$w = \log z$ is analytic on $D = \{z : \alpha < \arg z < \alpha +2\pi k\}$ $w = \log(z)$ is analytic on $D = {z : \alpha < \arg z = \theta< \alpha +2\pi k}$.
Now I have been told, conceptually and without proof, that this is true. 
I have been shown that taking the limit of $w$ as $\arg z$ approaches $\alpha$ from above and below when the domain is closed results in
$$\lim_{\theta \to \alpha^- } \ln|z| + i\theta = \ln|z| + i\alpha \\
\lim_{\theta \to -\alpha^+ } \ln|z| + i\theta = \ln|z| - i\alpha
$$
So the function is not analytic because it is not continuous at $\alpha$ and $\alpha + 2\pi k$ because at those points the limit is not unique.
So is the idea that if we cut out those points only, we now have a domain that if we take the limit from the left and right approaching any $\alpha \pm  \epsilon$ it will exist and be unique? Since we always have wiggle room between a given $\epsilon$ and $\alpha$?
If the above words are true, would they be considered a "rigorous" explanation or proof of why the opening statement is true?
 A: Define $Arg:\Bbb C-\{(x,0) \in \Bbb C:x≥0\}\rightarrow (0,2\pi)$ as follows:
$$Arg(x,y)=tan^{-1}(|\frac{y}{x}|); \ y>0,x>0$$
$$Arg(0,y)=\frac{\pi}{2};\ y>0$$
$$Arg(x,y)=\pi-tan^{-1}(|\frac{y}{x}|);\ y≥0,x<0$$
$$Arg(x,y)=\pi+tan^{-1}(|\frac{y}{x}|);\ y≤0,x<0$$
$$Arg(0,y)=\frac{3\pi}{2};\ y<0$$
$$Arg(x,y)=2\pi-tan^{-1}(|\frac{y}{x}|);\ y<0,x>0$$
Where $tan^{-1}:\Bbb R\rightarrow (-\frac{\pi}{2},\frac{\pi}{2})$ is the inverse of $tan_{|(-\frac{\pi}{2},\frac{\pi}{2})}\rightarrow \Bbb R$.
Define $Log:\Bbb C-\{(x,0) \in \Bbb C:x≥0\}\rightarrow \Bbb C$ as follows:$$Log(z)=ln(|z|)+\iota Arg(z)$$
Here $ln:(0,\infty)\rightarrow \Bbb R$ is the inverse of  exponential map $e:\Bbb R \rightarrow (0,\infty)$.
Then for each $z\in \Bbb C-\{(x,0) \in \Bbb C:x≥0\}$ we have $exp(Log(z))=z$. To prove this use $cos(tan^{-1}(\theta))=\frac{1}{\sqrt {1+\theta^2}}, sin(tan^{-1}(\theta))=\frac{\theta}{\sqrt {1+\theta^2}}$ for $\theta≥0$.
Define $arg:\Bbb C-\{r(cos\alpha+\iota sin\alpha )\in \Bbb C:r≥0\}\rightarrow (\alpha,\alpha+2\pi)$ by $arg(z)=\alpha + Arg(exp(-\iota \alpha) z)$.
By definition both $arg,Arg$ are continuos.
Finally define

$$log:\Bbb C-\{r(cos\alpha+\iota sin\alpha )\in \Bbb C:r≥0\}\rightarrow \Bbb C$$ $$log(z)=ln(|z|)+\iota arg(z)$$

One can again check that $exp(log(z))=z$ for $z\in \Bbb C-\{r(cos\alpha+\iota sin\alpha)\ :\ r≥0\}$ using $exp(Log(z))=z,z\in \Bbb C-\{(x,0) \in \Bbb C:x≥0\}$.
Now since $z\rightarrow z$ is holomorphic on $\Bbb C-\{r(cos\alpha+\iota sin\alpha )\in \Bbb C:r≥0\}$ and $log$ is continuous logarithm of 
 $z\rightarrow z$ in $\Bbb C-\{r(cos\alpha+\iota sin\alpha )\in \Bbb C:r≥0\}$ we can say $log$ is holomorphic on $\Bbb C-\{r(cos\alpha+\iota sin\alpha )\in \Bbb C:r≥0\}$.
The last statement, namely continuous logarithm is holomorphic is followed from a general theorem given in the book An Introduction to Classical Complex Analysis, Volume 1 By R.B. Burckel on page 79 (Exercise 3.41)
You can also prove holomorphicity of the the bijective continuous function $$log:\Bbb C-\{r(cos\alpha+\iota sin\alpha)\ :\ r≥0\}\rightarrow \{x+\iota y:x\in \Bbb R,y\in (\alpha,\alpha+2\pi)\}$$ using the function $$E:=exp|_{\{x+\iota y\ :\ x\in \Bbb R,\ y\in (\alpha,\alpha+2\pi)\}}\rightarrow \Bbb C-\{r(cos\alpha+\iota sin\alpha)\ :\ r≥0\}$$
Since $$E(log(z))=z,\ z\in \Bbb C-\{r(cos\alpha+\iota sin\alpha)\ :\ r≥0\}$$
And $$log(E(z))=z,\ z\in \{x+\iota y\ :\ x\in \Bbb R,\ y\in (\alpha,\alpha+2\pi)\}$$
Now apply Proposition 2.20 given on page 39 in the book - Functions of One Complex Variables by John B. Conway.
