# Question on the Proof of Casorati Weierstrass

The Casorati Weierstrass Theorem: Let $f:X\subseteq \mathbb{C}\to \mathbb{C}$ have an essential singularity at $w\in \mathbb{C}$. Then, $$\forall \epsilon,\delta>0,\zeta\in \mathbb{C}\ \exists z\in B(w,\delta)\setminus\left\{w\right\}: f(z)\in B(\zeta,\epsilon)$$

The proof of this I found here pg 215 goes something like this:

Let $\epsilon,\delta>0$ and $\zeta\in \mathbb{C}$. Suppose that $$0<\left|z-w\right|<\delta\Rightarrow\left|f(z)-\zeta\right|\ge\epsilon$$ The function $g:B(w,\delta)\setminus\left\{w\right\}\to \mathbb{C}$, $$g(z)=\frac{1}{f(z)-\zeta}$$ is holomorphic and bounded. Thus, $w$ is a removable singularity of $g$ and so, $$\hat{g}(z)=\phi(z)(z-w)^k$$ for some holomorphic function $\phi:B(w,r)\to \mathbb{C}$ so that $\phi(w)\neq 0$. Therefore the function $$f(z)=\zeta+\frac{1}{\phi(z)(z-w)^k}$$ either has a removable singularity at $w$ ($k=0$) or a pole ($k>0$) at $w$ which is a contradiction.

My questions on the bold face sentences: 1)"And so". Don't we also need $\lim_{z\to w}g(z)=0$ for that? (this can be derived easily, I am just asking if it is neccessary).

2) How can $k=0$? Since $\lim_{z\to w}g(z)=0$ the holomorphic extension $\hat{g}$ must have $\hat{g}(w)=0$ and so $k>0$. Thank you in advance

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You do not need $\lim_{z\to w} g(z) = 0$ to use the Riemann extension theorem. You just need that $g$ be bounded near $w$ to conclude that $g$ extends to a holomorphic function $\hat{g}$ at $w$. In fact, there is no way to conclude that $\lim_{z\to w} g(z) = 0$ (I'm not sure what your derivation for this is, but I would check it). This limit will hold if and only if $k > 0$, but there is a possibility that $\lim_{z\to w} g(z)\neq 0$, and in this case $k = 0$.