TITLE: Riemann's theorem on removable singularities
QUESTION [8 upvotes]: Theorem
Let $\Omega\subseteq \mathbb{C}$ open , $ a\in\Omega,\ f\in H(\Omega\backslash \{a\})$ and there is $r>0$ with
$f$ is bounded on $C(a,r)\backslash \{a\}$ ($C(a,r)$ is the circle with origin $a$ and radius $r$),
then $a$ is a removable singularity.
Proof
Let $h:\Omega\rightarrow \mathbb{C}$ be defined as:
$$
h(z)=\left\{\begin{array}{l}
0,\ \ \ \ \ \ \ \ \  \  \ \ \ \ \ \ \ \ \ \ z=a \\
(z-a)^{2}f(z), \ z\in\Omega\backslash \{a\}
\end{array}\right.
$$
Then we have:
$$
\lim_{z\rightarrow a}\frac{h(z)-h(a)}{z-a}=\lim_{z\rightarrow a}(z-a)f(z)=0\Rightarrow h'(a)=0. 
$$
$\color{red}{\text{ Why is} \lim\limits_{z\rightarrow a}(z-a)f(z)=0?}$
So we have $h\in H(\Omega)$ and therefore $(h(a)=h'(a)=0)$.
$$
h(z)=\sum_{n=2}^{\infty}c_{n}(z-a)^{n}\ (z\in K(a,\ r)).
$$
Letting $f(a):=c_{2}$, it follows:
$\color{red}{\text{ Why do we have} f(a):=c_{2}?}$
$$
f(z)=\sum_{n=0}^{\infty}c_{n+2}(z-a)^{n}\ (z\in K(a,\ r)),
$$
so $f\in H(\Omega).\ \square $

REPLY [5 votes]: The first red line follows because $|(z-a)f(z)|\le M|(z-a)|$ near $a$, where $M$ is the bound for $f(z)$. This clearly goes to zero as $z$ goes to $a$.
The coefficient $c_2$ is $h''(z)/2!$ evaluated at $a$. We see $h'(z)=(z-a)^2f'(z)+2(z-a)f(z)$ and $h''(z)=(z-a)^2f''(z)+2(z-a)f'(z)+2(z-a)f'(z)+2f(z)$. Plugging in $a$ and dividing by 2 gives the the value for $c_2$ above.

REPLY [3 votes]: Since $f(z)$ is bounded in $C(a,r)$ and $\lim_{z\to a}(z-a)=0$ you have $\lim_{z\to a}(z-a)f(z)=0$.  
You want the function to be continuous at $z=a$, so you need $\lim_{z\to a}f(z)$. Observe that $h'(z)=2(z-a)f(z)+(z-a)^2f'(z)$ for all $z\neq a$. Hence: 
$$h''(z)=2f(z)+2(z-a)f'(z)+(z-a)^2f''(z)$$
Plugging in $z\to a$, you have $\lim_{z\to a}f(z)=\frac{h''(a)}{2}=c_2$