March 30th
Today I learned the proof that meromorphic functions are conformal (i.e., angle-perserving) almost everywhere. Fix $f:\CC\to\CC$ analytic and, abusing notation, we will label $(u,v):=f:\RR^2\to\RR^2$ by the composition\[(x,y)\mapsto x+yi\mapsto f(x+yi)=(\op{Re}f(x+yi),\op{Im}f(x+yi)).\]Then the Cauchy-Riemann equations say that\[\frac{\del u}{\del x}=\frac{\del v}{\del y}\quad\text{and}\quad\frac{\del u}{\del y}=-\frac{\del v}{\del x}.\]At a high level, this means that the Jacobian of $f$ looks like\[\begin{bmatrix} \del u/\del x & \del u/\del y \\ \del v/\del x & \del v/\del y\end{bmatrix}=\begin{bmatrix} \del v/\del y & -\del v/\del x \\ \del v/\del x & \del v/\del y\end{bmatrix}\]is a rotation matrix when nonzero, so provided that $f'$ is nonzero, $f$ is conformal.
For formality reasons, we will stay in $\RR^2$ and just assume that our $f=(u,v):\RR^2\to\RR^2$ satisfy the Cauchy-Riemann equations as our way of using the fact $f$ is meromorphic. To show all angles are preserved, it suffices to show that angles with one side horizontal are preserved because we can just subtract to get all angles.
Let $\arg:\RR^2\to(\RR/2\pi\ZZ)$ be the function sending a point to its angle. Fixing a point $x$ and direction vector $v,$ we want to know that\[\arg\big(f(x+v)-f(x)\big)=\arg v\]as $v\to0.$ Note that we have to be a bit precise with $v\to0$ because we want to maintain direction, so we really want\[\arg\big(f(x+vdx)-f(x)\big)=\arg v\]as $dx\to0$ in $\RR.$
To use the derivative of $f,$ we would like to use the total derivative (equal to the Jacobian from earlier!), which by definition satisfies\[\lim_{dx\to0}\frac{\lVert f(x+vdx)-f(x)-(Df)(x)(vdx)\rVert}{\lVert vdx\rVert}=0.\]\todo{}