Proof of Casorati-Weierstrass A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Ch9.2
I have questions on the proof of Casorati-Weierstrass Theorem (Thm 9.7)
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If $z_0$ is an essential singularity of $f$ and $r$ is any positive real
  number, then every $w \in \mathbb C$ is arbitrarily close to a point in $f(\{0<|z-z_0|<r\})$. That is, for any $w \in \mathbb C$ and any
  $\varepsilon > 0$ there exists $z \in \{0<|z-z_0|<r\}$ such that $|w − f(z)| < \varepsilon$.

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Note: The book uses $D\!\!\!\cdot$ $[z_0,r]$ to denote a punctured disc $\{0<|z-z_0|<r\}$.
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*($\color{blue}{\text{blue}}$ box) Why do we have $$\lim_{z \to z_0} \frac{z-z_0}{f(z)-w} = 0$$? I tried:
  



*

*Pf (1): Let $\varepsilon > 0,$ we must find $\delta > 0$ s.t. $$|\frac{z-z_0}{f(z)-w}| < \varepsilon \ \text{whenever} \ |\frac{z-z_0}{1}| < \delta.$$

*We're given that there exists $M, \delta_1 > 0$ s.t. $$|\frac{1}{f(z)-w}| \le M \ \text{whenever} \ |\frac{z-z_0}{1}| < \delta_1$$
Therefore, for all $\varepsilon > 0$, by choosing $\delta := \min\{\frac{\varepsilon}{M},\delta_1\}$, we have 
$$|\frac{z-z_0}{f(z)-w}| \le |\frac{z-z_0}{1}\frac{1}{f(z)-w}|$$
$$ \le |\frac{z-z_0}{1}|M < \varepsilon \ \text{whenever} \  |\frac{z-z_0}{1}| < \delta$$
QED


  
*($\color{green}{\text{green}}$ box) That $z_0$ a singularity of $g$: Where is this used in the rest of the proof?
  

I suspect none and hence the round brackets.


  
*(1st $\color{red}{\text{red}}$ box) Why is $z_0$ a singularity of $g$?
  

I understand that if $z_0$ is a singularity, $z_0$ is isolated and removable. I just want to know why $z_0$ is a singularity of $g$ in the first place.


  
*(2nd $\color{red}{\text{red}}$ box) Why is $z_0$ a singularity of $f(z)-w$?
  

I (think I) understand that if $z_0$ is a singularity, $z_0$ is isolated and then either removable or pole. I just want to know why $z_0$ is a singularity of $f(z)-w$ in the first place.


  
*($\color{yellow}{\text{yellow}}$ box) Why is $z_0$ a removable singularity of $f(z)-w$ if $z_0$ is not a pole of $f(z)-w$? I tried:
  

Pf (5): By Prop 9.5b (*), $z_0$ is removable or for all $p \in \mathbb N$, $$\lim (z-z_0)^{p+1}(f(z)-w) \ne 0$$ However, for $p=n-1$, we have $$\lim (z-z_0)^{p+1}(f(z)-w)=0$$ Therefore, $z_0$ is removable. QED (5)


  
*($\color{black}{\text{black}}$ box) Why is $z_0$ a pole of $f(z)-w$ if $z_0$ is not a removable singularity of $f(z)-w$? I tried:
  

Pf (6): Observe that $$\lim (z-z_0)^n(f(z)-w) = 0$$
By assumption, $z_0$ is not removable. Therefore, by Prop 9.5b (*), $z_0$ is a pole of order $n-1$. QED (6)


(*) (Prop 9.5) Suppose $z_0$ is an isolated singularity of $f$. Then
(a) $z_0$ is removable iff $\lim_{z \to z_0} (z-z_0) f(z) = 0$
(b) $z_0$ is a pole iff $z_0$ is not removable and $\exists n \in \mathbb N: \lim_{z \to z_0} (z-z_0)^{n+1} f(z) = 0$

 A: Let's consider the proof step by step


*

*Please avoid writing implication in the converse order: it's always better to specify the assumption first and then proceed with the proof, for the sake of readability. Nonetheless, the proof is correct.

*Indeed, nowhere. It's just an info.

*Because if $f$ has a singularity at $z_0$ then $g\circ f$ must also have one there.

*An isolated singularity is an isolated point where the function is not defined.

*and 6. Yes. Indeed removable is so to say a "pole of order zero", hence just a special case.

A: 1)bounded $\times$ infinitesimal = infinitesimal
2)Not used eslewhere; just an additional info
3)-4)because if $z_0$ is a singularity for $f$ then it is such also for composition with $f$ like $\frac{z-z_0}{f(z)-w}$ or $f(z)-w$.
5-6)if it is clear that $f(z)-w$ has a singularity at $z_0$, then you know a singularity can be of three types; since $\lim_{z\to z_0}(z-z_0)^n(f(z)-w)=0$ this mean that, if $n=0$ then $f$ has a removable singularity at $z_0$; if $n\ge1$ what's happen?
