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 1

Given we want to show \( f_n(z) \neq 0 \) for all \( |z| \leq 1 \).

Rewrite \( f_n(z) \) as Multiply by \( 1-z \): Shift index in the second sum: Write out terms: Combine sums for \( k=1 \) to \( n-1 \): which simplifies to So Multiply both sides by \( 1-z \): Suppose \( f_n(z) = 0 \) with \( |z| \leq 1 \) and \( z \neq 1 \). Then Rearranged: 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

Solution 2 (Fast)

Write

\( f_n(z) = \sum_{k=0}^{n-1} (n-k) z^k \). \( f_n(z) = \sum_{k=0}^{n-1} (n-k) z^k = \sum_{k=0}^{n-1} \sum_{j=k}^{n-1} z^k \),

because for each fixed \( k \), the coefficient \( (n-k) \) counts how many \( j \ge k \) there are from \( k \) to \( n-1 \).

Interchange the sums:

\( f_n(z) = \sum_{j=0}^{n-1} \sum_{k=0}^{j} z^k = \sum_{j=0}^{n-1} \frac{1-z^{j+1}}{1-z} = \frac{1}{1-z} \sum_{j=0}^{n-1} (1-z^{j+1}) \).

Simplify the inner sum:

\( \sum_{j=0}^{n-1} 1 = n \), \( \sum_{j=0}^{n-1} z^{j+1} = z \frac{1-z^n}{1-z} \).

So

\( f_n(z) = \frac{1}{1-z} \left( n - z \frac{1-z^n}{1-z} \right) = \frac{n(1-z) - z(1-z^n)}{(1-z)^2} \).

Suppose \( f_n(z) = 0 \) for some \( |z| \le 1 \) and \( z \ne 1 \), then

\( n(1-z) = z(1-z^n) \).

Rewrite as

\( n = \frac{z(1-z^n)}{1-z} \).

Note that the right side is a sum of \( n \) complex numbers:

\( \frac{z(1-z^n)}{1-z} = z + z^2 + \cdots + z^n \),

which is a geometric sum without the first term (since \( z \ne 1 \)).

For \( |z| \le 1 \), the magnitude of this sum is at most \( n \), and equal to \( n \) only if all terms \( z, z^2, ..., z^n \) lie on the same ray and have magnitude 1.

That requires \( |z| = 1 \) and all terms aligned, so \( z = e^{2\pi i k/n} \) for some integer \( k \).

Check for these roots:

If \( z = 1 \), \( f_n(1) = \frac{n(n+1)}{2} \ne 0 \).

For \( z = e^{2\pi i k/n} \), \( 1-z^n = 0 \), so from the equation \( n(1-z) = z(1-z^n) \) we get \( n(1-z) = 0 \), so \( z = 1 \), contradiction.

Therefore no such root \( z \ne 1 \) exists with \( |z| \le 1 \).

Hence \( f_n(z) \ne 0 \) is not in the closed unit disk.

~Pinotation

See Also

2018 Putnam B Entire Test

2018 Putnam B Problem 1

2018 Putnam B Problem 3

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