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Let $0 < r < 1$. I need to prove that $p(z) = 1 + 2z + 3z^2 + \dots +(n+1)z^n$ has no zeroes in the disk $|z| < r$ if $n$ is sufficiently large. I'm thinking Rouche's theorem might be helpful, but I'm not exactly sure how to apply it. Help would be appreciated.

user66825
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2 Answers2

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Let $p_n(z) = \sum_{k=0}^n (k+1)z^k$, and $p(z) = \frac{1}{(1-z)^2}$. Note that $p$ is analytic on $B(0,1)$ and has no zeros or poles in $B(0,1)$.

Find a lower bound for $|p(z)|$ on $B(0,1)$.

Hint:

If $|z|<1$, we have $|1-z| < 2$, hence $|p(z)|> \frac{1}{4}$.

Choose $r \in (0,1)$, and $r < s <1$. Then show that $p_n \to p$ uniformly on $B(0,s)$. Find a suitable upper bound for $|p_n(z) -p(z)|$ in $B(0,s)$.

Hint:

Now choose $N$ such that if $n \ge N$, we have $|p_n(z) -p(z)| \le \frac{1}{4}$ for all $z \in B(0,s)$.

Finally, apply Rouché's theorem on $|z|=r$.

Hence if $|z|=r$, we have $|p_n(z) -p(z)| \le \frac{1}{4} < |p(z)|$. We conclude from Rouché's theorem that $p_n$ the same number of zeros as $p$ in $B(0,r)$ (that is, zero zeros).

copper.hat
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  • Can you explain why $p_n \to p$ uniformly? Is it because $p_n \to p$ uniformly on the compact disk $|z| \le s$? – user66825 Mar 15 '13 at 05:13
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    Rouché is a bit overkill here: if $p_n(z) = 0$ we immediately get the contradiction $|p(z)| < |p(z)|$. – Erick Wong Mar 15 '13 at 05:14
  • @user66825: The power series of $p$ converges inside the radius of convergence ($R=1$), and uniformly on compact subsets of $B(0,R)$. And as you noted, $\overline{B}(0,s)$ is compact. – copper.hat Mar 15 '13 at 05:19
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    @ErickWong: This is true, but the OP mentioned Rouché. Also, to some extent, this is the nature of Rouché's theorem... – copper.hat Mar 15 '13 at 05:23
  • @copper.hat You're right, I hadn't noticed the mention! – Erick Wong Mar 15 '13 at 05:24
  • @ErickWong: For some reason, I miss fairly obvious applications of Rouché from time to time, so am sympathetic with requests involving its use :-). – copper.hat Mar 15 '13 at 05:27
  • I mistakenly thought that pointwise convergence on a compact set $\implies$ uniform convergence on that set, so I'm not completely sure now why $p_n \to p$ uniformly. – user66825 Mar 15 '13 at 05:54
  • @user66825: A power series has the property of converging uniformly on compact subsets contained inside the radius of convergence (excuse my loose wording). – copper.hat Mar 15 '13 at 05:56
  • Does this argument for uniform convergence on $B(0, s)$ look right? On $B(0, s)$ we have $|(k+1)z^k| \le (k+1)s^k$, and $\sum (k+1)s^k$ converges since (applying the ratio test), $\lim \dfrac{(k+2)s^{k+1}}{(k+1)s^k} = s < 1$. Therefore, the $\sum (k+1)z^k$ converges uniformly by the Weierstrass $M$ test. – user66825 Mar 15 '13 at 06:06
  • @user66825: Yes, that works. (However, the uniform convergence on compact subsets works for any power series inside its radius of convergence.) – copper.hat Mar 15 '13 at 06:17
  • In the last sentence, is something else being used besides the fact that $p$ is analytic in $B(0,1)$ and has no zeros? – Tiberio Jan 10 '20 at 20:31
  • @Tiberio: I'm not sure why I mentioned poles, the only relevant fact is that $p$ has no zeros on or in $|z|=r$ from which we get that $p_n$ has no zeros on or in $|z|=r$. – copper.hat Jan 10 '20 at 21:42
  • Perfect, thank you for the quick response! – Tiberio Jan 10 '20 at 23:13
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Hint:

When $n\to \infty$, note that $p(z)=\displaystyle(\sum_{i=0}^{n+1}z^i)'\to(\cfrac{1}{1-z})'=\cfrac{1}{(1-z)^2}$ for $|z|<1$, which cannot be $0$.

NECing
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