View source for Complex Analysis/Zeros and poles counting integral

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

==Zero of order n==
Let <math>U\subseteq \mathbb C</math> be open, <math>f: U \to  \mathbb{C}</math> a holomorphic function, and <math>z_o \in U</math>. The function <math>f</math> has <math>z_o</math>  a zero of order <math>n\in\mathbb{N}</math> at <math>z_o</math> if there exists a holomorphic function <math>g: U \to  \mathbb{C}</math>, such that:
:<math>g(z_o)\not= 0 \, \wedge \, f(z)=(z-z_o)^n \cdot g(z)</math>.
==Pole of order n==
Let <math>U\subseteq \mathbb C</math> be open, <math>f: U \setminus \{z_o\} \to  \mathbb{C}</math> a holomorphic function, and <math>z_o \in U</math>. The function <math>f</math> has <math>z_o</math> a pole of order <math>n\in\mathbb{N}</math> at <math>z_o</math> if there exists a holomorphic function <math>g: U \to  \mathbb{C}</math>, such that:
:<math>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\}</math>.

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

=== Task 1: Zero of order n ===
Using the definition of the order of a zero, compute the expression for <math>z \in U</math>:
:<math>\frac{f'(z)}{f(z)}=</math>

=== Task 2: Zero of order n ===
Explain why for the term <math>\frac{g'(z)}{g(z)}</math>, a neighborhood <math>D_\varepsilon(z_o) \subset U</math> exists where <math>\frac{g'}{g}</math> has no singularities.

=== Task 3: Zero of order n ===
Explain why <math>\frac{g'(z)}{g(z)}</math> does not necessarily need to be defined on the entire set <math>U\subseteq \mathbb C</math>.

=== Task 4: Zero of order n ===
What can you conclude for the following integrals:
:<math>\displaystyle \int_{\partial D_\varepsilon(z_o)} \frac{g'(z)}{g(z)}\, dz = ...</math>
and 
:<math>\displaystyle \int_{\partial D_\varepsilon(z_o)} \frac{f'(z)}{f(z)}\, dz  = ... </math>
=== Task 5: Pole of order n ===
Apply the calculations and explanations to poles of order <math>n\in \mathbb{N}</math> and compute the integrals:
:<math>\displaystyle \int_{\partial D_\varepsilon(z_o)} \frac{g'(z)}{g(z)}\, dz = ...</math>
and 
:<math>\displaystyle \int_{\partial D_\varepsilon(z_o)} \frac{f'(z)}{f(z)}\, dz  = ... </math>

==Statement==
Let <math>U\subseteq \mathbb C</math> be open, and <math>f \in \mathcal M(U)</math>. Let <math>N(f)</math> be the set of zeros and <math>P(f)</math> the set of poles of <math>f</math>. Let <math>\Gamma \in C(U)</math> be a [[en:Complex Analysis/Chain|Chain]] that encircles each zero and each pole of <math>f</math> exactly once in the positive orientation [[Winding number]] , i.e., <math>n(\Gamma, z) = 1</math> for each <math>z\in N(f)\cup P(f)</math>. For <math> z \in N(f) \cup P(f)</math>, we set:
:<math> 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. 
 </math>
then 
<center><math> \frac 1{2\pi i}\int_\Gamma \frac{f'(z)}{f(z)}\, dz = \sum_{z\in N(f) \cup P(f)} o_z(f). </math></center>
==Proof==
For each <math>z_0 \in N(f)\cup P(f)</math>, there exists a neighborhood <math>U_{z_0}</math> and a holomorphic function <math>g_{z_0}\colon U_{z_0} \to \mathbb C</math> such that <math>g_{z_0}(z_0) \ne 0</math>, <math>U_{z_0} \cap (N(f) \cup P(f)) = {z_0}</math>, and
:<math> f(z) = (z-z_0)^{o_{z_0}(f)}g_{z_0}(z) \qquad(\star)</math>
holds.

=== Proof 1: Holomorphicity and Application of Residue Theorem ===
The integrand is holomorphic everywhere in <math>U</math>, except possibly at <math>N(f) \cup P(f)</math>. By the [[Complex Analysis/Residue Theorem|Residue Theorem]], it suffices to compute the [[Residue|residues]] of <math>f</math> at the points of <math>N(f) \cup P(f)</math>.

=== Proof 2: Residue for Zeros/Poles ===
Let <math>z_0 \in N(f) \cup P(f)</math>. Differentiating <math>(\star)</math>, we obtain:
:<math> 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} </math>
Thus, for <math>z</math> near <math>z_0</math>:
:<math> \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} </math>with

=== Proof 3: Application of Residue Theorem ===
The second term is holomorphic, so <math>z_0</math> is a simple pole of <math>f'/f</math>, and 
:<math> \mathrm{res}_{z_0} \frac{f'}{f} = o_{z_0}(f) </math>
The claim follows by the Residue Theorem.

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