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 $z\in {\gamma }^{\ast }$, $\mathrm{\mid }g(z)+t(f(z)-g(z))-g(z)\mathrm{\mid }\le \mathrm{\mid }t\mathrm{\mid }\mathrm{\mid }f(z)-g(z)\mathrm{\mid }<(1+\epsilon )q\mathrm{\mid }g(z)\mathrm{\mid }<\mathrm{\mid }g(z)\mathrm{\mid },$ 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. 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 real $t\in [0,1]$, step 1.1 and [L1] give $J(t)=n(h_t\circ \gamma ,0)$, and [L4] makes that an integer. Hence $J$ is an integer-valued holomorphic function on a connected open disc, so it is constant.

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 $\gamma$.