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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.


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 $

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up vote 3 down vote accepted

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.

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Thank you very much! – Chris Jun 6 '12 at 16:00
  1. Since $f(z)$ is bounded in $C(a,r)$ and $\lim_{z\to a}(z-a)=0$ you have $\lim_{z\to a}(z-a)f(z)=0$.
  2. 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 $\lim_{z\to a}f(z)=\frac{h''(a)}{2}=c_2$
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Thank you very much! – Chris Jun 6 '12 at 16:00

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