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I'm having a bit of trouble with this question, I don't have much of an idea where to start and any help would be appreciated!

"Suppose that $P(z) = a_0 + a_1z + a_2z^2 + ... + a_nz^n$ is a polynomial of degree n ($a_n \neq 0$). Show that for every $\epsilon \in \mathbb{R}$, with $0 < \epsilon < 1$, there exists $R > 0$ such that

$$(1-\epsilon)|a_n||z^n| \leq |P(z)| \leq (1+\epsilon)|a_n||z^n|$$

for all $z \in \mathbb{C}$ satisfying $|z| > R$."

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1 Answer 1

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Hint

$$\frac{P(z)}{z^n}=a_n+\left(\frac{a_{n-1}}{z}+...+\frac{a_0}{z^n}\right).$$

Since $\frac{a_{n-1}}{z},...,\frac{a_0}{z^n}\underset{z\to \infty }{\longrightarrow }0$, if $\varepsilon>0$, there is an $R$ s.t. $$\left|\frac{a_{n-1}}{z}+...+\frac{a_0}{z^n}\right|\leq |a_n|\varepsilon,$$ for all $|z|>R$.

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  • $\begingroup$ So we get $P(z)/z^n \leq a_n + |a_n|/2$ If we multiply by $z^n$ we almost get the right hand side part, but I still am unsure of where to go after that. We can't bound the 3/2 by $1+\epsilon$ because $\epsilon$ can be less than 1/2. Any more hints possible? Thanks! $\endgroup$
    – Sorey
    Commented Nov 8, 2015 at 15:05
  • $\begingroup$ I corrected my answer. I didn't see that you wanted a solution with $\varepsilon$. (I guess you want to prove that $p(z)=0$ has at least one solution). $\endgroup$
    – Surb
    Commented Nov 8, 2015 at 15:23
  • $\begingroup$ Ok one last question if you don't mind! That will work for the upper bound, however, $1-\epsilon$ isn't always less than $\epsilon$. So how do you get the lower bound? $\endgroup$
    – Sorey
    Commented Nov 8, 2015 at 15:34
  • $\begingroup$ No problem :-) $$\left|\frac{P(z)}{z^n}-a_n\right|<|a_n|\varepsilon\implies \left|\left|\frac{P(z)}{z^n}\right|-|a_n|\right|<|a_n|\varepsilon\implies -\varepsilon|a_n|<\left|\frac{P(z)}{z^n}\right|-|a_n|<|a_n|\varepsilon.$$ Got it ? $\endgroup$
    – Surb
    Commented Nov 8, 2015 at 15:47
  • $\begingroup$ Yes, thank you so much!! Great help $\endgroup$
    – Sorey
    Commented Nov 8, 2015 at 15:58

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