Past exam of the metric spaces and complex analysis course of the University of Oxford / 2023 / 4 / b

We prove the Casorati-Weierstrass theorem by contradiction. Fix an allowed $\rho$. If the stated image were not a dense set in $\mathbb{C}$, there would be a disc $B(w,ϵ)$ disjoint from it. HenceThe holomorphic function $h(z)=1/(f(z)-w)$ is bounded by $1/ϵ$ on that punctured disc. By the permitted Riemann removable singularity theorem, it extends to a holomorphic function $H$ on the full disc.

If $H(a)\ne 0$, then $w+1/H$ is holomorphic near $a$ and agrees with $f$ away from $a$. This makes $a$ a removable singularity of $f$.

If $H(a)=0$, then $H$ is not identically zero, since $h$ is nowhere zero away from $a$. Its Taylor series therefore has a finite first nonzero coefficient: otherwise the identity theorem would make it identically zero. Write $H(z)=(z-a{)}^{m}K(z)$ with $m\ge 1$ and $K(a)\ne 0$. Thenhas a pole of order $m$ at $a$, since multiplying by $(z-a{)}^m$ gives the nonzero limit $1/K(a)$.

Both cases contradict an essential singularity. Thus