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$.

Question

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?

Answer 1

Your proof is correct, but you can prove it without using the concept of removable singularity.

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 $k<m$. 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 $0$.