Suppose the function $f(z)$ has a removable singularity at $z_o$. Why can I rewrite this function as $f(z)=(z-z_0)^kg(z)$ for some $k\in \mathbb{N}$ and a holomorphic function $g(z)$ where $g(z_0)\neq0$? Is there a simple reason? I try to deduce it from the fact $lim_{z\rightarrow z_0}(z-z_0)f(z)=0$, but don't get any approach.

  • $\begingroup$ How do you define “removable singularity”? $\endgroup$ Sep 11 '19 at 12:48
  • $\begingroup$ Removable singularity $z_o$ that fulfills $lim_{z\rightarrow z_0}(z-z_0)f(z)=0$ or moreover where I can reduce a fraction by factoring out the common zeroes in the denominator and counter $\endgroup$
    – Thesinus
    Sep 11 '19 at 12:59

If $z_0$ is removable, then $f$ admits a Laurent series representation in some punctured disc $D^*(z_0,r)$ with singular part equal to zero, so:

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

Suppose the first $k-1$ terms of the Laurent series are zero, so $a_k$ is the first non zero term. Then:

$$f(z)=(z-z_0)^k\sum_{n=0}^{\infty} {a_{n+k}}(z-z_0)^n$$

We then have that: $$f(z)=(z-z_0)^k(a_k+a_{k+1}(z-z_0)+...) =(z-z_0)^kg(z)$$

With $g(z_0)=a_k\neq 0$ by hypothesis.


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