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:

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

${\displaystyle g(z_{o})\not =0\,\wedge \,f(z)=(z-z_{o})^{n}\cdot g(z)}$.

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

${\displaystyle g(z_{o})=w_{o}\in \mathbb {C} \,\wedge \,f(z)=(z-z_{o})^{-n}\cdot g(z){\mbox{ mit }}z\in U\setminus \{z_{o}\}}$.

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

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

${\displaystyle {\frac {f'(z)}{f(z)}}=}$

Explain why for the term ${\displaystyle {\frac {g'(z)}{g(z)}}}$, a neighborhood ${\displaystyle D_{\varepsilon }(z_{o})\subset U}$ exists where ${\displaystyle {\frac {g'}{g}}}$ has no singularities.

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

What can you conclude for the following integrals:

${\displaystyle \displaystyle \int _{\partial D_{\varepsilon }(z_{o})}{\frac {g'(z)}{g(z)}}\,dz=...}$

and

${\displaystyle \displaystyle \int _{\partial D_{\varepsilon }(z_{o})}{\frac {f'(z)}{f(z)}}\,dz=...}$

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

${\displaystyle \displaystyle \int _{\partial D_{\varepsilon }(z_{o})}{\frac {g'(z)}{g(z)}}\,dz=...}$

and

${\displaystyle \displaystyle \int _{\partial D_{\varepsilon }(z_{o})}{\frac {f'(z)}{f(z)}}\,dz=...}$

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

${\displaystyle o_{z}(f):=\left\{{\begin{array}{rr}m&z\ {\text{is}}\ {\text{zero order}}\ m{\text{-ter}}\ {\text{Order}}\\-m&z\ {\text{is}}\ {\text{Pole}}\ m{\text{-ter}}\ {\textrm {Order}}\end{array}}\right.}$

then

${\displaystyle {\frac {1}{2\pi i}}\int _{\Gamma }{\frac {f'(z)}{f(z)}}\,dz=\sum _{z\in N(f)\cup P(f)}o_{z}(f).}$

For each ${\displaystyle z_{0}\in N(f)\cup P(f)}$, there exists a neighborhood ${\displaystyle U_{z_{0}}}$ and a holomorphic function ${\displaystyle g_{z_{0}}\colon U_{z_{0}}\to \mathbb {C} }$ such that ${\displaystyle g_{z_{0}}(z_{0})\neq 0}$, ${\displaystyle U_{z_{0}}\cap (N(f)\cup P(f))={z_{0}}}$, and

${\displaystyle f(z)=(z-z_{0})^{o_{z_{0}}(f)}g_{z_{0}}(z)\qquad (\star )}$

holds.

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

Let ${\displaystyle z_{0}\in N(f)\cup P(f)}$. Differentiating ${\displaystyle (\star )}$, we obtain:

${\displaystyle f'(z)=o_{z_{0}}(f)(z-z_{0})^{o_{z_{0}}(f)-1}g_{z_{0}}(z)+(z-z_{0})^{o_{z_{0}}(f)}g_{z_{0}}'(z),\qquad z\in U_{z_{0}}}$

Thus, for ${\displaystyle z}$ near ${\displaystyle z_{0}}$:

${\displaystyle {\frac {f'(z)}{f(z)}}={\frac {o_{z_{0}}(f)}{z-z_{0}}}+{\frac {g'{z_{0}}(z)}{g{z_{0}}(z)}},z\in U_{z_{0}}}$with

The second term is holomorphic, so ${\displaystyle z_{0}}$ is a simple pole of ${\displaystyle f'/f}$, and

${\displaystyle \mathrm {res} _{z_{0}}{\frac {f'}{f}}=o_{z_{0}}(f)}$

The claim follows by the Residue Theorem.

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• Source: Kurs:Funktionentheorie/Null- und Polstellen zählendes Integral - URL:

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• Date: 01/07/2024