Prove that for every \(n\in \ZZ^{\geq 0}\) the following polynomial has no roots in the open unit disc: \[ f_n(z) \definedas \sum_{k=0}^n {z^k \over k!} .\]
Hint: check \(n=1,2\) directly.
Solution
For \(n=0\), \(f_0\equiv1\). For \(n=1\), \(f_1(z)=1+z\) has its only zero at \(-1\), on the boundary of the unit disk.
Now suppose \(n\ge2\). On \(|z|=1\), \[ |e^z-f_n(z)| \le \sum_{k=n+1}^\infty {1\over k!} \le \sum_{k=3}^\infty {1\over k!}. \] For \(k\ge3\), \(k!\ge 2\cdot3^{k-2}\), so \[ \sum_{k=3}^\infty {1\over k!} \le {1\over2}\sum_{j=1}^\infty {1\over3^j} ={1\over4}. \] Also \[ |e^z|=e^{\Re z}\ge e^{-1}>{1\over3}>{1\over4} \] on \(|z|=1\). Hence \[ |e^z-f_n(z)|<|e^z| \] there. By Rouché’s theorem, \(f_n\) and \(e^z\) have the same number of zeros in \(\DD\). Since \(e^z\) has none, neither does \(f_n\).