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so I currently read about isolated singularities and residue.

If we have a complex function $f(z)$ with a simple pole at $z_0$ we can use

$$\operatorname{Res}(f;z_0)=\lim_{z\to z_0}(z-z_0)f(z) \tag{1}$$ to find its residue for $z_0$. Now, I read that we can also use $(1)$ to analyze another isolated singularity $z_1$. Basically I read that we can say the following:

  • $(1)=c, c$ finite: we have a simple pole
  • $(1)=\infty$: Pole of higher order
  • $(1)=0$: $f(z)$ is analytic at $z_1$

Now, all three cases make sense to me, if we assume, that $z_1$ is a pole. Sadly, it wasn't clearly stated so I started to think about "What if $z_1$ is a removable or essential singularity? Can we still use $(1)$ to gather information and if so, what kind of information?"

From how I understand $(1)$ and how it is derive and if one thinks about the properties of the Laurent-Series of removable and essential isolated singularities I'd say we can't use $(1)$ to really get more information about $z_1$ if we assume it to be any kind of isolated singularity.

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You should just examine the other two cases. If we have a removable singularity at $w\in \mathbb{C}$, then the Laurent Expansion of $f$ at $w$ is just its Taylor expansion $$ f(z)=\sum_{k=0}^\infty c_k(z-w)^k.$$ To this end,
$$ \lim_{z\to w} f(z)(z-w)=\lim_{z\to w} \left(\sum_{k=0}^\infty c_k(z-w)^k\right)(z-w)=0.$$ On the other hand, if there is an essential singularity, we Laurent expand as $$ f(z)=\sum_{k=-\infty}^\infty c_k(z-w)^k.$$ Then, $$ \lim_{z\to w} f(z)(z-w)=\lim_{z\to w} \sum_{k=-\infty}^\infty c_k(z-w)^{k+1}$$ which does not exist by the characterization of isolated essential singularities.

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  • $\begingroup$ So basically I can only use it to analyze the situation further if I already know that the problem point is a pole, right? $\endgroup$ – xotix Dec 13 '18 at 11:24
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If $f$ has an isolated singularity at $z_0$, then the limit $(1)$ doesn't exist. This follows from the Casorati-Wierstrass theorem. On the other hand, the limit $(1)$ doesn't allow you to distinguish the case in which $f$ is actually defined at $z_0$ (and it's analytic there) from the case in which $z_0$ is a removable singularity of $f$ (but, of course, you have the definition of $f$ in order to distinguish these two cases).

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