The Argument Principle and Rouché's Theorem

Let $\mathrm{\Omega }\subseteq \mathbb{C}$ be a complex domain, let $f$ be meromorphic on $\mathrm{\Omega }$, let $Z(f)$ be the zero set of $f$, and let $P(f)$ be its pole set. On the open set

$\mathrm{\Omega }\setminus (Z(f)\cup P(f))$

the function $f$ is holomorphic and nonzero, so the quotient

$\frac{f^{\mathrm{\prime }}}{f}$

is holomorphic there. This quotient is the logarithmic derivative of $f$.

The logarithmic derivative is not defined at a zero or a pole of $f$ by the displayed quotient itself. At every zero of finite positive order and at every pole, the next lemma identifies its principal part and shows that the resulting singularity is simple. The identically zero function is excluded from that local conclusion because its quotient $f^{\mathrm{\prime }}\mathrm{/}f$ is defined nowhere.

Let $\mathrm{\Omega }\subseteq \mathbb{C}$ be open, let $f$ be meromorphic on $\mathrm{\Omega }$, and let $\mathrm{\Gamma }$ be admissible for the residue theorem in $\mathrm{\Omega }$. Assume also that $f$ is not identically zero on any connected component of $\mathrm{\Omega }$ and has no zeros on ${\mathrm{\Gamma }}^{\ast }$, so every zero has finite positive order and every index $n(\mathrm{\Gamma },a)$ of Integration over a complex chain and the index of a chain is defined at every zero or pole of $f$.

Here, and in the argument-principle results that use this definition, meromorphic on an open set means meromorphic on every connected component in the sense of Meromorphic functions on a plane domain. The zero and pole sets are the unions of the corresponding componentwise sets.

Whenever only finitely many zeros and poles of $f$ have nonzero index with respect to $\mathrm{\Gamma }$, define the weighted zero count

$Z(f,\mathrm{\Gamma }):={\sum }_{a\in Z(f)}n(\mathrm{\Gamma },a){\mathrm{ord}}_a(f)$

and the weighted pole count

$P(f,\mathrm{\Gamma }):={\sum }_{b\in P(f)}n(\mathrm{\Gamma },b){\mathrm{ord}}_b^{\mathrm{p}\mathrm{o}\mathrm{l}\mathrm{e}}(f),$

where ${\mathrm{ord}}_a(f)$ is the zero order from The order of a zero of a holomorphic function and ${\mathrm{ord}}_b^{\mathrm{p}\mathrm{o}\mathrm{l}\mathrm{e}}(f)$ is the positive pole order at $b$.

These are finite sums only after a finiteness argument. Under the argument principle hypotheses, that finiteness comes from applying the residue theorem to the logarithmic derivative.

Let $f$ be meromorphic on a neighbourhood of $a\in \mathbb{C}$.

1. If $a$ is a zero of $f$ of order $m\ge 1$, then $\mathrm{Res}\text{\hspace{0.17em}⁣}(\frac{f^{\mathrm{\prime }}}{f},a)=m\mathrm{.}$
2. If $a$ is a pole of $f$ of order $m\ge 1$, then $\mathrm{Res}\text{\hspace{0.17em}⁣}(\frac{f^{\mathrm{\prime }}}{f},a)=-m\mathrm{.}$

In either case $f^{\mathrm{\prime }}\mathrm{/}f$ has a simple pole at $a$.

Let $\mathrm{\Omega }\subseteq \mathbb{C}$ be open, let $f$ be meromorphic on $\mathrm{\Omega }$, and let $\mathrm{\Gamma }$ be admissible for the residue theorem in $\mathrm{\Omega }$. Suppose in addition that $f$ is not identically zero on any connected component of $\mathrm{\Omega }$ and that $f(z)\mathrm{\ne }0$ for every $z\in {\mathrm{\Gamma }}^{\ast }$. Then

$\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}\frac{f^{\mathrm{\prime }}(z)}{f(z)}dz=Z(f,\mathrm{\Gamma })-P(f,\mathrm{\Gamma }),$

where the weighted zero and pole counts are those of Zero and pole counts weighted by multiplicity and winding number.

Only finitely many terms in those weighted counts are nonzero.

As in Zero and pole counts weighted by multiplicity and winding number, meromorphicity on the possibly disconnected open set $\mathrm{\Omega }$ is understood componentwise.

Let $\gamma :[a,b]\to \mathbb{C}$ be a closed complex contour, let $f$ be meromorphic on a neighbourhood of ${\gamma }^{\ast }$, and suppose $f(z)\mathrm{\ne }0$ for every $z\in {\gamma }^{\ast }$. Then $f\circ \gamma$ is a closed complex contour with $0\mathrm{\notin }(f\circ \gamma {)}^{\ast }$, and

$\frac{1}{2\pi i}{\int }_{\gamma }\frac{f^{\mathrm{\prime }}(z)}{f(z)}dz=n(f\circ \gamma ,0)\mathrm{.}$

Equivalently, if $\theta$ is any continuous argument of $f\circ \gamma$, then

$\frac{1}{2\pi i}{\int }_{\gamma }\frac{f^{\mathrm{\prime }}(z)}{f(z)}dz=\frac{\theta (b)-\theta (a)}{2\pi }\mathrm{.}$

If $\gamma$ is also admissible and null-homologous in a larger open set on which $f$ is meromorphic, then the same integer equals $Z(f,\gamma )-P(f,\gamma )$ by The argument principle for an admissible null-homologous cycle.

Let $\mathrm{\Omega }\subseteq \mathbb{C}$ be open, let $f$ be meromorphic on $\mathrm{\Omega }$, let $\mathrm{\Gamma }$ be admissible for the residue theorem in $\mathrm{\Omega }$, and let $w\in \mathbb{C}$ satisfy $f(z)\mathrm{\ne }w$ for every $z\in {\mathrm{\Gamma }}^{\ast }$. Then

$\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}\frac{f^{\mathrm{\prime }}(z)}{f(z)-w}dz=N_w(f,\mathrm{\Gamma })-P(f,\mathrm{\Gamma }),$

where

$N_w(f,\mathrm{\Gamma }):={\sum }_{f(a)=w}n(\mathrm{\Gamma },a){\mathrm{ord}}_a(f-w)$

is the weighted multiplicity count of the preimages of $w$, and $P(f,\mathrm{\Gamma })$ is the weighted pole count of $f$.

In particular, if $f$ is holomorphic on $\mathrm{\Omega }$, then the pole term vanishes and the integral counts the preimages of $w$ with multiplicity.

Let $\mathrm{\Omega }\subseteq \mathbb{C}$ be open, let $f$ be meromorphic on $\mathrm{\Omega }$, let $\mathrm{\Gamma }$ be admissible for the residue theorem in $\mathrm{\Omega }$, suppose $f(z)\mathrm{\ne }0$ for every $z\in {\mathrm{\Gamma }}^{\ast }$, and let $g$ be holomorphic on $\mathrm{\Omega }$. Then

$\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}g(z)\frac{f^{\mathrm{\prime }}(z)}{f(z)}dz={\sum }_{a\in Z(f)}n(\mathrm{\Gamma },a){\mathrm{ord}}_a(f)g(a)-{\sum }_{b\in P(f)}n(\mathrm{\Gamma },b){\mathrm{ord}}_b^{\mathrm{p}\mathrm{o}\mathrm{l}\mathrm{e}}(f)g(b),$

and only finitely many terms are nonzero.

Let $\mathrm{\Omega }\subseteq \mathbb{C}$ be open, let $\gamma$ be a closed complex contour that is null-homologous in $\mathrm{\Omega }$, and let $f,g$ be holomorphic on $\mathrm{\Omega }$. If

$\mathrm{\mid }f(z)-g(z)\mathrm{\mid }<\mathrm{\mid }g(z)\mathrm{\mid }(z\in {\gamma }^{\ast }),$

then $f$ and $g$ have the same weighted number of zeros with respect to $\gamma$.

In particular, if $\gamma$ is the positively oriented boundary of a Jordan domain, then $f$ and $g$ have the same number of zeros inside $\gamma$, counted with multiplicity.

Let

$p(z)=a_n{z}^n+a_{n-1}{z}^{n-1}+\cdots +a_0,a_n\mathrm{\ne }0.$

On the circle $\mathrm{\mid }z\mathrm{\mid }=R$ one has

$\mathrm{\mid }{a}_{n-1}{z}^{n-1}+\cdots +a_0\mathrm{\mid }\le \mathrm{\mid }{a}_{n-1}\mathrm{\mid }{R}^{n-1}+\cdots +\mathrm{\mid }{a}_0\mathrm{\mid },$

so for sufficiently large $R$ the lower-degree tail is strictly smaller than $\mathrm{\mid }{a}_n\mathrm{\mid }{R}^n=\mathrm{\mid }{a}_n{z}^n\mathrm{\mid }$. Rouché's theorem therefore gives the same number of zeros for $p$ and its leading term $a_n{z}^n$ inside $\mathrm{\mid }z\mathrm{\mid }<R$, namely $n$ counting multiplicity.

This is exactly the standard leading-term proof that every nonconstant complex polynomial has roots and, more sharply, that a degree-$n$ polynomial has exactly $n$ roots counted with multiplicity, agreeing with the canonical result A complex polynomial of degree $n$ has exactly $n$ roots counted with multiplicity.

Let $f$ be holomorphic on a neighbourhood of the closed disc $\overline{D(a,r)}$, and suppose $f$ has no zero on the circle $\mathrm{\mid }z-a\mathrm{\mid }=r$. Let $m$ be the total multiplicity of the zeros of $f$ in $D(a,r)$. If $g$ is holomorphic on a neighbourhood of $\overline{D(a,r)}$ and

$\mathrm{\mid }g(z)-f(z)\mathrm{\mid }<\mathrm{\mid }f(z)\mathrm{\mid }(\mathrm{\mid }z-a\mathrm{\mid }=r),$

then $g$ has exactly $m$ zeros in $D(a,r)$, counted with multiplicity.

In particular, if $a$ is an isolated zero of $f$ of order $m$ and the disc is chosen so that $a$ is the only zero of $f$ in $\overline{D(a,r)}$, then every such perturbation $g$ has exactly $m$ zeros in $D(a,r)$ counted with multiplicity.

Let $\mathrm{\Omega }\subseteq \mathbb{C}$ be open, let $f_n:\mathrm{\Omega }\to \mathbb{C}$ be holomorphic, and suppose $f_n\to f$ locally uniformly on $\mathrm{\Omega }$. Let $a\in \mathrm{\Omega }$ be an isolated zero of $f$ of multiplicity $m\ge 1$. Then there is $r>0$ such that $\overline{D(a,r)}\subseteq \mathrm{\Omega }$, the function $f$ has no zero on $\mathrm{\mid }z-a\mathrm{\mid }=r$, and for all sufficiently large $n$ the function $f_n$ has exactly $m$ zeros in $D(a,r)$ counted with multiplicity.

Let $\mathrm{\Omega }\subseteq \mathbb{C}$ be a complex domain, let each $f_n:\mathrm{\Omega }\to \mathbb{C}$ be holomorphic and nowhere zero, and suppose $f_n\to f$ locally uniformly on $\mathrm{\Omega }$. Then either $f\equiv 0$ on $\mathrm{\Omega }$, or $f$ is nowhere zero on $\mathrm{\Omega }$.

Let $\mathrm{\Omega }\subseteq \mathbb{C}$ be a complex domain, let each $f_n:\mathrm{\Omega }\to \mathbb{C}$ be holomorphic and injective, and suppose $f_n\to f$ locally uniformly on $\mathrm{\Omega }$. Then $f$ is injective or constant.

If $f$ is nonconstant and holomorphic on a complex domain and $a\in \mathrm{\Omega }$, then $f-f(a)$ has a zero of positive multiplicity at $a$. Choose a small circle around $a$ on which $f-f(a)$ does not vanish. The preimage-count corollary says that for every $w$ sufficiently close to $f(a)$ the function $f-w$ has the same positive number of zeros inside that circle. So every value near $f(a)$ is attained nearby, and the image is open.

This reproduces the already-published open mapping theorem Open mapping theorem for holomorphic functions. The earlier page keeps its local-normal-form proof because the reading order needs that theorem before the argument principle exists.

Fix a nonconstant holomorphic map $f$ and a point $a$ of local degree $m={\mathrm{deg}}_{a}f$. The earlier local-sheet theorem A local degree-m holomorphic map has m nearby sheets says that every nearby value other than $f(a)$ has exactly $m$ nearby preimages. On the other hand, the argument-principle preimage count on a small circle around $a$ counts the zeros of $f-w$ in that same disc, with multiplicity, and therefore gives the same integer $m$.

So the multiplicity seen analytically by the argument principle agrees with the local degree already built from the normal form. This is an agreement remark, not a replacement of the earlier construction.

Let $\mathrm{\Omega }\subseteq \mathbb{C}$ be open, let $f$ be holomorphic on $\mathrm{\Omega }$, let $\mathrm{\Gamma }$ be a closed complex contour null-homologous in $\mathrm{\Omega }$, and let $w\in \mathbb{C}$ satisfy $f(z)\mathrm{\ne }w$ for every $z\in {\mathrm{\Gamma }}^{\ast }$. Suppose $f(z)=w$ has exactly one solution $a$ inside $\mathrm{\Gamma }$, that solution is simple, and $n(\mathrm{\Gamma },a)=1$. Then

$a=\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}\frac{\zeta f^{\mathrm{\prime }}(\zeta )}{f(\zeta )-w}d\zeta \mathrm{.}$

Thus, on a contour enclosing exactly one simple preimage branch, the inverse value is recovered by a contour integral.