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!!)
Best Answer
There is a geometric way of seeing that inversion works properly, which is based on considering not $1/z$ but $\bar{1/z}$. Then your $(x,y)$ gets mapped to $(\frac x{x^2+y^2},\frac y{x^2+y^2})$, which can be interpreted geometrically as follows. Let $O$ be the origin, and let $P$ be a point. Then $P$ gets sent to the unique point $P'$ on the ray $\vec{OP}$ such that $OP\cdot OP'=1$ (where $OP$ is the length of the segment joining $O$ and $P$). If $R$ is any point on the unit circle between $P$ and $P'$, then we can rewrite the equation as $\frac{OP}{OR}=\frac{OR}{OP'}$. Hence, triangles $OPR$ and $ORP'$ will be similar. You can then prove that triangles $OPQ$ and $OQ'P'$ are similar for any points $P$ and $Q$. (multiply to get $\frac{OP}{OR}\frac{OR}{OQ}=\frac{OQ'}{OR}\frac{OR}{OP'}$).
Once you have that it is easy to show that if a line $\ell$ does not go through the origin, it gets mapped to a circle that goes through the origin as follows. Suppose that $P$ is on $\ell$ such that $\vec{OP}$ is perpendicular to $\ell$. Then for any other point $Q$ on $\ell$ we have that triangle $OPQ$ has right angle $OPQ$, and hence triangle $OQ'P'$ has right angle $OQ'P'$ for all points $Q'$. But that implies that the points $Q'$ trace out a circle with diameter $OP'$. Hence, $\vec{OP}$, which was perpendicular to $\ell$ is perpendicular to the circle that $\ell$ got sent to. This shows full conformality since if you have two lines $\ell_1$ and $\ell_2$, you have three cases.
There is also the analytic way of doing this.
If we consider $\mathbb C$ as the real plane $\mathbb R^2$, then what you're trying to show is that certain functions $f\colon \mathbb R^2\to\mathbb R^2$ are angle-preserving.
To be angle preserving at a particular point means that the derivative $df(z_0)$, which is a $2\times 2$ real matrix, preserves the angles between vectors. To see this, take two (smooth) curves $c_1\colon\mathbb R\to\mathbb R^2$ and $c_2\colon\mathbb R\to\mathbb R^2$. The angle between them at point $z_0=c(t_1)=c(t_2)$ is, as you said, the angle between the tangent lines, which is of course the angle between the tangent vectors $dc_1(t_1)$ and $dc_2(t_2)$ -- let's call them $v_1$ and $v_2$.
The function $f$ will map the two curves to the curves $f\circ c_1$ and $f\circ c_2$, whose tangent vectors will be $df(z_0)v_1$ and $df(z_0)v_2$. Thus, we want $df(z_0)$ to be a matrix $M$ that preserves the angles between vectors.
Fortunately, you're not asking how to classify all conformal mappings, so we just need to check that the derivatives of $z\to az+b$, $z\to 1/z$ and $z\to\bar z$ preserve angles at every $z_0$.
It turns out that the Cauchy-Riemann equations tell you that complex-differentiable functions (such as $z\to az+b$ and $z\to 1/z$) have derivates that look like $M=\left[\begin{matrix}a&b\\-b&a\\\end{matrix}\right]$ which since $\det M=a^2+b^2$ look exactly like $a^2+b^2\left[\begin{matrix}\cos(\theta)&\sin(\theta)\\-\sin(\theta)&\cos(\theta)\end{matrix}\right]$ for some angle $\theta$, which is exactly dilation by a factor of $a^2+b^2$ and rotation by angle $\theta$, which evidently preserve angles between vectors (for a quick proof, the rotation matrix is orthonormal and so actually preserves dot products, while dilation merely scales dot products by some constant).
To get the action of $\bar z$, note that it sends $x+iy$ to $x-iy$ and so its derivative is the matrix $\left[\begin{matrix}1&0\\0&-1\end{matrix}\right]$. This matrix does not preserve the dot product -- a short computation shows that it reverses the sign of the dot product. But that is ok since reversing the sign of the dot product is a reversal of orientation, i.e. the angle changes from $\theta$ to $180^o-\theta$, which for the purposes of being conformal is the same thing.