If $z = re^{i\theta}$, how does a branch of the logarithm restrict the interval for $\theta$? I am following Stein and Shakarchi on Complex analysis. On page 98 they define a branch of the logarithm to be a choice of domain for the logarithm. I suspect they mean a choice of interval for the complex argument $\theta$. I cite:

Theorem 6.1 Suppose $\Omega$ is simply connected with $1 \in \Omega$ and $0 \not \in \Omega$. Then in $\Omega$, there is a branch of the logarithm $F(z) = \log_\Omega (z)$ such that (i) $F$ is holomarphic in $\Omega$, (ii) $e^{F(z)} = z$ for all $z \in \Omega$ and (iii) $F(r) = \log(r)$ for real numbers $r$ sufficiently close to $1$.
[Proof omitted]
For example in the slit plane $\mathbb{C} \setminus (- \infty,0]$ we have the principal branch of the logarithm [with the formula] $\log (z) = \log r + i \theta$ for $\theta \in (- \pi, \pi)$

So the authors have found
(a) an interval for $\theta$ and
(b) a formula for the principal branch of the logarithm.
How did the authors do this (in particular the interval for $\theta$)?
If $a \in \mathbb{R}$, to me it seems that $(a\pi,(a+2)\pi)$ would have been an equally good choice and that this choice does not depend on theorem 6.1.
Can you guide me on how to work out the other branches of the logarithm and also the appropriate formulas for them?
 A: In short, the reason for this particular choice of an interval is condition (iii). What you said about taking any $(a\pi,(a+2)\pi)$ is not entirely correct (at least if you want the principal branch of the logarithm), because if you take any $a$ that is not an odd integer, your interval will contain a multiple of $\pi$ of the form $(2n+1)\pi$ with $n\in \mathbb{Z}.$ But such an argument would mean that the associated complex number lies on the negative real axis which we removed from the domain $\Omega.$ Hence, the only options for the interval are $((2k+1)\pi,(2k+3)\pi)$ with $k\in\mathbb{Z}.$
In principle any $k$ is sufficient to fulfill conditions (i) and (ii), but let us now also consider condition (iii). Take any $k$ other than $-1,$ then the interval for $\theta$ will not contain $0.$ Therefore, the corresponding branch of the logarithm can never take real values so it cannot fulfill condition (iii). We are therefore left with only this particular choice of interval.
You can apply similar reasoning to any domain $\Omega$ of the form
$$\mathbb{C}\setminus \{re^{i\varphi}:r\in \mathbb{R},r\geq 0\}$$ where you take the entire complex plane and remove just the ray starting at the origin with angle $\varphi\in\mathbb{R}.$ As an exercise, you can think about which choices for $\phi$ are allowed to fulfill the conditions of Theorem 6.1. You could also remove any other curve starting at the origin and going to infinity (like an Archimedian or logarithmic spiral) but then your argument in the formula of the logarithm will also depend on $r.$
