Is $\mathbb{Ln}(z)$ holomorphic on a circle defined by $\Gamma=\{z\in\mathbb{C}:\left|z\right|=1\}$ I know that the principal value of $\mathbb{Ln}(z)$ does not include the negative $x$ axis and thus its input consists of only complex numbers with argument in the interval $(-\pi,\pi)$. 
However does this mean that $\mathbb{Ln}(z)$ is not holomorphic on a circle $\Gamma$ defined as:
$\Gamma=\{z\in\mathbb{C}:\left|z\right|=1\}$
I do not have intuition about the complex logarithm. Any suggestions ?
 A: As @Daniel pointed in the comments the principal value of the complex logarithm can have slightly different definitions
$$\ln:\Bbb C\setminus\{0\}\to \Bbb R+i(-\pi,\pi],\quad z\mapsto \ln|z|+\arg(z)\tag{1}$$
where $\arg(z)\in(-\pi,\pi]$. Or it can also be defined as
$$\ln:\Bbb C\setminus\Bbb R_{\ge 0}\to \Bbb R+i(-\pi,\pi),\quad z\mapsto \ln|z|+\arg(z)\tag{2}$$
where $\arg(z)\in(-\pi,\pi)$. These definitions can also vary depending of the codomain chosen. 
So if we choose $(2)$ (or some codomain-variation of it) as the definition of the complex logarithm (as @Noah did) then $\ln(-1)$ is not defined but $\ln$ is holomorphic. If we choose $(1)$ (or some codomain-variation of it) then $\ln(-1)=i\pi$ but $\ln$ is no longer holomorphic.

Regardless the definition we use we can see that for some $z\in\Bbb C\setminus\Bbb R_{\ge 0}$ if we fix $|z|$ then
$$\lim_{\arg(z)\to -\pi^+}\ln z=\ln|z|-i\pi\neq \lim_{\arg(z)\to \pi^-}\ln z=\ln|z|+i\pi\implies \lim_{z\to-|z|}\ln z\,\text{ does not exists}$$
because $-|z|=|z|e^{-i\pi}=|z|e^{i\pi}$ (the notation of $-\pi^+$ and $\pi^-$ is redundant with the definition of the principal argument of $z$ but I think it make clear the limits). 
Hence (the principal value of) the complex logarithm cannot be continuous when is defined in $\Bbb C\setminus\{0\}$, so it cannot be holomorphic in this domain, neither in $\Gamma$ because $-1\in\Gamma$.
