2018 Putnam B Problems/Problem 2

Problem

Let $ n $ be a positive integer, and let $ f_n(z) = n + (n-1)z + (n-2)z^2 + \dots + z^{n-1} $. Prove that $ f_n $ has no roots in the closed unit disk $ \{z \in \mathbb{C}: |z| \leq 1 \} $.

Solution

Given $f_n(z) = n + (n-1)z + (n-2)z^2 + \dots + z^{n-1},$ we want to show $ f_n(z) \neq 0 $ for all $ |z| \leq 1 $.

Rewrite $ f_n(z) $ as $f_n(z) = \sum_{k=0}^{n-1} (n-k)z^k.$ Multiply by $ 1-z $: $(1-z)f_n(z) = \sum_{k=0}^{n-1} (n-k)z^k - \sum_{k=0}^{n-1} (n-k)z^{k+1}.$ Shift index in the second sum: $= n + \sum_{k=1}^{n-1} (n-k)z^k - \sum_{k=1}^{n} (n-(k-1))z^k.$ Write out terms: $= n + \sum_{k=1}^{n-1} (n-k)z^k - \sum_{k=1}^{n} (n-k+1)z^k.$ Combine sums for $ k=1 $ to $ n-1 $: $= n + \sum_{k=1}^{n-1} [(n-k) - (n-k+1)]z^k - (n-n+1)z^n,$ which simplifies to $= n + \sum_{k=1}^{n-1} (-1)z^k - z^n = n - \sum_{k=1}^{n} z^k.$ So $(1-z)f_n(z) = n - \frac{z(1-z^n)}{1-z} \quad \text{if } z \neq 1.$ Multiply both sides by $ 1-z $: $(1-z)^2 f_n(z) = n(1-z) - z(1-z^n).$ Suppose $ f_n(z) = 0 $ with $ |z| \leq 1 $ and $ z \neq 1 $. Then $n(1-z) = z(1-z^n).$ Rearranged: $n = \frac{z(1-z^n)}{1-z}.$ Consider $ |z| \leq 1 $. If $ |z| = 1 $, then $ |1-z^n| \leq 2 $ and denominator $ |1-z| $ is at most 2, but the right side stays bounded by something less than $ n $ (except at $ z=1 $, excluded). This can't equal $ n $ exactly.

If $ |z| < 1 $, the geometric sum $ \frac{1-z^n}{1-z} $ has magnitude less than $ \frac{1}{1-|z|} $, which is finite but smaller than $ n $. So the equation can't hold for any $ z \neq 1 $.

At $ z=1 $, $ f_n(1) = n + (n-1) + \dots + 1 = \frac{n(n+1)}{2} \neq 0 $.

Thus, $ f_n(z) \neq 0 $ for $ |z| \leq 1 $. :-)

~Pinotation

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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2018 Putnam B Problems/Problem 2

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