Let \(p(z)\) be a polynomial of degree \(d\ge2\) with distinct roots \(r_1,\ldots,r_d\). Show that \[ \sum_{i=1}^d\frac{1}{p'(r_i)}=0. \]
Solution
1. For each root \(r_i\), the meromorphic function \(1/p(z)\) has a simplepole at \(r_i\) with residue\[ \operatorname{Res}_{z=r_i}\frac1{p(z)}=\frac1{p'(r_i)}. \]
Since the roots are distinct, \(p'(r_i)\neq0\). Thus \[ p(z)=(z-r_i)q_i(z) \] with \(q_i(r_i)=p'(r_i)\). Therefore \[ \operatorname{Res}_{z=r_i}\frac1{p(z)} =\lim_{z\to r_i}\frac{z-r_i}{p(z)} =\frac1{q_i(r_i)} =\frac1{p'(r_i)}. \]
2. If \(R\) is large enough that all roots of \(p\) lie inside\(|z|<R\), then\[ \int_{|z|=R}\frac{dz}{p(z)}\longrightarrow0 \qquad(R\to\infty). \]
Write \[ p(z)=a_dz^d+O(z^{d-1}), \qquad a_d\neq0. \] Hence for all sufficiently large \(R\) there is a constant \(C>0\) such that \[ \left|\frac1{p(z)}\right|\le \frac{C}{R^d} \qquad(|z|=R). \] The ML estimate then gives \[ \left|\int_{|z|=R}\frac{dz}{p(z)}\right| \le 2\pi R\frac{C}{R^d} =2\pi C R^{1-d}\longrightarrow0, \] because \(d\ge2\).
3. Therefore\[ \sum_{i=1}^d\frac1{p'(r_i)}=0. \]
For every sufficiently large \(R\), the residue theorem and 1 give \[ \int_{|z|=R}\frac{dz}{p(z)} =2\pi i\sum_{i=1}^d\frac1{p'(r_i)}. \] The right-hand side is independent of \(R\), while by 2 the left-hand side tends to zero as \(R\to\infty\). Hence the sum must vanish.