# Bound the complex roots of a polynomial above

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.

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Did you ask 7 questions in the last 12 hours, including 5 in the last 2 hours? You might wish to slowdown a little... By the way, mentioning where you are stuck, what you tried and where you failed is well considered on this site. – Did Jan 11 '12 at 10:54
@Didier: The OP asked another one: math.stackexchange.com/questions/98138/… – Paul Jan 11 '12 at 10:58
I wanted to post the questions before trying to solve them... This is outrageous, if you ask me. – Did Jan 11 '12 at 11:16
@Didier: the beauty of it is that he's trying to sell us the hints in his exercise sheet as his guesses. – t.b. Jan 11 '12 at 11:31
And I did not sell them as guesses... Obviously, you have a lot to learn about the generally accepted practices of attribution in mathematics. You copy verbatim several sentences from a document without mentioning the source, as if these were your own, and you do not even see the problem?! Since you ask, probably rhetorically: Is there an issue with that? let me answer: yes there is! – Did Jan 11 '12 at 12:12

## 1 Answer

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 $|z|=\max\{1,S\}+\delta$, $$|P(z)-a_nz^n|=\left|\sum_{j=0}^{n-1}a_jz^j\right|\leqslant \sum_{j=0}^{n-1}|a_j||z|^j<\sum_{j=0}^{n-1}|a_j|(\max\{1,S\}+\delta)^n\leqslant |a_nz^n|.$$

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