 z^2 $$ invariant as a whole?

Answer 1

A circle (or in fact any non-degenerate conic) is isomorphic to a projective line. So a circle in $\mathbb{CP}^2$ is isomeorphic to $\mathbb{CP}^1$, and the projective transformations which leave it fixed are isomorphic to the projective transformations of $\mathbb{CP}^1$, i.e. the Möbius transformations.

A projective transformation of the line is uniquely determined by three points and their images. So if you fix three distinct points $A,B,C$ on your circle, you can describe any element of your group by specifying three distinct points $A',B',C'$ on that circle which are the images of $A,B,C$.

To find the resulting transformation in $\mathbb{CP}^2$, remember that any such transformation is uniquely determined by four points and their images, while no three points in either quadruple may be collinear. You already have three suitable pairs of points, so all you need is a fourth point. I suggest you take the one point on the circle in harmonic position. To find that, construct the tangents to the circle in $A$ and $B$, connect their point of intersection with $C$ and label the other intersection of that line with the circle as $D$. Do the same for the image points.

For details on computing a transformation give pairs of preimage and image points, see this post of mine. For details of computing $D$ resp. $D'$, feel free to post that as a separate question and notify me in a comment.