If $\abs{f-g}<\abs{f}+\abs{g}$ on the circle $\abs{z-z_0}=R$, then $f$ and $g$ have the same number of zeros in $\abs{z-z_0}<R$.Suggest a slogan
Let \(z_0\in\CC\) and \(R>0\), and let \(f\) and \(g\) be holomorphic on an open set containing the closed disc \(\abs{z-z_0} \leq R\). Suppose \(f\) and \(g\) have no zeros on the circle \(\abs{z-z_0} = R\) and \[ \abs{f(z)-g(z)} < \abs{f(z)} + \abs{g(z)} \quad\text{for all } z \text{ with } \abs{z-z_0}=R. \] Then \(f\) and \(g\) have the same number of zeros in the open disc \(\abs{z-z_0}<R\), counted with multiplicity.
State Rouche's theorem on a closed disc $\abs{z - z_0} \leq R$.