Bound the complex roots of a polynomial above

Question

We consider $P(z)=a_{0}+a_{1}z+\cdot+a_{n-1}z^{n-1}+a_{n}z^n$, with $a_{0},\ldots,a_{n-1},a_{n} \in \mathbb{C}$ and $a_{n}\neq0$. Let $R=\max_{0\leq k\leq n-1}\left | \frac{a_k}{a_n} \right |$ and $S=\sum_{k=0}^{n-1}\left | \frac{a_k}{a_n} \right |$.

Can you help me establish the two following ?

a) Any complex root of $P$ has modulus less than or equal to $\max(1,S)$.

b) Any complex root of $P$ has modulus less than or equal to $1+R$.

It is worth noting that the approximation in b) is often better than that in a). Thank you for any hint or answer.

Answer 1

Fix $\delta>0$. To solve the first question, we apply Rouché's theorem to the circle of center $0$ and radius $\max\{1,S\}+\delta$. Indeed, if $|z|=\max\{1,S\}+\delta$, $$|P(z)-a_nz^n|=\left|\sum_{j=0}^{n-1}a_jz^j\right|\leqslant \sum_{j=0}^{n-1}|a_j||z|^j<\sum_{j=0}^{n-1}|a_j|(\max\{1,S\}+\delta)^n\leqslant |a_nz^n|.$$