Zeros, Singularities, Residues: Zeros
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In the above expression, observe that MATH. If we set MATH then $g(z)$ is analytic at $z_{0}$ and $g(z_{0}) \neq 0$ such that MATH. Thus, the function $f(z)$ has a zero of order $m$ at $z_{0}$ if and only if $f(z)$ can be written as MATH where $g(z)$ is analytic at $z_{0}$ and $g(z_{0}) \neq 0$ .

If $f(z)$ is a non-constant analytic function in a neighborhood $N(z_{0})$ of a point $z_{0}$ and if $f(z_{0}) = 0$ then there exists a punctured disk MATH for some $r > 0$ such that $f(z)$ is analytic in $D_{r}(z_{0})$ and $f(z) \neq 0$ for all $z \in D_{r}(z_{0})$. This shows that zeros of nonconstant analytic functions are isolated . If the zeros of an analytic function $f$ is not isolated then $f$ will be either identically equal to $0$ or it will not be analytic at the limit point of the zeros of $f$. It is summarized as follows:

Theorem: Let $G$ be a connected open set and let MATH be an analytic function in $G$. Then, the following statements are equivalent.

1. $f \equiv 0$.
2. There is a point $z^{*} \in G$ such that $f^{(n)}(z^{*}) = 0$ for each $n \ge 0$.
3. The set MATH has a limit point in $G$.

   
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