Show that for every polynomial \(p(z)\) there exists \(z\) with \(|z|=1\) such that \[ \left|p(z)-\frac1z\right|\ge 1. \]
Solution
1. Suppose, for contradiction, that\[ \left|p(z)-\frac1z\right|<1 \]for every \(|z|=1\). Then\[ |zp(z)-1|<1 \]on the unit circle.
For \(|z|=1\), \[ |zp(z)-1| =|z|\left|p(z)-\frac1z\right| =\left|p(z)-\frac1z\right|. \] Thus the assumed strict inequality gives the displayed estimate.
2. Rouché’s theorem applied on \(|z|=1\) gives a contradiction.
Set \[ q(z)=zp(z). \] On \(|z|=1\), 1 says \[ |q(z)-1|<1=|1|. \] Hence Rouché’s theorem implies that \[ q(z)=1+(q(z)-1) \] and the constant function \(1\) have the same number of zeros in the unit disk, counted with multiplicity. The constant function has no zeros, so \(q\) should have none.
But \[ q(0)=0\cdot p(0)=0, \] so \(q\) has a zero at the origin. This contradiction proves that the strict inequality cannot hold at every point of the unit circle. Therefore some \(|z|=1\) satisfies \[ \left|p(z)-\frac1z\right|\ge1. \]