The winding number jumps by one across a regular planar arc

Write the regular arc near $0$ as $z(t)=x(t)+iy(t)$ on a parameter interval $[-\delta ,\delta ]$, with $t=0$ corresponding to $0$, so $x(0)=y(0)=0$ and after an orientation-preserving affine change of parameter the positive real tangent gives $x^{\mathrm{\prime }}(0)=1$, $y^{\mathrm{\prime }}(0)=0$.

Shrink $\delta$ so that $x^{\mathrm{\prime }}\ge 1\mathrm{/}2$ on the interval; then [F6] makes the real part strictly increasing, [F7] shows its image contains a symmetric interval $[-a,a]$ about zero; restrict the arc to the preimage of this interval, and the inverse $x\mapsto t(x)$ is $C^1$ with derivative $1\mathrm{/}{x}^{\mathrm{\prime }}(t(x))$ by the difference quotient and the positive lower bound; hence the arc is a graph $z(x)=x+if(x)$ for $\mathrm{\mid }x\mathrm{\mid }\le a$ with $f(0)=f^{\mathrm{\prime }}(0)=0$, and after shrinking $a$ further one has $\mathrm{\mid }{z}^{\mathrm{\prime }}(x)\mathrm{\mid }\le 2$ and $\mathrm{\mid }f(x)\mathrm{\mid }\le \eta \mathrm{\mid }x\mathrm{\mid }$ for a fixed $0<\eta <1\mathrm{/}2$.

The parameter pieces outside the local arc form a compact set disjoint from $0$; for each parameter $t$ in it continuity of the contour gives a relative interval on which $\mathrm{\mid }z\mathrm{\mid }$ exceeds half its positive value at $t$, the family of all these intervals covers the compact parameter set, so finitely many cover it, and the minimum of the finitely many positive half-values is a number $d>0$ with $\mathrm{\mid }z\mathrm{\mid }\ge d$ on the remainder.

If $\epsilon <d\mathrm{/}2$ then $\mathrm{\mid }z\mathrm{\mid }\ge d>\epsilon$ on the remainder, so $i\epsilon$ and $-i\epsilon$ avoid the remainder, and they avoid the arc because its real coordinate vanishes only at $x=0$, where $z(0)=0$; hence both points lie off the trace of $\mathrm{\Gamma }$.

By the winding definition, linearity and orientation-preserving reparametrization, $n(\mathrm{\Gamma },i\epsilon )-n(\mathrm{\Gamma },-i\epsilon )=\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}\frac{2i\epsilon }{z^2+{\epsilon }^2}dz$, the integrand being continuous on the trace of $\mathrm{\Gamma }$ for these values of $\epsilon$.

On the remainder $\mathrm{\mid }{z}^2+{\epsilon }^2\mathrm{\mid }=\mathrm{\mid }z-i\epsilon \mathrm{\mid }\mathrm{\mid }z+i\epsilon \mathrm{\mid }\ge d^2\mathrm{/}4$, so the ML estimate bounds the contribution of the remainder to the integral of step 4.1 by a constant times $\epsilon$.

On the local graph one has $\mathrm{\mid }z(x)-i\epsilon \mathrm{\mid }\mathrm{\mid }z(x)+i\epsilon \mathrm{\mid }\ge c(x^2+{\epsilon }^2)$ for a constant $c>0$: for $\mathrm{\mid }x\mathrm{\mid }\ge \epsilon$ each factor is at least $\mathrm{\mid }x\mathrm{\mid }$, while for $\mathrm{\mid }x\mathrm{\mid }<\epsilon$ the bound $\mathrm{\mid }f(x)\mathrm{\mid }\le \eta \mathrm{\mid }x\mathrm{\mid }<\epsilon \mathrm{/}2$ gives $\mathrm{\mid }z(x)\pm i\epsilon \mathrm{\mid }\ge \epsilon \mathrm{/}2$; substituting $x=\epsilon u$ turns the local contribution into ${\int }_{-a\mathrm{/}\epsilon }^{a\mathrm{/}\epsilon }\frac{2i{z}^{\mathrm{\prime }}(\epsilon u)}{(z(\epsilon u)\mathrm{/}\epsilon {)}^2+1}du$, whose integrand converges uniformly on bounded $u$-intervals to $\frac{2i}{1+u^2}$ and is dominated by $C\mathrm{/}(1+u^2)$, so the tails are uniformly of order $1\mathrm{/}R$ outside $[-R,R]$ and the integral tends to ${\int }_{\mathbb{R}}\frac{2idu}{1+u^2}=4i{\lim }_{R\to \mathrm{\infty }}\arctan R=2\pi i$ by [F8]; hence the index difference tends to $1$.

Each winding number is an integer by [F5], so the difference $n(\mathrm{\Gamma },i\epsilon )-n(\mathrm{\Gamma },-i\epsilon )$ is an integer for every sufficiently small $\epsilon >0$; since it tends to $1$ by steps 5.1 and 5.2, it equals $1$ for all sufficiently small $\epsilon$.