Winding, the argument principle and Rouché's theorem

1. Counting a turn continuously

Let $\gamma:[0,1]\to\mathbb C\setminus\{a\}$ be a closed piecewise $C^1$ curve. Along the interval, one can choose a continuous argument $\theta(t)$ of $\gamma(t)-a$: start in a small angle chart and continue through finitely many overlapping charts covering the compact parameter interval, adding multiples of $2\pi$ at joins. This does not contradict the impossibility of a global argument on the entire punctured plane. The parameter interval has no loop identification until its endpoints are compared.

Set $r(t)=|\gamma(t)-a|>0$. On smooth pieces, logarithmic differentiation gives

$$\frac{\gamma'(t)}{\gamma(t)-a}=\frac{r'(t)}{r(t)}+i\theta'(t).$$

Integrating and using $r(1)=r(0)$ yields

$$\frac1{2\pi i}\int_\gamma\frac{dz}{z-a}=\frac{\theta(1)-\theta(0)}{2\pi}\in\mathbb Z.$$

This integer is the winding number $\operatorname{Ind}(\gamma,a)$. A positively oriented simple circle winds once about each interior point and zero times about exterior points. Clockwise orientation reverses the sign. A contour traversed twice counts twice; self-intersecting curves must be analysed by winding, not by a single informal “inside”.

Winding is invariant under a continuous deformation avoiding $a$. For smooth contour families this follows from continuity of the displayed integral and its integer values. In general the continuous-argument or local-grid argument supplies the same result. It is locally constant as a function of $a$ off the curve and zero on the unbounded component, since the integral tends to zero as $|a|\to\infty$ while staying integer valued.

2. The logarithmic derivative detects multiplicity

A meromorphic function on an open set is holomorphic there except for isolated poles. Suppose it is not identically zero on a domain. Near a zero of order $m$, write $f(z)=(z-a)^mg(z)$ with $g(a)\ne0$. Then

$$\frac{f'(z)}{f(z)}=\frac m{z-a}+\frac{g'(z)}{g(z)},$$

so its residue is $m$. Near a pole of order $m$, the same formula with exponent $-m$ gives residue $-m$. At a nonzero finite value, the logarithmic derivative is holomorphic. Multiplicities, not merely distinct points, are what it records.

Argument principle, simple-boundary form. Let $f$ be meromorphic on a neighbourhood of the closure of a bounded Jordan domain $\Omega$ with piecewise smooth positively oriented boundary. Assume there are no zeros or poles on the boundary. Then

$$\frac1{2\pi i}\int_{\partial\Omega}\frac{f'(z)}{f(z)}\,dz=N-P,$$

where $N$ and $P$ count zeros and poles inside with multiplicity.

The zeros and poles in this compact region are finite: otherwise an accumulation point in the neighbourhood would violate zero isolation or the local definition of meromorphy, and the boundary has none. The residue theorem applied to $f'/f$ and the local calculation prove the formula. For regions with holes, orient all components as in C05; for more general cycles, insert winding multiplicities with valid domain hypotheses.

The same integral is the winding of the image boundary $f\circ\gamma$ about zero, because $d(f(\gamma(t)))/dt=f'(\gamma(t))\gamma'(t)$. Consequently the total argument change of $f$ around the contour is $2\pi(N-P)$. A plotted image can suggest this change, but a sampled plot can miss small loops and is not by itself a proof.

Worked example. For $f(z)=(z-1)^3/(z+2)^2$ on $|z|=3$, there are three zeros and two poles counted with multiplicity, hence $N-P=1$. The image curve winds once around zero. On $|z|=3/2$, the count is three, since the pole at $-2$ is outside. On $|z|=1$ the theorem is inapplicable because a zero lies on the contour.

3. Rouché's theorem: a zero count stable under a smaller perturbation

Theorem. Let $f,g$ be holomorphic on a neighbourhood of the closure of a bounded Jordan domain with piecewise smooth boundary. If $|g(z)|<|f(z)|$ on its boundary, then $f$ and $f+g$ have the same number of interior zeros, counted with multiplicity.

Proof. For $0\le t\le1$, put $h_t=f+tg$. On the boundary,

$$|h_t|\ge|f|-t|g|\ge|f|-|g|>0.$$

Compactness gives a common positive lower bound there. Thus

$$n(t)=\frac1{2\pi i}\int_{\partial\Omega}\frac{f'+tg'}{f+tg}\,dz$$

is continuous in $t$. By the argument principle, $n(t)$ is the nonnegative integer counting zeros of $h_t$ inside. A continuous integer-valued function on an interval is constant. Evaluating at $t=0,1$ proves the result. ∎

Strict inequality excludes boundary zeros throughout the homotopy. Replacing it by a non-strict comparison without extra information is not justified.

Worked example: a small perturbation of $z^5$. For $p(z)=z^5+2z+1$ on $|z|=2$,

$$|2z+1|\le5<32=|z^5|.$$

Therefore $p$ has five zeros in $|z|<2$. On $|z|=1/3$, compare $2z+1$ with $z^5$: $|2z+1|\ge1/3>1/243$. The linear comparison function has its root at $-1/2$, outside this disk, so $p$ has no zero there. The annulus $1/3<|z|<2$ contains all five zeros counted with multiplicity.

Worked example: count without solving. For $q(z)=z^4+4z+1$ on $|z|=1$, take $f=4z$ and $g=z^4+1$. The bound $|g|\le2<4$ gives exactly one zero in the unit disk. Solving a quartic explicitly is unnecessary.

4. Zero location and the danger of numerical pictures

When searching for a contour comparison, choose a term or a simpler polynomial that dominates everywhere on the boundary. Bounds must be uniform. A few sampled boundary points do not prove strict inequality, especially near cancellation. Triangle and reverse-triangle inequalities often provide certified margins.

If the derivative and a root approximation are available, a small circle around the approximation can sometimes separate the linear Taylor term from its remainder, yielding an existence-and-uniqueness proof through Rouché. The remainder needs a rigorous bound, not a floating-point residual alone. A tiny value of $p(z_0)$ does not by itself imply a nearby zero without derivative or conditioning information.

The argument principle also distinguishes zeros from poles. If $N-P=0$, the function need not have neither: it may have equal total numbers of zeros and poles. Nor does a zero net winding mean the image curve never approaches or encloses zero in subloops.

5. Exercises

P01 — Foundation. Compute the winding number about zero of $\gamma(t)=2e^{-4\pi it}$, $0\le t\le1$.

P02 — Core. Find $\frac1{2\pi i}\int_{|z|=2}f'/f\,dz$ for $f(z)=z^2(z-1)/(z-3)^4$.

P03 — Core. Count the zeros of $z^6+3z+1$ inside $|z|<1$ using Rouché.

P04 — Core. Show that every zero of $z^4+z+1$ lies in $|z|<2$.

P05 — Diagnosis. Explain why the argument principle cannot be applied on $|z|=1$ to $f(z)=z^2-1$ without modifying the contour.

P06 — Proof. For a zero of order $m$, prove that the logarithmic derivative has residue $m$ without explicitly differentiating an infinite Laurent series.

P07 — Extension. Show that $z^5-5z+1$ has exactly one zero in $|z|<1/2$ and exactly five in $|z|<2$.

P08 — Proof. Give a meromorphic function with $N-P=0$ inside the circle $|z|=2$ but with both a zero and a pole there. Compute its logarithmic-derivative integral.

6. Full solutions

P01. A continuous argument is $-4\pi t$, with change $-4\pi$, so the winding is $-2$. The radius does not affect the count.

P02. The zero at zero has order two and the zero at one has order one; the pole at three lies outside. Thus $N-P=3$ and the normalized integral is three. There are no boundary zeros or poles.

P03. On the unit circle, $|z^6+1|\le2<3=|3z|$. Comparing with $3z$ gives one zero inside, counted with multiplicity.

P04. On $|z|=2$, $|z+1|\le3<16=|z^4|$. Thus there are four zeros inside, which exhaust the degree-four polynomial's four zeros counted with multiplicity. The strict comparison also excludes boundary roots.

P05. The roots $1$ and $-1$ are on the contour; $f'/f$ is undefined there. The required contour integral and no-boundary-zero hypothesis fail. A slightly smaller or larger radius gives a different valid count, respectively zero or two; neither count applies to the boundary itself.

P06. Factor $f=(z-a)^mg$ with $g$ holomorphic and $g(a)\ne0$. Shrink the neighbourhood so $g\ne0$. Product and quotient rules give $f'/f=m/(z-a)+g'/g$, and the second term is holomorphic, proving the residue statement.

P07. On $|z|=1/2$, compare $-5z$ with $z^5+1$: $|z^5+1|\le1+1/32=33/32<5/2$. Therefore exactly one zero lies inside. On $|z|=2$, compare $z^5$ with $-5z+1$: its modulus is at most eleven, below thirty-two. Thus exactly five zeros lie inside the larger disk. The remaining four zeros counted with multiplicity lie in the annulus $1/2<|z|<2$.

P08. Take $f(z)=z/(z-1)$. The disk contains one simple zero at zero and one simple pole at one, so $N-P=0$. Directly $f'/f=1/z-1/(z-1)$, whose two residues are $1$ and $-1$; the contour integral is zero. Cancellation of counts is not absence of singularities.

7. Summary

The logarithmic derivative converts local zero and pole orders to residues, and its boundary integral measures an integer winding. A smaller boundary perturbation cannot change that integer, which is the mechanism behind Rouché. The method counts roots without solving their equations, provided all boundary and multiplicity hypotheses are checked.