# Suppose $g(z)$ is analytic at $z_0$. Proof $g(z)$ has zero of order $m$ at $z_0$ iff $lim_{z\rightarrow z_0} \frac{g(z)}{(z-z_0)^m} \neq 0$.

Suppose $$g(z)$$ is analytic at $$z_0$$. Proof $$g(z)$$ has zero of order $$m$$ at $$z_0$$ iff $$lim_{z\rightarrow z_0} \frac{g(z)}{(z-z_0)^m} \neq 0$$.

Sketch:

We have that $$g(z)=\sum_{n\rightarrow\infty} a_n(z-z_0)^n$$ If g(z) has zero of order m at $$z_0$$ then $$\frac{g(z)}{(z-z_0)^m}=a_m+...$$ Therefore $$lim_{z\rightarrow z_0} \frac{g(z)}{(z-z_0)^m} = a_m \neq 0$$

If $$lim_{z\rightarrow z_0} \frac{g(z)}{(z-z_0)^m}=l \neq 0$$

Def: $$h(z)=\frac{g(z)}{(z-z_0)^m}$$ at $$z\neq z_o$$ and $$h(z_0)=l$$

Since $$lim_{z\rightarrow z_0} h(z)$$ exists $$h(z)$$ has removable singularity at $$z_0$$

Hence $$h(z)=\sum_{n\rightarrow\infty} a_n(z-z_0)^n$$ where $$a_1=l$$ which implies that $$g(z)=\sum_{n\rightarrow\infty} a_n(z-z_0)^{m+n}$$

Hence $$g(z)$$ has zero of order m at $$z_0$$

Is the proof valid?

• How do you define a zero of order $m$? – José Carlos Santos Oct 20 '18 at 6:31
• the smallest m such that a_m is not zero – Jhon Doe Oct 20 '18 at 6:36

Suppose that $$z_0$$ is not a zero of order $$m$$. There are then $$3$$ possibilities:
1. $$z_0$$ is not a zero of $$g$$. Then, since $$\lim_{z\to z_0}(z-z_0)^m=0$$, the limit $$\lim_{z\to z_0}\frac{g(z)}{(z-z_0)^m}$$ doesn't exist (in $$\mathbb C$$).
2. $$z_0$$ is a zero of order $$k. Then$$\lim_{z\to z_0}\frac{g(z)}{(z-z_0)^m}=\lim_{z\to z_0}\frac{g(z)}{(z-z_0)^k(z-z_0)^{m-k}}=\lim_{z\to z_0}\frac{\frac{g(z)}{(z-z_0)^k}}{(z-z_0)^{m-k}},$$ which, again, doesn't exist (in $$\mathbb C$$).
3. $$z_0$$ is a zero of order $$k>m$$. Then a similar computation shows that the limit does exist, but it is equal to $$0$$.