18 - Riemann's Removable Singularity Theorem

Riemann's Removable Singularity Theorem

Suppose $U$ is an open subset of $\mathbb{C}$ and $z_0\in U$
Suppose $f:U\setminus \{z_0\}\to \mathbb{C}$ is holomorphic then

$z_0$ is a removable singularity if and only if $f$ is bounded near $z_0$

Proof

Proof $⟹$
If $z_0$ is removable then there is holomorphic $g$ such that $g(z)=f(z)$ on $B(z_0,r)$ for $r>0$
So $f(z)=g(z)\to g(z_0)$ as $z\to z_0$ so $f$ is bounded

Proof $⟸$
Assume that $f$ is bounded and define $h(z)$ by

$$h(z)=\{\begin{array}{ll}(z-z_0{)}^{2}f(z)& \text{ if }z\ne z_0\\ & \\ 0& \text{ if }z=z_0\end{array}$$

so $h(z)$ is holomorphic on $U\setminus \{z_0\}$ using $f$ is complex differentiable
If $z=z_0$ then

$$\frac{h(z)-h(z_0)}{z-z_0}=(z-z_0)f(z)\to 0$$

as $z\to z_0$ since $f$ is bounded near $z_0$ by assumption

So $f$ is holomorphic everywhere in $U$
However for $r>0$ such that $\overline{B}(z_0,r)\subset U$ so by Taylor series for holomorphic functions
Then it is equal to its Taylor Series centred at $z_0$ thus

$$h(z)=\sum _{k=0}^{\infty }{a}_k(z-z_0{)}^k$$

Since $h(z_0)=h^{\prime }(z_0)=0$ then $a_0=a_1=0$ hence

$$\sum _{k=0}^{\infty }{a}_{k+2}(z-z_0{)}^k$$

defines a holomorphic function in $B(z_0,r)$

Since it agrees with $f(z)$ on $B(z_0,r)\setminus \{0\}$ then extend $f$ to holomorphic function on $U$