I am trying to show that elements of the general Möbius group generated by an affine transformation $f(z) = az+b$, the inversion map $f(z)=\frac{1}{z}$ and complex conjugation $f(z)=\overline{z}$, where $a,b \in \mathbb{C}$, are conformal maps.
So if I have a curve lying in the $x$-$y$ plane, then its image in the $u$-$v$ plane is another curve, and I want to show that the angles between the curves are preserved under the three transformations above. Because we speak of angles between curves, the problem reduces to proving that the angle between their tangents is preserved and hence to the simplest problem f showing that angles between straight lines are preserved under these maps.
Now let $X_1$ and $X_2$ be straight lines. Then under an affine transformation and complex conjugation it is not hard to show that the angle between $X_1$ and $X_2$, let's call it $angle(X_1 , X_2)$ is preserved (Under complex conjugation the sign of the angle is reversed however).
Now the inversion map is a bit more tricky. If my two lines in $x-y$ space pass through the $y$-axis at 1, viz. that $X_k$ has equation $\beta_k z + \overline{\beta_k z} + 1 = 0$ where $ k \in {0,1}$, then under inversion these straight lines get mapped to circles through the origin, and hence it is plain that the angles between the tangents of the circles at $(0,0)$ is preserved.
However, what if my lines $X_1$ and $X_2$ are such that under inversion $X_1$ maps to a line while $X_2$ maps to a circle? This is where I am getting stuck.
On the other hand, if I try to prove the conformality of the inversion map by saying that my $x$ and $y$ coordinate curves in $x-y$ space get mapped to $\frac{x}{x^2 + y^2}$ and $\frac{-y}{x^2 + y^2}$ in $u-v$ space, and then write vector equations of lines and compute partial derivates, the maths becomes ugly and the method is certainly not elegant.
Anyone got any ideas? (Maybe involving Cauchy - Riemann Equations!!)