The Argument Principle

The Argument Principle

Lemma

Suppose with holomorphic in a neighborhood of and . Then has a pole of order at with residue .

Proof The function is holomorphic in a neighborhood of , thus it can be expressed as a power series, thus the principle part of the Laurent expansion at is . This implies that the residue of at is .

Argument Principle

Let be a bounded Jordan domain, and let . Let , and assume that every is a pole of , with order ). Let be the zeroes of , with orders . Let and . Suppose that f has no zeroes on . Then

Proof We have . From the definition of orders of poles and zeroes, we can write where is holomorphic and non-zero near , and similarly . Thus by above lemma, we have and . By the Residue Theorem, we then obtain $$\int_{\partial D}\frac{f^{\prime}(z)}{f(z)}\dd{z}=2\pi i\left(\sum_{j=1}^{k}K_{j}-\sum_{j=1}^{n}N_{j}\right)=2\pi i(K-N)$$$\square$

A Geometric Understanding of

The function wants to be the derivative "" if were holomorphic on the domain. What happens is that the integral picks up the change in value of as you follow the boundary path, i.e. you get a factor of for every time you loop counterclockwise around the origin.

Winding Numbers

Winding Number (Index)

Let be a closed piecewise path in . We define the index (or winding number) of about as

Winding Number is Integer

By the homotopy theorem, , where is CP-homotopic to . This is capturing the idea of how many times winds about the origin.

Lemma

Let , let be a closed path in , and assume for all (so that is a closed path in . Then

Proof We compute from the definitions:$$\begin{aligned}\ind_0f\circ\gamma&=\frac1{2\pi i}\int_{f\circ\gamma}\frac1z\dd{z}\&=\frac1{2\pi i}\int_0^1\frac{f^{\prime}(\gamma(t))\gamma^{\prime}(t)}{f(\gamma(t))}\dd{t}\&=\frac1{2\pi i}\int_\gamma\frac{f^{\prime}(z)}{f(z)}\dd{z}\end{aligned}$$$\square$

Definition

If is a bounded Jordan domain and is holomorphic and non-zero in a neighborhood of , then we define where are boundary components of . And the change in argument of along to be

Argument Principle in Alternative Form

Let be a bounded Jordan domain, and let . Let , and assume that every is a pole of , with order ). Let be the zeroes of , with orders . Let and . Suppose that f has no zeroes on . Then

Proof By previous version of argument principle, we have $$2\pi(K-N)=\frac{2\pi}{2\pi i}\int_{\partial D}\frac{f^{\prime}(z)}{f(z)}\dd{z}=2\pi\ind_0f(\partial D)=\Delta_{\partial D}f$$$\square$

e.g. Let , , and .

This function has simple zeroes at . So the image winds counterclockwise around the origin twice, and winds counterclockwise around the origin once.

Rouché’s Theorem

Definition

Let be a bounded Jordan domain, and let with no zeroes on the boundary. Let be the zeroes of in , with orders . Then we define Immediately by the argument principle, we have that

Rouché's Theorem

Let be a bounded Jordan domain, let , and suppose that for . Then .

Proof It suffices to show that for any boundary component with Jordan parameterization , holds. By definition, and similarly Let . Since on , we have for all , and so by the triangle inequality, for all . Thus the range of lies in , and so is a CP-homotopy from to in , therefore by homotopy invariance, we have .

e.g. Yet another proof of the fundamental theorem of algebra: Let be a polynomial of degree . Then let and . If we choose large enough such that when , then by Rouché’s theorem, This does show that has roots (including multiplicity) in .