Suppose we wanted to show that the principal branch - from here on denoted by $\operatorname{Log}(z)$- of the complex logarithm is holomorphic on its domain $\mathbb{C} \smallsetminus \mathbb{R}^{\leq 0}$.
Here is one approach, I believe.
We have $\operatorname{Log}(z) = \log(|z|) + i \arg(z)$, where $\arg(z) \in (- \pi, \pi]$.
It suffices to show that $L_{1}(x+iy)=\Re(\operatorname{Log}(z)) = \log(|z|) = \log(x^2 + y^2)$ and $L_2(x + iy)=\Im(\operatorname{Log}(z)) = \arg(z) = \arctan(\frac{y}{x})$ are $C^2$ functions and that the respective partials $ \partial_{x} L_{1} $, $ \partial_{y} L_{1}$, $ \partial_{x} L_{2}$ and $\partial_{y} L_{2}$ satisfy the Cauchy-Riemann equations. Indeed,
$ \partial_{x} L_{1} = \frac{x}{x^2+y^2} = \partial_{y} L_{2}$ and $ \partial_{y} L_{1} =\frac{y}{x^2+y^2} = - \partial_{x} L_{2}$
and the partials are clearly continuous. We conclude $\operatorname{Log}(z)$ is holomorphic on its domain.
Here is my question: Where did we use the specific properties of the branch. That is, it is clear to me that we have shown complex differentiability, but I don't see that we have shown this specifically for this branch of the complex logarithm.
Towards answering my own question: Is it that we are allowed to write $\arg(z) = \arctan( \frac{y}{x})$?
When you write $\arg z = \arctan (y/x)$, you have restricted $\arg z \in (-\pi, \pi)$, so you have chosen the branch to be $\mathbb{C} \smallsetminus \{z \in \mathbb{R} : z \leq 0\}$, i.e., discarding the points with argument $-\pi$. (There's actually more to say here. "$y/x$" passes through two periods of the tangent as $x+\mathrm{i}y$ wanders over the plane. It would be better to write $\arctan(y,x)$, the version of the arctangent that tracks which quadrant contains $x+\mathrm{i}y$. But this version allows the angle $\pi$. So what we want is a function like the normal arctangent which has an upper and lower horizontal asymptote, to chop out the non-positive real axis, and which spans from $-\pi$ to $\pi$.)
Notice that you could choose a different ray for the branch cut from $0$ by setting $\arg z = \theta + \arctan(y/x)$, which puts the branch cut at $-\pi + \theta$. (One can even make the branch cut be a curve different from a ray by letting $\theta$ be a continuous function depending on the radius. In general, all one needs from the cut to get holomorphicity is that the cut runs from $0$ to complex infinity and does not self-intersect.)
We also use the specific branch cut when we do not check the compatibility of the partial derivatives along $\{z \in \mathbb{R} : z \leq 0\}$, where such a check would fail. In particular, the $\frac{\partial \ln(x+\mathrm{i}y)}{\partial y}$ derivative does not exist on the cut because the logarithm has a jump discontinuity as you cross the cut in the $y$-direction. For instance, along the circle of radius $r$, \begin{align*} \lim_{\theta \rightarrow \pi^-} \ln (r \mathrm{e}^{\mathrm{i} \theta}) &= \mathrm{i} \pi \ln r \quad \text{ and } \\ \lim_{\theta \rightarrow -\pi^+} \ln (r \mathrm{e}^{\mathrm{i} \theta}) &= -\mathrm{i} \pi \ln r \text{.} \end{align*}