Proving a polynomial in the complex plane has no zeros in a certain radius

Question

The question is as follows:

For $0<r<1$, prove for sufficiently large n, the polynomial $Q_n(z)=\sum_{k=1}^n kz^{k-1}$ has no zeros for $|z|<r$

I tried using Rouche's theorem multiple times on different terms of the sum and this didn't really lead anywhere.

I also tried to use the argument principle but again, this didn't really lead anywhere. I think the maximum principle might be involved somewhere, but again, I'm not gaining much progress trying to use this either.

I noticed this sum tends to $1/(1-z)^2$ for large $n$, so intuitively the polynomial having no zeros makes sense, however I can't seem to put pen to paper following this logic.

Any tips? Thanks :)

Answer 1

One possible approach is to observe that $Q_n(z)$ has zeros in the open ball $B_r(0)$ if and only if $(1-z)^2 Q_n(z)$ does. On the other hand, we have $$(1-z)^2 Q_n(z) = 1 - (n+1) z^n + n z^{n+1}.$$ From here, you should be able to find a proof using Rouche's theorem.

Answer 2

$Q_n(z)$ converges uniformly to $f(z)=1/(1-z)^2$ on $|z| \le r$, so that $$ |Q_n(z)| \ge |f(z)| - |f(z) - Q_n(z)| \\ \ge \min_{|z|\le r }|f(z)| - \max_{|z|\le r }|f(z) - Q_n(z)| > 0 $$ for sufficiently large $n$.

Concretely: For $|z| \le r < 1$ is $$ \begin{align} |Q_n(z)| &= \left| \frac{1}{(1-z)^2} - \sum_{k={n+1}}^\infty k z^{k-1}\right| \\ &\ge \frac{1}{(1+r)^2} - \sum_{k={n+1}}^\infty k r^{k-1} \\ &= \frac{1}{(1+r)^2} - r^n\sum_{k=0}^\infty (k+n+1) r^{k} \\ &\ge \frac{1}{(1+r)^2} - (n+1)r^n\sum_{k=0}^\infty (k+1) r^{k} \\ &= \frac{1}{(1+r)^2} - \frac{(n+1)r^n}{(1-r)^2} \end{align} $$ and that is strictly positive for sufficiently large $n$, since $$ \lim_{n \to \infty} (n+1)r^n = 0 \, . $$

Answer 3

Let $Q(z) = {1 \over (1-z)^2}$, then $Q_n \to Q$ uniformly on $\overline{B}(0,r)$. Note that for $|z|=r$ we have $|Q(z)| = {1 \over |1-z|^2} \ge {1\over (1+r)^2}$. Choose $n$ such that $|Q_n(z)-Q(z)| < {1\over (1+r)^2}$, and then $|Q_n(z)-Q(z)| < |Q(z)|$ for $|z|=r$ and so Rouché lets us conclude that $Q_n, Q$ have the same number of zeros in $B(0,r)$.