Let $a\in(-\pi,\pi]$ and $f:G\to\mathbb C$, $G = \{ z\in\mathbb C\setminus\{0\},\operatorname{Arg}z\neq a \}$ $$f(z)=\ln|z|+\imath \arg_a z,\quad a<\arg_az<a+2\pi$$ Prove that $f$ is analytic and that $f'(z)= 1/z$
Attempt
I am trying to use a theorem which dictates that if $L(z)-f(z)= 2k\pi$ for $k=1,2,\dots$, then $f(z)$ is a branch and therefore it is analytic, but I have a hard time with $\arg_a(z)$ and its relation to $\operatorname{Arg} z$.
If anyone could help me with this exercise , I would be grateful. Please excuse my English.
For simplicity let $Arg(z):=\theta $ then we have $arg_a z = \theta $ if $\pi \geq \theta >a $ and $arg_a z= \theta + 2\pi $ otherwise. So if $\theta \neq \pi $ we have an open nhd. of $z$ with $f(z)-L(z)=$some constant hence $f'(z)=L'(z)=1/z$.
For $\theta=\pi$ , you may use Cauchy-Riemann polar equations or inverse function theorem..