The Holomorphic Function That Knows Its Zeros Are Real

Let $f$ be an entire function satisfying:

$f$ has only real zeros,
$\overline{f(z)} = f(\bar{z})$ for all $z \in \mathbb{C}$ (the symmetry condition),
$f$ has infinitely many zeros $x_1, x_2, \ldots \in \mathbb{R}$ .

Now consider a much simpler question:

Suppose $f$ is a polynomial with real coefficients, and all zeros of $f$ lie on the unit circle $|z|=1$ . Prove that all zeros of $f'$ also lie on the unit circle.

Wait — is that even true? Find a counterexample or prove it.

Actually, let's make the problem precise and beautiful:

Problem. Let $p(z)$ be a polynomial of degree $n \geq 2$ with all roots on the unit circle $|z|=1$ . Must all roots of $p'(z)$ also lie on the unit circle?

If false, give a counterexample. If true, prove it. What is the correct statement?

Answer: The Holomorphic Function That Knows Its Zeros Are Real

Key Idea / Intuition

The answer is no in general — roots of the derivative need not stay on the unit circle. The key insight is the Gauss–Lucas theorem: the roots of $p'$ lie in the convex hull of the roots of $p$ . For roots on the unit circle, the convex hull is the closed unit disk $|z| \leq 1$ . So the correct statement is that all roots of $p'$ lie in $|z| \leq 1$ , with equality (on the circle itself) only in degenerate cases. A single clean example demolishes the stronger claim, and then Gauss–Lucas clinches the sharp bound.

Formal Proof / Solution

Step 1: Counterexample showing $p'$ need not have roots on $|z|=1$

Take $p(z) = z^n - 1, \quad n \geq 3.$ The roots of $p$ are the $n$ -th roots of unity, all on $|z|=1$ . ✓

Now $p'(z) = nz^{n-1}$ , which has only the root $z = 0$ (with multiplicity $n-1$ ).

Since $|0| = 0 < 1$ , the root of $p'$ is strictly inside the unit disk, not on the unit circle.

So the naive conjecture is false.

Step 2: The correct statement via Gauss–Lucas

Theorem (Gauss–Lucas). If $p$ is a polynomial with roots $z_1, \ldots, z_n \in \mathbb{C}$ , then every root of $p'$ lies in the convex hull $\operatorname{conv}(z_1, \ldots, z_n)$ .

Proof of Gauss–Lucas. Write $\frac{p'(z)}{p(z)} = \sum_{k=1}^n \frac{1}{z - z_k}.$ Suppose $p'(w) = 0$ but $p(w) \neq 0$ . Then $\sum_{k=1}^n \frac{1}{w - z_k} = 0.$ Taking complex conjugates and writing $\frac{1}{w-z_k} = \frac{\overline{w-z_k}}{|w-z_k|^2}$ , we get $\sum_{k=1}^n \frac{\bar{w} - \bar{z}_k}{|w-z_k|^2} = 0 \implies \bar{w} = \frac{\sum_k \frac{\bar{z}_k}{|w-z_k|^2}}{\sum_k \frac{1}{|w-z_k|^2}}.$ This says $\bar{w}$ (hence $w$ ) is a weighted average of the $\bar{z}_k$ (hence of the $z_k$ ) with positive weights $\lambda_k = \frac{1}{|w-z_k|^2}$ . So $w \in \operatorname{conv}(z_1,\ldots,z_n)$ . $\square$

Step 3: Application to roots on $|z|=1$

If all roots of $p$ lie on $|z|=1$ , then $\operatorname{conv}(z_1,\ldots,z_n) \subseteq \overline{\mathbb{D}}$ (the closed unit disk), since the unit disk is convex and contains the unit circle.

Therefore, every root of $p'$ satisfies $|z| \leq 1$ .

The example $p(z)=z^n-1$ shows this bound is sharp: roots of $p'$ can be strictly inside $|z|<1$ .

Summary

| Claim | Truth | |---|---| | All roots of $p'$ lie on $\|z\|=1$ | ❌ False ( $p(z)=z^n-1$ gives $p'$ vanishing only at $0$ ) | | All roots of $p'$ lie in $\|z\|\leq 1$ | ✅ True by Gauss–Lucas |

The beautiful moral: Gauss–Lucas is a convexity theorem in disguise, and the "logarithmic derivative = weighted barycenter" identity is the one-line proof.