# rewrite function with removable singularity

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.

• How do you define “removable singularity”? Sep 11 '19 at 12:48
• 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 Sep 11 '19 at 12:59

## 1 Answer

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.