Complex Analysis MT23, Riemann's removable singularity theorem

Suppose

• $U$ is an open subset of $\mathbb C$
• $z _ 0 \in U$
• $f : U \setminus \{z _ 0\} \to \mathbb C$ holomorphic
• $f$ bounded near $z _ 0$

Then

• $f$ extends to a holomorphic function on all of $U$.

Overall idea: create a new function $(z - z _ 0)^2 f(z)$ which, by assumption on the boundedness of $f$, is $0$ at $z _ 0$ and has $0$ derivative at $z _ 0$. Then we can recover a Taylor expansion for $f$ by dividing the Taylor expansion for $(z-z _ 0)^2 f(z)$ by $(z - z _ 0)^2$.

Consider the function

\[h(z) = \begin{cases} (z - z _ 0)^2 f(z) &z \ne z _ 0 \\\\ 0 &z = z _ 0 \end{cases}\]

we aim to show $h(z) / (z-z _ 0)^2$ is holomorphic at $z = z _ 0$. Note $h(z _ 0) = 0$ and

\[h'(z _ 0) = \lim _ {z \to z _ 0} \frac{h(z) - h(z _ 0)}{z - z _ 0} = \lim _ {z \to z _ 0} (z-z _ 0)f(z) = 0\]

(this is where we use the fact that $f(z)$ is bounded near $z _ 0$) Hence $h$ is holomorphic and so on some disc $B(0, r)$ in $U$

\[h(z) = \sum^\infty _ {k=0} a _ k (z-z _ 0)^k = \sum^\infty _ {k = 2} a _ k(z-z _ 0)^k\]

Then

\[f(z) = \frac{h(z)}{(z-z _ 0)^2} = \sum^\infty _ {k=0} a _ {k+2} (z-z _ 0)^k\]

which must also be holomorphic on $B(0, r)$. Hence letting $f(z _ 0) = a _ 2$ defines a holomorphic function.