# Question on the proof of Casorati-Weierstrass theorem

Theorem. (Casorati-Weierstrass; Lang Complex analysis) Let $$0$$ be an essential singularity of the function $$f$$ and let $$D$$ be a disc centered at $$0$$ on which $$f$$ is holomorphic except at $$0$$. Let $$U$$ be the complement of $$0$$ in $$D$$. Then $$f(U)$$ is dense in the complex numbers.

Proof. Suppose there is $$\alpha\in\Bbb C$$ and a positive number $$s>0$$ such that $$|f(z)-\alpha|>s,\ \forall z\in U.$$ The function $$g(z) = \frac{1}{f(z)-\alpha}$$ is then holomorphic on $$U$$, and bounded on the disc $$D$$. Hence $$0$$ is a removable singularity of $$g$$, and $$g$$ may be extended to a holomorphic function on all of $$D$$. It then follows that $$1/g(z)$$ has at most a pole at $$0$$, which means that $$f(z)-\alpha$$ has at most a pole, contradicting the hypothesis that $$f(z)$$ has an essential singularity.

Question. Why can we conclude that $$1/g(z)$$ has at most a pole at $$0$$?

When you remove the singularity at $$0$$ you will get a holomorphic function on the disk which is not identically $$0$$. If the value at $$0$$ is $$0$$ then $$g$$ has a zero of some finite order $$N$$. [Theorem 10.18 in Rudin's RCA]. This implies that $$f(z)-\alpha=\frac 1 {g(z)}$$ has a pole of order $$N$$ at $$0$$.
• Why $g$ has a zero of some finite order? Jan 20 at 6:34
• Any analytic function vanishing at a point $c$ which is not identically $0$ has a zero of finite order at that point. For a proof write down the power series $\sum a_n(z-c)^{n}$ at the point and note that there is a smallest $N$ such that $a_n \neq 0$. @love_sodam Jan 20 at 6:38