Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 3 / 13G (Further Analysis)

For $0\le t\le 1$, put $F_t=f+tg$. On the curve, $|F_t|\ge |f|-t|g|>0$, so no member of this homotopy has a boundary zero. The argument principle counts the interior zeros, with multiplicity, by its winding number $N(t)=\mathrm{wind}(F_t\circ \gamma ,0)$. For a piecewise smooth curve this isThere are no poles, and the null-homotopy of the Jordan curve in the domain permits this interior count. The loops $F_t\circ \gamma$ form a continuous homotopy avoiding zero, so their integer winding number stays constant. Equivalently, for a piecewise smooth curve the integral varies continuously with $t$ and is integer-valued. In particularThis proves Rouché's theorem. The counting is with multiplicity and assumes the usual positive orientation; reversing orientation reverses both winding counts without changing the conclusion.

The function is holomorphic on a neighborhood of the closed unit disk. On $|z|=1$, the mapping assumption gives $|g(z)|<1=|z|$. Apply Rouché's theorem to $z$ and $-g(z)$. Their sum $z-g(z)$ has the same number of interior zeros as $z$, namely one counting multiplicity. ThusThe zero is simple since its total multiplicity is one. No boundary fixed point is possible because $g$ maps the boundary strictly inside.

For nonnegative integers $m,n$, put $B(z)=z^m[(z-a)/(1-\bar{a}z){]}^n$. Since $|a|<1$, the denominator has no zero on or inside the unit circle. On $|z|=1$,so $|B(z)|=1$. Consequently $|-a|<|B(z)|$, and Rouché's theorem gives the same interior zero count for $B-a$ as for $B$. The latter has $m$ zeros at zero and $n$ at $a$, with the multiplicities combining when $a=0$. ThereforeThis is the interior zero count for a finite Blaschke product minus a constant. If $m+n=0$, the nonzero constant $1-a$ has zero zeros, so the same conclusion includes that endpoint case.