Find the number of zeros in \(\abs{z} < R\) of \[ p(z) \da z^d + a_1z^{d-1} + \cdots + a_d ,\] supposing that \(\abs{a_k}< {R^k \over d}\) for every \(k\) (noting the strict inequality).
Solution
Strategy: bound the difference. Find the big and small term:
- Big: \(F(z) = z^d\), so \(\abs{F(z)} = R^d\) on \(\abs{z} = R\).
- Small: \(g(z) = p(z) - F(z) = a_1 z^{d-1} + \cdots + a_d\), so \[ \abs{g(z)} &\leq \abs{a_1} R^{d-1} + \abs{a_2} R^{d-2} + \cdots + \abs{a_d} \\ &< {R\over d} \cdot R^{d-1} + {R^2 \over d} \cdot R^{d-2} + \cdots + {R^{d-1} \over d} \cdot R + {R^{d} \over d} \\ &= d {R^d\over d} = R^d ,\] so \(\abs{g} < R^d = \abs{F}\), meaning \(Z_{p-F} = Z_F = d\) in \(R\DD\).