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I have the polynomial $P_n(z)=1-\sum_{k=1}^{n}z^k$. We know that this polynomial has exactly $n$ roots in $\mathbb{C}$. Let $\rho$ be the number of roots of $P_n$, thence if $n\to\infty$ then $\rho$ must tend to $\infty$ too. Though, if we interpret the sum as a geometric series, we get that $$P_n(z)=1-\sum_{k=1}^{n}z^k=\frac{z^{n+1}-2z+1}{1-z}$$ And if we make $n\to\infty$ it only converges for $|z|<1$, becoming $P_\infty(z)=-\frac{2z-1}{1-z}$, that has only one real root for $z=\frac{1}{2}$. So, where did the other roots go? I plotted $P_n$ variating $n$ and noted that the roots tend to accumulate on the unit disk. (Interactive Mapping). So, can we say that all the roots will accumulate on the unit disk as $n\to\infty$ for $z\neq1$? How can we prove this? Thanks!

Here I have some screenshots of the mapping. Respectively, $n=5$, $n=10$, $n=100$. You can see that the roots tend to accumulate towards the unit circle.

| n=5 n=10 n=100 |

Thanks.

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    $\begingroup$ Thanks for drawing our attention to samuelj.li/complex-function-plotter $\endgroup$
    – lhf
    Commented Nov 26, 2021 at 14:05
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    $\begingroup$ The limit expression is only valid for $|z|<1$. Also note that in the complex numbers, there is no order, so "$-1<z<1$" makes only sense for real numbers. $\endgroup$
    – Peter
    Commented Nov 26, 2021 at 14:40
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    $\begingroup$ You can simplify this example by removing a few superfluous terms: $z^n-1$ converges to $-1$. Where do the roots go? $\endgroup$
    – Arthur
    Commented Nov 26, 2021 at 14:40
  • $\begingroup$ @Arthur it converges to -1 iff $|z|<1$. For $|z|=1$ we have $1^\infty$, but can we manage to solve the limit as $n\to\infty$ to get that the infinite roots lie on the unit disk? $\endgroup$ Commented Nov 26, 2021 at 14:47
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    $\begingroup$ @Peter you're right. Edited in $|z|<1$ $\endgroup$ Commented Nov 26, 2021 at 14:50

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This can be proved using Rouché's theorem.

Let $0 < \epsilon < 1/2$, and take $N$ large enough that $$ (1-\epsilon)^{N+1} < 1-2\epsilon \quad\text{and}\quad (1+\epsilon)^{N+1} > 3 + 2 \epsilon . $$ Then for any $n \ge N$ we have $$ (1-\epsilon)^{n+1} < 1-2\epsilon \quad\text{and}\quad (1+\epsilon)^{n+1} > 3 + 2 \epsilon $$ too, and I will show that this implies that the annulus $1-\epsilon \le |z| < 1+\epsilon$ contains all roots of $f(z) = 2z-1-z^{n+1}$ except $z=1/2$.

Let $$ g(z) = 2z-1 = 2(z-\tfrac12) ,\qquad h(z) = -z^{n+1} . $$ Then on the circle $|z|=1-\epsilon$, the point closest to $1/2$ is $z=1-\epsilon$, so we have $$ |g(z)| = 2 |z-\tfrac12| \ge 2 \bigl((1-\epsilon) - \tfrac12 \bigr) = 1 - 2 \epsilon , $$ and $|h(z)|=(1-\epsilon)^{n+1}$, so according to the inequalities above we have $|h(z)|<|g(z)|$ on the circle, which according to Rouché means that $f=g+h$ has equally many zeros inside the circle as $g$, namely one (the only zero of $g$ is at $1/2$).

On the other hand, on the circle $|z|=1+\epsilon$, the point furthest from $1/2$ is $z=-1-\epsilon$, so we have $$ |g(z)| = 2 |z-\tfrac12| \le 2 \bigl|(-1-\epsilon) - \tfrac12 \bigr| = 3 + 2 \epsilon , $$ and $|h(z)|=(1+\epsilon)^{n+1}$, so according to the inequalities above we have $|h(z)|>|g(z)|$ on the circle, which according to Rouché means that $f=g+h$ has equally many zeros inside the circle as $h$, namely $n+1$ (since $h$ has a zero of that multiplicity at the origin).

Thus, for any $n \ge N$, the function $f$ has $n$ of its $n+1$ zeros in the annulus between those two circles, of radius $1 \pm \epsilon$. And $\epsilon$ can be chosen arbitrarily small to begin with, which shows that the zeros do accumulate on the unit circle.

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    $\begingroup$ This is great! My numerical testing suggests that apart from that single zero $\approx 1/2$ the rest of them are actually outside of the unit circle (unless you leave the $z-1$ uncancelled which then produces a single zero on the unit circle). Any thoughts about how to see that (if true)? $\endgroup$ Commented Nov 26, 2021 at 20:54
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    $\begingroup$ Those zeros satisfy $$z^n+\frac1z=2.$$ For real $x$ the function $x^n+1/x$ is increasing when $x>n^{-1/(n+1)}$. If $\epsilon$ rules out zeros below that threshold, it follows that $|z|\ge1$. $\endgroup$ Commented Nov 26, 2021 at 21:22
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    $\begingroup$ @JyrkiLahtonen: If you try Rouché on the circle $|z|=1$ instead of $|z|=1-\epsilon$, you still get $|h|<|g|$ on the whole circle, except at the point $z=1$, and I think (although I haven't thought very carefully about it) that it might be possible to get $|h|<|g|$ on the contour consisting of the unit circle except with an arbitrary small indentation going just inside $z=1$. If that's correct, the $n$ “large” zeros would indeed be one at $z=1$ and the rest outside the unit circle. $\endgroup$ Commented Nov 26, 2021 at 22:00

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