Putnam 2020, Problem B-5

For \( j \in \{1, 2, 3, 4\} \), let \( z_j \) be a complex number with \( |z_j| = 1 \) and \( z_j \neq 1 \). Prove that
\[
3 - z_1 - z_2 - z_3 - z_4 + z_1 z_2 z_3 z_4 \neq 0.
\]

Solution:
**First solution.** (by Mitja Mastnak) It will suffice to show that for any $z_1,z_2,z_3,z_4 \in \mathbb{C}$ of modulus $1$ such that $|3 - z_1 - z_2 - z_3 - z_4| = |z_1z_2z_3z_4|$, at least one of $z_1,z_2,z_3$ is equal to $1$.

To this end, let $z_1 = e^{\alpha i}$, $z_2 = e^{\beta i}$, $z_3 = e^{\gamma i}$ and
$$f(\alpha,\beta,\gamma) = |3 - z_1 - z_2 - z_3|^2 - |1 - z_1z_2z_3|^2.$$

A routine calculation shows that
$$f(\alpha,\beta,\gamma) = 10 - 6\cos(\alpha) - 6\cos(\beta) - 6\cos(\gamma)$$
$$+ 2\cos(\alpha+\beta+\gamma) + 2\cos(\alpha-\beta)$$
$$+ 2\cos(\beta-\gamma) + 2\cos(\gamma-\alpha).$$

Since the function $f$ is continuously differentiable, and periodic in each variable, $f$ has a maximum and a minimum and it attains these values only at points where $\nabla f = (0,0,0)$. A routine calculation now shows that
$$\frac{\partial f}{\partial \alpha} + \frac{\partial f}{\partial \beta} + \frac{\partial f}{\partial \gamma} = 6(\sin(\alpha) + \sin(\beta) + \sin(\gamma) - \sin(\alpha+\beta+\gamma))$$
$$= 24 \sin\left(\frac{\alpha+\beta}{2}\right) \sin\left(\frac{\beta+\gamma}{2}\right) \sin\left(\frac{\gamma+\alpha}{2}\right).$$

Hence every critical point of $f$ must satisfy one of $z_1z_2 = 1$, $z_2z_3 = 1$, or $z_3z_1 = 1$. By symmetry, let us assume that $z_1z_2 = 1$. Then
$$f = |3 - 2\operatorname{Re}(z_1) - z_3|^2 - |1 - z_3|^2;$$

since $3 - 2\operatorname{Re}(z_1) \geq 1$, $f$ is nonnegative and can be zero only if the real part of $z_1$, and hence also $z_1$ itself, is equal to $1$.

**Remark.** If $z_1 = 1$, we may then apply the same logic to deduce that one of $z_2,z_3,z_4$ is equal to $1$. If $z_1 = z_2 = 1$, we may factor the expression
$$3 - z_1 - z_2 - z_3 - z_4 + z_1z_2z_3z_4$$

as $(1 - z_3)(1 - z_4)$ to deduce that at least three of $z_1,\ldots,z_4$ are equal to $1$.

**Second solution.** We begin with an "unsmoothing" construction.

Turning to the original problem, consider $z_1,\ldots,z_4 \in \mathbb{C}$ with $|z_j| = 1$ and
$$3 - z_1 - z_2 - z_3 - z_4 + z_1z_2z_3z_4 = 0;$$

we must prove that at least one $z_j$ is equal to $1$. Let $z$ be any fourth root of $z_1z_2z_3z_4$, put $w_j = z_j/z$, and put $s = w_1 + \cdots + w_4$. In this notation, we have
$$s = z^3 + 3z^{-1},$$

where $s \in S_4$ and $z^3 + 3z^{-1} \in H_4$. That is, $s$ is a boundary point of $S_4$, so in particular it is the extremal point of $S_4$ on the ray emanating from the origin through $s$. By Lemma 1, this implies that $w_1,\ldots,w_4$ take at most two distinct values. As in the proof of Lemma 2, we distinguish two cases.

- If $w_1 = w_2 = w_3$, then
$$w_1^{-3} + 3w_1 = z^3 + 3z^{-1}.$$

From the geometric description of $H_n$, we see that this forces $w_1^{-1} = z$ and hence $z_1 = 1$.

- If $w_1 = w_2$ and $w_3 = w_4$, then $s \in [-4,4]$ and hence $s = \pm 4$. This can only be achieved by taking $w_1 = \cdots = w_4 = \pm 1$; since $s = z^3 + 3z^{-1}$ we must also have $z = \pm 1$, yielding $z_1 = \cdots = z_4 = 1$.

**Remark.** With slightly more work, one can show that Lemma 2 remains true for all positive integers $n$. The missing extra step is to check that for $m = 1,\ldots,n-1$, the hypocycloid curve
$$\{mz^{n-m} + (n-m)z^{-m} : z \in \mathbb{C}, |z| = 1\}$$

is contained in the filled interior of $H_n$. In fact, this curve only touches $H_n$ at points where they both touch the unit circle (i.e., at $d$-th roots of unity for $d = \gcd(m,n)$); this can be used to formulate a corresponding version of the original problem, which we leave to the reader.