Complex Analysis/Zero and Pole counting integral

The integral counting zeros and poles counts, as the name suggests, the zeros and poles of a meromorphic function along with their multiplicities. More precisely:

Zero of order n

Let $U\subseteq ℂ$ be open, $f:U\to ℂ$ a holomorphic function, and $z_o\in U$. The function $f$ has $z_o$ a zero of order $n\in \mathbb{N}$ at $z_o$ if there exists a holomorphic function $g:U\to ℂ$, such that:

$g(z_o)=0\wedge f(z)=(z-z_o{)}^n\cdot g(z)$.

Pole of order n

Let $U\subseteq ℂ$ be open, $f:U\setminus \{z_o\}\to ℂ$ a holomorphic function, and $z_o\in U$. The function $f$ has $z_o$ a pole of order $n\in \mathbb{N}$ at $z_o$ if there exists a holomorphic function $g:U\to ℂ$, such that:

$g(z_o)=w_o\in ℂ\wedge f(z)=(z-z_o{)}^{-n}\cdot g(z)\text{ mit }z\in U\setminus \{z_o\}$.

Tasks

Let $U\subseteq ℂ$ be open, $f:U\to ℂ$ a holomorphic function, and $z_o\in U$. Furthermore, let $f$ have $z_o$ a zero of order $n\in \mathbb{N}$ at $z_o$.

Task 1: Zero of order n

Using the definition of the order of a zero, compute the expression for $z\in U$:

$\frac{f^{\prime }(z)}{f(z)}=$

Task 2: Zero of order n

Explain why for the term $\frac{g^{\prime }(z)}{g(z)}$, a neighborhood $D_{\epsilon }(z_o)\subset U$ exists where $\frac{g^{\prime }}{g}$ has no singularities.

Task 3: Zero of order n

Explain why $\frac{g^{\prime }(z)}{g(z)}$ does not necessarily need to be defined on the entire set $U\subseteq ℂ$.

Task 4: Zero of order n

What can you conclude for the following integrals:

$\underset{\partial D_{\epsilon }(z_o)}{\int }\frac{g^{\prime }(z)}{g(z)}dz=...$

and

$\underset{\partial D_{\epsilon }(z_o)}{\int }\frac{f^{\prime }(z)}{f(z)}dz=...$

Task 5: Pole of order n

Apply the calculations and explanations to poles of order $n\in \mathbb{N}$ and compute the integrals:

$\underset{\partial D_{\epsilon }(z_o)}{\int }\frac{g^{\prime }(z)}{g(z)}dz=...$

and

$\underset{\partial D_{\epsilon }(z_o)}{\int }\frac{f^{\prime }(z)}{f(z)}dz=...$

Statement

Let $U\subseteq ℂ$ be open, and $f\in ℳ(U)$. Let $N(f)$ be the set of zeros and $P(f)$ the set of poles of $f$. Let $\mathrm{\Gamma }\in C(U)$ be a Chain that encircles each zero and each pole of $f$ exactly once in the positive orientation Winding number , i.e., $n(\mathrm{\Gamma },z)=1$ for each $z\in N(f)\cup P(f)$. For $z\in N(f)\cup P(f)$, we set:

$o_z(f):=\{\begin{array}{cc}m& z\text{is}\text{zero order}m\text{-ter}\text{Order}\\ -m& z\text{is}\text{Pole}m\text{-ter}\text{Order}\end{array}$

then

$\frac{1}{2\pi i}\underset{\mathrm{\Gamma }}{\int }\frac{f^{\prime }(z)}{f(z)}dz=\sum _{z\in N(f)\cup P(f)}{o}_z(f).$

Proof

For each $z_0\in N(f)\cup P(f)$, there exists a neighborhood $U_{z_0}$ and a holomorphic function $g_{z_0}:U_{z_0}\to ℂ$ such that $g_{z_0}(z_0)\ne 0$, $U_{z_0}\cap (N(f)\cup P(f))=z_0$, and

$f(z)=(z-z_0{)}^{o_{z_0}(f)}{g}_{z_0}(z)(\star )$

holds.

Proof 1: Holomorphicity and Application of Residue Theorem

The integrand is holomorphic everywhere in $U$, except possibly at $N(f)\cup P(f)$. By the Residue Theorem, it suffices to compute the residues of $f$ at the points of $N(f)\cup P(f)$.

Proof 2: Residue for Zeros/Poles

Let $z_0\in N(f)\cup P(f)$. Differentiating $(\star )$, we obtain:

$f^{\prime }(z)=o_{z_0}(f)(z-z_0{)}^{o_{z_0}(f)-1}{g}_{z_0}(z)+(z-z_0{)}^{o_{z_0}(f)}{g}_{{z_0}^{\prime }}(z),z\in U_{z_0}$

Thus, for $z$ near $z_0$:

$\frac{f^{\prime }(z)}{f(z)}=\frac{o_{z_0}(f)}{z-z_0}+\frac{g^{\prime }{z}_0(z)}{g{z}_0(z)},z\in U_{z_0}$with

Proof 3: Application of Residue Theorem

The second term is holomorphic, so $z_0$ is a simple pole of $f^{\prime }/f$, and

${\mathrm{r}\mathrm{e}\mathrm{s}}_{z_0}\frac{f^{\prime }}{f}=o_{z_0}(f)$

The claim follows by the Residue Theorem.

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• Date: 01/07/2024

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