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I have observed, from my complex analysis course and Wikipedia respectively, that a function $f: \mathbb{C} \to \mathbb{C}, \, z \mapsto f(z)$ having a removable singularity at $z_0$ is equivalent to the following statements :

\begin{equation} f(z) = \sum_{n=0}^{\infty} a_n (z-z_0)^n \end{equation}

and \begin{equation} \lim_{z\to z_0}f(z)(z-z_0) = 0 \end{equation}

where the first comes from the Laurent series for which $a_n \in \mathbb{C}$ and, more specifically here, $a_n = 0$ when $n < 0$. I want to prove the equivalence between these two criteria but I'm not sure how to end the proof and if it's valid. Let's begin with the implication:

\begin{align} \lim_{z\to z_0}f(z)(z-z_0) &= \lim_{z\to z_0}\left(\left(\sum_{n=0}^{\infty} a_n (z-z_0)^n\right)(z-z_0)\right)\\ &= \lim_{z\to z_0}\sum_{n=0}^{\infty} a_n (z-z_0)^{n+1}\\ &= \sum_{n=0}^{\infty} \lim_{z\to z_0} a_n (z-z_0)^{n+1}\\ &= \sum_{n=0}^{\infty} a_n \lim_{z\to z_0}(z-z_0)^{n+1}\\ &= \sum_{n=0}^{\infty} a_n \cdot 0\\ &= 0 \end{align}

Now the reciprocal :

\begin{align} 0&=\lim_{z\to z_0}f(z)(z-z_0)\\ &=\lim_{z\to z_0}\left(\left(\sum_{n=-\infty}^{\infty} a_n (z-z_0)^n\right)(z-z_0)\right)\\ &=\lim_{z\to z_0}\sum_{n=-\infty}^{\infty} a_n (z-z_0)^{n+1}\\ &=\lim_{z\to z_0}\sum_{n=-\infty}^{-1} a_n (z-z_0)^{n+1} + \lim_{z\to z_0}\sum_{n=0}^{\infty} a_n (z-z_0)^{n+1} \end{align} We have seen in the above development that this last term is $0$. Therefore we have \begin{align} 0&=\lim_{z\to z_0}\sum_{n=-\infty}^{-1} a_n (z-z_0)^{n+1} \end{align} But I fail to see how that implies $a_n = 0 \quad \forall n < 0$, is it just a matter of showing that $\forall n <0$ we have \begin{equation} \lim_{z \to z_0} (z-z_0)^{n+1} = \pm \infty \end{equation} and therefore the $a_n$s should compensate by being equal to zero ?

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    $\begingroup$ This should help you: en.wikipedia.org/wiki/Removable_singularity#Riemann.27s_theorem $\endgroup$
    – Bruno B
    Commented Aug 19, 2023 at 17:08
  • $\begingroup$ Yes it's a nice proof too, but is it possible to finish mine the way I started it ? $\endgroup$
    – Sileo
    Commented Aug 20, 2023 at 8:16
  • $\begingroup$ Hmm I don't really see a way to finish your proof, I'm afraid. Not to say it's impossible though. $\endgroup$
    – Bruno B
    Commented Aug 20, 2023 at 10:20

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