Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 4 / 14E / i / Solution

The form of Rouché's theorem used here is as follows. Suppose $f$ and $g$ are holomorphic on a neighborhood of the closure of a bounded Jordan domain, and $|g-f|<|f|$ on its positively oriented boundary. Then $f$ and $g$ have the same number of zeros inside, counted with multiplicity.

To prove this form, put $f_t=f+t(g-f)$ for $0\le t\le 1$. The strict inequality prevents $f_t$ from vanishing on the boundary. The argument principle counts its interior zeros by $(2\pi i{)}^{-1}\oint f_t^{\prime }/f_t$. This integral varies continuously with $t$: the integrand is continuous and its denominator stays uniformly away from zero on the compact boundary times $[0,1]$. Its values are integers, so the integral is constant. Its values at $t=0,1$ are the respective zero counts. This proves Rouché's theorem with the multiplicities and boundary hypotheses required below.