Determine the winding number of the map \[ f:S^1\to\CC\setminus\{0\}, \qquad f(z)=8z^4+4z^3+2z^2+z^{-1}. \]
Solution
1. Set\[ g(z)=zf(z)=8z^5+4z^4+2z^3+1. \]On \(|z|=1\),\[ |4z^4+2z^3+1|\le7<8=|8z^5|. \]
This is the triangle inequality on the unit circle.
2. The polynomial \(g\) has exactly five zeros in the open unit disk, counted with multiplicity, and no zeros on \(S^1\).
By 1, Rouché’s theorem applies to \[ g(z)=8z^5+(4z^4+2z^3+1) \] on the unit circle. Hence \(g\) and \(8z^5\) have the same number of zeros in the disk, namely five. The strict boundary inequality also implies \(g(z)\ne0\) for \(|z|=1\).
3. As a meromorphic function on the unit disk, \(f=g/z\) has five zeros and one pole, counted with multiplicity.
By 2, \(g\) has five zeros in the disk. Since \[ g(0)=1, \] none of them is at \(0\). Thus dividing by \(z\) introduces one simple pole at \(0\) and removes no zeros.
4. The winding number of the loop \(f|_{S^1}\) about \(0\) is\[ \boxed{4}. \]
By the argument principle, the winding number of \(f(S^1)\) about \(0\) equals the number of zeros minus the number of poles of \(f\) inside the unit circle, counted with multiplicity. By 3 this is \[ 5-1=4. \]