Q:

Let \(n\) be a positive integer and let

\[ f_n(z) = n + (n-1)z + (n-2)z^2 + \cdots + z^{n-1} \]

Prove that \(f_n\) has no roots on the closed unit disk \(\{ z \in \mathbb{C} : \lvert z \rvert \leq 1 \}\).

A:

Personally I think my solution is a bit underwhelming. The problem statement seems to be taunting us to do some complex roots of unity magic, but I managed to bypass all of that, which is disappointing.

Remark 1

\(f_n\) satisfies the following recurrence relation:

\[ f_{n+1}(z) = n + 1 + zf_n(z) \]

This is obtained by some simple rearrangements of the \(f_{n+1}\) formula:

\[ f_{n+1}(z) = n+1 + nz + (n-1)z^2 + \cdots + 2z^{n-1} + z^n \]

\[ = n+1 + z \left[ n + (n-1)z + \cdots + 2z^{n-2} + z^{n-1} \right] \]

\[ = n+1 + zf_n(z)\]

Remark 2

Secondly, we note that for all \(z\), the difference between successive pairs of functions is

\[ f_{n+1}(z) - f_n(z) = 1 + z + z^2 + z^3 + \cdots + z^n \]

This is the result of simply subtracting corresponding terms in \(f_{n+1}\) and \(f_n\), and is, as the intro comments, highly suggestive of roots of unity.

Remark 3

However, we won’t need to consider any zeros of the above function. Instead, we just note that for \(z\) in the closed unit disk,

\[ 1 + z + z^2 + \cdots + z^n = n+1 \]

if and only if \(z = 1\).

Proof Remark 3:

The backward direction of the iff is obvious, since if we plug in \(z = 1\), the series will clearly sum to \(n+1\).

For the forward direction, suppose \(z\) satisfies \(1 + z + z^2 + \cdots + z^n = n+1\).

We note that since \(z\) is in the closed unit disk, we have \(\lvert z \rvert \leq 1\). This implies that each of \(\lvert z \rvert\), \(\lvert z^2 \rvert\), …, \(\lvert z^n \rvert\) is \(\leq 1\).

If we take the real component of both sides of the equation, we have

\[ n+1 = \operatorname{Re}(n+1) = \operatorname{Re}(1 + z + z^2 + \cdots + z^n) = \operatorname{Re}(1) + \operatorname{Re}(z) + \operatorname{Re}(z^2) + \cdots + \operatorname{Re}(z^n) \]

Here we note that on the LHS, we have \(n+1\), and on the RHS we have the sum of \(n+1\) terms with each term having absolute value \(\leq 1\). Thus for the equality to hold, we need every term on the RHS to have value \(1\), i.e.

\[ \operatorname{Re}(z) = \operatorname{Re}(z^2) = \cdots = \operatorname{Re}(z^n) = 1 \]

and the facts that \(\lvert z \rvert \leq 1\) and \(\operatorname{Re}(z) = 1\) immediately imply that \(z = 1\).

Solving the Problem

Suppose (for contradiction) that the complex number \(z\) satisfies \(\lvert z \rvert \leq 1\) and \(f_n(z) = 0\).

From Remark 1, we have

\[ f_{n+1}(z) = n+1 + zf_n(z) = n+1 \]

since \(f_n(z) = 0\). Now if we subtract \(f_n(z) = 0\) from both sides, we see

\[ f_{n+1}(z) - f_n(z) = n+1 \]

From Remark 2, we also know

\[ f_{n+1}(z) - f_n(z) = 1 + z + z^2 + z^3 + \cdots + z^n \]

Thus

\[ n+1 = 1 + z + z^2 + z^3 + \cdots + z^n \]

which, by Remark 3, implies that \(z = 1\). But this is a contradiction, since \(f_n(1) \neq 0\). \(\square\)