A positively oriented convex polygon has winding index one inside and zero outside

Fix exterior points $q_1,q_2\mathrm{\notin }P$ and choose $R>\max \{{\max }_{u\in P}\mathrm{\parallel }u{\mathrm{\parallel }}_2,\mathrm{\parallel }{q}_1{\mathrm{\parallel }}_2,\mathrm{\parallel }{q}_2{\mathrm{\parallel }}_2\}$, which exists because $P$ is compact. For each $j\in \{1,2\}$, compactness gives a nearest point $p_j\in P$. For every $u\in P$, convexity puts $p_j+t(u-p_j)$ in $P$ for $0\le t\le 1$; minimality of $p_j$ gives $\mathrm{\parallel }{q}_j-p_j{\mathrm{\parallel }}_2^2\le \mathrm{\parallel }{q}_j-p_j-t(u-p_j){\mathrm{\parallel }}_2^2$. Expanding, dividing by $t>0$, and letting $t\downarrow 0$ yields $⟨q_j-p_j,u-p_j⟩\le 0$. Hence the ray $q_j+s(q_j-p_j)$, $s\ge 0$, avoids $P$ and, being unbounded, meets the circle $C_R=\{w:\mathrm{\mid }w\mathrm{\mid }=R\}$. The circle $C_R$ is disjoint from $P$ and connected, so these two rays and an arc of $C_R$ join $q_1$ to $q_2$ in $\mathbb{C}\setminus P$. Thus $\mathbb{C}\setminus P$ is path-connected and unbounded.

By [F1], $n(\mathrm{\Gamma },q)$ is the winding index of the contour. Since $\mathrm{\Gamma }\subset P$, the exterior $\mathbb{C}\setminus P$ lies in the unbounded component of $\mathbb{C}\setminus \mathrm{\Gamma }$; [F3] gives $n(\mathrm{\Gamma },q)=0$ for every $q\mathrm{\notin }P$.

Fix $z\in \mathrm{int}P$. Choose a ray from $z$ whose direction is not aimed at a vertex. Its intersection with the convex polygon is a bounded interval starting at $z$, so the ray exits at the interior $p$ of a single edge with directed tangent $v\mathrm{\ne }0$. The polygon interior lies locally to the left of this positively oriented edge. Apply the orientation-preserving similarity $\rho (w)=\bar{v}(w-p)\mathrm{/}\mathrm{\mid }v\mathrm{\mid }$, which sends $p$ to $0$ and the edge tangent to the positive real direction. For every $a\mathrm{\notin }\mathrm{\Gamma }$, put $c=\bar{v}\mathrm{/}\mathrm{\mid }v\mathrm{\mid }$. Parametrizing the transformed contour and using [F1] gives ${\int }_{\rho \circ \mathrm{\Gamma }}\frac{d\zeta }{\zeta -\rho (a)}={\int }_{\mathrm{\Gamma }}\frac{cdw}{c(w-a)}={\int }_{\mathrm{\Gamma }}\frac{dw}{w-a}$, so $n(\rho \circ \mathrm{\Gamma },\rho (a))=n(\mathrm{\Gamma },a)$. For sufficiently small $\epsilon >0$, the points ${\rho }^{-1}(i\epsilon )$ and ${\rho }^{-1}(-i\epsilon )$ lie respectively on the interior and exterior local sides of the edge. Applying [F4] to $\rho \circ \mathrm{\Gamma }$ gives $n(\rho \circ \mathrm{\Gamma },i\epsilon )-n(\rho \circ \mathrm{\Gamma },-i\epsilon )=1$; the just-proved invariance rewrites this as $n(\mathrm{\Gamma },{\rho }^{-1}(i\epsilon ))-n(\mathrm{\Gamma },{\rho }^{-1}(-i\epsilon ))=1$. The exterior-side index is zero by step 2.1. Thus an interior point near that edge has winding number one.

The interior of a convex polygon is connected. Its points avoid $\mathrm{\Gamma }$, so [F2] makes the winding number constant throughout the interior. Step 3.1 identifies this constant as one. Together with step 2.1 this proves both cases.