= Solution
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 $B(w,\epsilon)$ disjoint from it. Hence
$$
|f(z)-w|\geq\epsilon\qquad(0<|z-a|<\rho).
$$
The <holomorphic function> $h(z)=1/(f(z)-w)$ is bounded by $1/\epsilon$ on that punctured disc. By the permitted <Riemann removable singularity theorem>, it extends to a <holomorphic function> $H$ on the full disc.
If $H(a)\ne0$, 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 $m\geq1$ and $K(a)\ne0$. Then
$$
f(z)=w+\frac{(z-a)^{-m}}{K(z)}
$$
has 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
$$
\boxed{\overline{f(B(a,\rho)\setminus\{a\})}=\mathbb C\quad\text{for every allowed }\rho>0}.
$$
Back to article page
