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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.2012-01-11
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    @Didier: The OP asked another one: http://math.stackexchange.com/questions/98138/the-graph-of-cot-is-the-image-of-the-graph-of-tan-by-a-simple-transformati2012-01-11
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    and one more: http://math.stackexchange.com/questions/98139/inequality-for-with-cot2012-01-11
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    @Paul: Indeed. *Sigh*.2012-01-11
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    I understand your amazement for so many questions but I'm doing this because I found some interesting problems about complex numbers on the internet (more precisely here : http://www.eleves.ens.fr/home/kortchem/cg/complexes_expbs.pdf). Maybe this is stupid, but I wanted to post the questions before trying to solve them so that I would have the solutions ready when I would be stuck. I hope this does not disturb much. Or does it ?2012-01-11
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    And for this problem, the case $\left | z \right |\leq1$ is trivial. Thus we can suppose $\left | z \right |>1$.When $z$ is big, thepreponderant term of $P(z)$ is $a_{n}z^n$ and this can help us get a majorization. For b) I guess we should use geometric series (with ratio $\left | z \right |$ or $\left | \frac{1}{z} \right |$?2012-01-11
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    DidierPiau : "By the way, mentioning where you are stuck, what you tried and where you failed is well considered on this site." I'll try to act accordingly in the future.2012-01-11
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    *I wanted to post the questions before trying to solve them*... This is outrageous, if you ask me.2012-01-11
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    @Didier: the beauty of it is that he's trying to sell us [the hints](http://i.stack.imgur.com/tZkrs.png) in [his exercise sheet](http://www.eleves.ens.fr/home/kortchem/cg/complexes_expbs.pdf) as his guesses.2012-01-11
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    And is there any issue with that ? Maybe this could have helped someone find the answer. Anyway, next time I'll try to think some time before asking questions. Sorry for that and thank you for the advice.2012-01-11
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    And I did not sell them as guesses. Just mentioned them here for you to know.2012-01-11
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    *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!2012-01-11
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    @t.b. Yes. And now, a propitious change of username... Maybe momo1729 is supposed to cover the tracks of IsmailLemhadri.2012-01-11
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    Just pardon me to say that I did mention the source in a previous comment (http://www.eleves.ens.fr/home/kortchem/cg/complexes_expbs.pdf). What I want to insist on is that you should not view my comment as an attempt to show you that I have made guesses about the problem ; this would simply be useless because I openly mentioned that I had not tackled it yet.2012-01-11
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    And by the way, the username change certainly does not come with the intention of hiding anything or whatsoever. I guess I should just not publish my real name on the Internet -that's something I should have done from the day I registered, even though M.SE encourages users to do so.2012-01-11

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