Rouche's theorem in the classical strict-inequality form

Because ${\gamma }^{\ast }$ is compact and $\mathrm{\mid }f-g\mathrm{\mid }<\mathrm{\mid }g\mathrm{\mid }$ there, the ratio $\mathrm{\mid }f-g\mathrm{\mid }\mathrm{/}\mathrm{\mid }g\mathrm{\mid }$ has a maximum $q<1$ on ${\gamma }^{\ast }$. Choose $\epsilon >0$ with $(1+\epsilon )q<1$. Then for every complex $t$ with $\mathrm{\mid }t\mathrm{\mid }<1+\epsilon$ and every $z\in {\gamma }^{\ast }$, $\mathrm{\mid }g(z)+t(f(z)-g(z))-g(z)\mathrm{\mid }=\mathrm{\mid }t\mathrm{\mid }\mathrm{\mid }f(z)-g(z)\mathrm{\mid }\le (1+\epsilon )q\mathrm{\mid }g(z)\mathrm{\mid }<\mathrm{\mid }g(z)\mathrm{\mid },$ so $h_t(z):=g(z)+t(f(z)-g(z))$ never vanishes on ${\gamma }^{\ast }$.

For fixed $z$, the function ${\phi }_z(t):=\frac{h_t^{\mathrm{\prime }}(z)}{h_t(z)}=\frac{g^{\mathrm{\prime }}(z)+t(f^{\mathrm{\prime }}(z)-g^{\mathrm{\prime }}(z))}{g(z)+t(f(z)-g(z))}$ is holomorphic on the disc $\mathrm{\mid }t\mathrm{\mid }<1+\epsilon$ by step 1.1. The numerator and denominator are jointly continuous in $(z,t)$ on the contour trace times that disc: [L5] supplies continuity of $f,g,f^{\mathrm{\prime }},g^{\mathrm{\prime }}$, and step 1.1 makes the denominator nonzero. Therefore $J(t):=\frac{1}{2\pi i}{\int }_{\gamma }\frac{h_t^{\mathrm{\prime }}(z)}{h_t(z)}dz$ is holomorphic there by [L3]. For every complex $t$ in this disc, $h_t$ is holomorphic in $z$ and nonzero on the contour by step 1.1, so [L1] gives $J(t)=n(h_t\circ \gamma ,0)$, and [L4] makes that an integer. By [L5], $J$ is continuous. At any $t_0$ in the disc, continuity supplies a neighbourhood where $\mathrm{\mid }J(t)-J(t_0)\mathrm{\mid }<1$; two distinct integers differ in modulus by at least $1$, so $J(t)=J(t_0)$ there. Thus $J$ is locally constant and $J\mathrm{\prime }(t_0)=0$. The disc is a nonempty connected open set (indeed convex), so [L6] makes $J$ constant on the entire disc.

Since $h_0=g$ and $h_1=f$, step 2.1 gives $\frac{1}{2\pi i}{\int }_{\gamma }\frac{g^{\mathrm{\prime }}(z)}{g(z)}dz=\frac{1}{2\pi i}{\int }_{\gamma }\frac{f^{\mathrm{\prime }}(z)}{f(z)}dz\mathrm{.}$ Applying [L2] to both sides shows that $f$ and $g$ have the same weighted zero count with respect to $\gamma$.