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For each positive integer $n$, define the polynomial $$ P_n(z) = z + 2^3z^2 + ... + n^3z^n. $$ Prove that the roots of $P_n(z)$ lie inside the unit circle.


Assuming the existence of $|\alpha| \geq 1, P_n(\alpha) = 0$, I tried using the triangle inequality: $$ |\alpha|^nn^3 = |\alpha + 8\alpha^3 + ... + (n-1)^3 \alpha^{n-1} | \leq |\alpha| + ... + (n-1)^3 |\alpha|^{n-1} $$ which seems to be too weak to create a contradiction.

Thoughts/solutions are appreciated.

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  • $\begingroup$ have you tried rouche's theorem? $\endgroup$ Sep 5 '17 at 2:49
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The Gauss-Lucas theorem says that the roots of the derivative of a polynomial are in the convex hull of the roots of the polynomial. So:

The roots of $f_n(z) = 1 + z + \ldots + z^n = (z^n-1)/(z-1)$ are roots of unity, and lie on the unit circle. The roots of $f'_n(z) = \sum_{j=1}^n j z^{j-1}$ lie in the convex hull of these. Since the roots of $f_n$ are all simple, they are not roots of $f'_n$, so the roots of $f'_n$ are (strictly) inside the unit circle.

The roots of $g_n(z) = z f'_n(z) = \sum_{j=1}^n j z^j$ are $0$ and the roots of $f'_n(z)$, so again these are strictly inside the unit circle.

Iterating this process, we find that for all positive integers $k$, the roots of $\sum_{j=1}^n j^k z^j$ are strictly inside the unit circle.

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