Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 4 / 2F / b

Put $z(t)=e^{2\pi it}$ and use the straight homotopy $H(s,t)=z(t)+s(\gamma (t)-z(t))$. For $0\le s<1$, the triangle inequality givesAt $s=1$ it is nonzero by the hypothesis. Thus this is a homotopy of closed loops avoiding the origin, even though the given distance bound is non-strict. The homotopy invariance of winding number gives

If the boundary loop meets zero, the assertion already holds. Otherwise suppose, for contradiction, that $f$ has no zero anywhere in the disc. Then $K(s,t)=f((1-s)e^{2\pi it})$ is a homotopy through nonvanishing closed loops from the boundary loop to the constant loop $f(0)$. Constant loops have winding number zero, whereas part (i) gives winding number one. This contradicts homotopy invariance of winding number. Hence $\boxed{f(z)=0\text{ for some }z\in B}$.