Let $f(z)=\ln\left|T(z)\right|$ where $T(z)=\frac{1+z}{1-z}$. What is the biggest domain in which $f(z)$ is harmonic?
My Approach: $f(z)$ is the real part of all analytic branches of $g(z)=\log(T(z))$. Therefore, if $g_0(z)$ is an analytic branch of $g(z)$ in $D$, then $f(z)$ is harmonic in $D$. Every analytic branch of $g(z)$ would be analytic in a domain with a branch cut: meaning $D$ would have the following form:
$$D_t=\mathbb{C}\setminus\left\{re^{it}\mid r\in[0,\infty)\right\}$$
So my initial intuition was that the biggest domain in which $f(z)$ is harmonic, is $\mathbb{C}$ with some branch cut.
However, a different approach is to notice that $f(z)$ is the real part of all of the analytic branches of $g(z)$. This means that for every $z\neq\left\{1,-1\right\}$ (since those points are obvious singluarities) we can find an analytic branch of $g(z)$ at some neighborhood of $z$, and then $f(z)$ would be harmonic in that neighborhood. Repeating this process for all $z\neq\left\{1,-1\right\}$ we will get eventually that the biggest domain is $\mathbb{C}\setminus\left\{-1,1\right\}$.
The main difference between the approaches, as I see it, is that the first approach assumes a global analytic branch, and the second approach discusses local analytic branches. Which is the right one? Are they both wrong?
Thanks!