14 - Zeros of Holomorphic Functions

Zeros of Holomorphic Functions

Let $U$ be an open set
Suppose that $g:U\to \mathbb{C}$ is holomorphic on $U$

Let $S=\{z\in U:g(z)=0\}$

If $z_0\in S$ then either $z_0$ is isolated in $S$ or $g=0$ on a neighbourhood of $z_0$
If $z_0$ is isolated in $S$ then there is unique integer $k>0$ and holomorphic function $g_1$ such that

$$g(z)=(z-z_0{)}^k{g}_1(z)$$

where $g_1(z_0)\ne 0$ and $k$ is the multiplicity of zero of $g$ at $z=z_0$
Also known as order of vanishing

Proof

Pick any $z_0\in U$ with $g(z_0)=0$

Since $g$ is analytic at $z_0$ there is $r>0$ such that $\overline{B}(z_0,r)\subseteq U$ then

$$g(z)=\sum _{k=0}^{\infty }{c}_k(z-z_0{)}^k\text{ for all }z\in B(z_0,r)\subseteq U$$

where coefficients $c_k$ are given by Taylor series for holomorphic functions

If $c_k=0$ for all $k$ then $g(z)=0$ for all $z\in \overline{B}(z_0,r)$

Otherwise let $k=\min \{n\in \mathbb{N}:c_n\ne 0\}$
As $g(z_0)=0$ then $c_0=0$ so $k\ge 1$

Let $g_1(z)=(z-z_0{)}^{-k}g(z)$
So $g_1(z)$ holomorphic on $U\setminus \{z_0\}$ and

$$g_1(z)=\sum _{n=0}^{\infty }{c}_{k+n}(z-z_0{)}^n\text{ for all }z\in B(z_0,r)$$

Hence

$$g_1(z_0)=c_k\ne 0$$

Hence $g_1(z)$ is holomorphic on all of $U$

Since $g_1$ is continuous at $z_0$ and $g_1(z_1)\ne 0$ there exists $ϵ>0$ such that

$$g_1(z)\ne 0\text{ for all }z\in B(z_0,ϵ)$$

However $(z-z_0{)}^k$ vanishes only at $z_0$ hence

$$g(z)=(z-z_0{)}^k{g}_1(z)\text{ is non zero on }B(z_0,ϵ)\setminus \{z_0\}$$

so $z_0$ is isolated

In order to show $k$ is unique
Suppose

$$g(z)=(z-z_0{)}^k{g}_1(z)=(z-z_0{)}^l{g}_2(z)$$

with $g_1(z_0)\ne 0$ and $g_2(z_0)\ne 0$

If $k<l$ then

$$\frac{g(z)}{(z-z_0{)}^k}=(z-z_0{)}^{l-k}{g}_2(z)\text{ for all }z\ne z_0$$

Hence as $z\to z_0$ then $\frac{g(z)}{(z-z_0{)}^k}\to 0$ which contradicts assumption that $g_1(z)\ne 0$
So by symmetry then $k\ngeqq l$ so $k=l$

Regular Point

Let $f:U\to \mathbb{C}$ be a function where $U$ is open

If $f$ is holomorphic at $z_0$ then

$$z_0\text{ is a regular point of }f$$

Singular Point

Let $f:U\to \mathbb{C}$ be a function where $U$ is open

If $f$ is not holomorphic at $z_0$ then

$$z_0\text{ is a singular}$$

Isolated Singularity

Let $f:U\to \mathbb{C}$ be a function where $U$ is open

If $f$ is holomorphic on $B(z_0,r)\setminus \{z_0\}$ for some $r>0$ then

$$z_0\text{ is an isolated singular}$$

Meromorphic

Let $f:U\to \mathbb{C}$ be a function where $U$ is open

$f$ is a meromorphic function on $U$ if

$$f\text{ has only isolated singularities all of which are poles}$$

Types of Isolated Singularities

Let $f:U\to \mathbb{C}$ be a function where $U$ is open

Let $z_0$ be an isolated singularity of function $f$ then

1. Removable Singularity - Exists function $g$ holomorphic in $B(z_0,r)$ with $r>0$ such that

$$f(z)=g(z)\text{ in }B(z_0,r)\setminus \{z_0\}$$

2. Pole of Order $n$ - Exists function $g$ holomorphic in $B(z_0,r)$ with $r>0$ such that

$$g(z_0)\ne 0\text{ and }f(z)=(z-z_0{)}^{-n}g(z)\text{ in }B(z_0,r)zzz\{z_0\}$$

3. Essential Singularity - Otherwise

Poles and Zeroes of Reciprocal lemma

Let $f$ be a holomorphic function in a neighbourhood of $z_0$ then

$z_0$ is a pole if and only if $|f(z)|\to \infty$ as $z\to z_0$

In this case function

$$h(z)=\{\begin{array}{ll}\frac{1}{f(z)}& \text{ if }z\ne z_0\\ 0& \text{ if }z=z_0\end{array}$$

is holomorphic on neighbourhood of $z_0$, multiplicity of zero at $z_0$ is equal to order of pole of $f$

Proof

Assume $f$ has pole of order $n$ so by Characterisation of Isolated Singularities then
$f(z)$ has a Laurent Expansion of

$$f(z)=\frac{c_{-n}}{(z-z_0{)}^n}+\cdots +c_0+c_1(z-z_0)+\cdots$$

which can be rewritten as

$$f(z)=\frac{1}{(z-z_0{)}^n}(c_{-n}+c_{-n+1}(z-z_0)+\cdots )$$

where the series converges to a function analytic at $z_0$ denoted by $g$ then

$$\frac{1}{f}=(z-z_0{)}^n\frac{1}{g}$$

Since $g(z_0)=c_{-n}\ne 0$ then function $h=\frac{1}{g}$ is holomorphic in neighbourhood of $z_0$ and

$$h(z_0)\ne 0$$

Thus $\frac{1}{f}$ has a removable singularity and after removing has a zero multiplicity of $n$

Assume $|f|\to \infty$ then

$$\frac{1}{f}\to 0$$

so by Characterisation of Isolated Singularities then
It can be extended to holomorphic function $h$ with $0$ at $z_0$

As the $0$ must be of finite order so $n\ge 1$ then

$$h(z)=(z-z_0{)}^{n}g(z)$$

where $g$ is holomorphic with $g(z_0)\ne 0$ then

$$f(z)=(z-z_0{)}^{-n}\cdot \frac{1}{g(z)}$$

so $f$ has a pole of order $n$