Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 86 / 1 / Solution

Let $f$ be holomorphic on a neighborhood of the closure of a bounded region with piecewise smooth positively oriented boundary, with no boundary zeros and with $f$ not identically zero. The argument principle stateswhere $m_a$ is the zero multiplicity. More generally a closed contour avoiding the zeros gives ${\sum }_a{m}_a\mathrm{Ind}(\Gamma ,a)$, and the integral is the winding number of $f\circ \Gamma$ about zero.

To prove it, factor near a zero $a$ as $f(z)=(z-a{)}^{m_a}h(z)$ with $h(a)\ne 0$. Then $f^{\prime }/f=m_a/(z-a)+h^{\prime }/h$. Remove small disjoint circles about the finitely many zeros in the region. On the remainder $f^{\prime }/f$ is holomorphic, so the Cauchy integral theorem reduces the boundary integral to the sum of the small-circle integrals. Each is $2\pi i{m}_a$ because $h^{\prime }/h$ is holomorphic on its small disc. This proves the formula. Along a contour, the imaginary part of $d\log f$ is the change in a continuous argument, giving the winding interpretation and the indexed version.

Now choose a closed disc $\overline{D(z_0,r)}\subset \Omega$ on which $z_0$ is the only zero of the nonconstant limit $f$. Its boundary has $\min |f|>0$. Local uniform convergence gives $\max |f_n-f|<\min |f|$ there for all sufficiently large $n$. The contour functions $f+t(f_n-f)$, $0\le t\le 1$, have no boundary zero. Their winding numbers are constant under this homotopy, so the argument principle shows that $f_n$ has exactly the same positive number $m$ of zeros, counted with multiplicity, inside the disc. This is the Rouche theorem argument, here derived from the contour count.

Choose a zero $z_n$ in that disc. For every $0<ϵ<r$, the compact annulus $ϵ\le |z-z_0|\le r$ contains no zero of $f$, so uniform convergence makes $f_n$ nonzero on it eventually. Therefore all of the chosen zeros eventually lie within $ϵ$ of $z_0$. ConsequentlyThe maps $f_n$ are eventually not identically zero, since they converge at any fixed point where $f$ is nonzero; thus their zero counts above are legitimate.

A double zero need not persist as a double zero. On the unit disc take $f(z)=z^2$, $f_n(z)=z^2+1/n$. Its zeros $\pm i/\sqrt{n}$ converge to zero but are simple, while $f_n^{\prime }$ vanishes only at zero and $f_n(0)=1/n\ne 0$. Hence the additional simultaneous derivative condition cannot always be imposed. Local uniform convergence preserves total nearby multiplicity, not its concentration at a single point.