Proof of Casorati-Weierstrass

Question

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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1. ($\color{blue}{\text{blue}}$ 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

2. ($\color{green}{\text{green}}$ 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.

3. (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.

4. (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.

5. ($\color{yellow}{\text{yellow}}$ 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)

6. ($\color{black}{\text{black}}$ 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)

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(*) (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 $\exists n \in \mathbb N: \lim_{z \to z_0} (z-z_0)^{n+1} f(z) = 0$

Answer 1

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?

Answer 2

Let's consider the proof step by step

1. 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.

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

3. Because if $f$ has a singularity at $z_0$ then $g\circ f$ must also have one there.
4. An isolated singularity is an isolated point where the function is not defined.
5. and 6. Yes. Indeed removable is so to say a "pole of order zero", hence just a special case.