HMMT 二月 2025 · 团队赛 · 第 6 题
HMMT February 2025 — Team Round — Problem 6
题目详情
- [40] Complex numbers ω , . . . , ω each have magnitude 1. Let z be a complex number distinct from 1 n ω , . . . , ω such that 1 n z + ω z + ω 1 n
- · · · + = 0 . z − ω z − ω 1 n Prove that | z | = 1.
解析
- [40] Complex numbers ω , . . . , ω each have magnitude 1. Let z be a complex number distinct from 1 n ω , . . . , ω such that 1 n z + ω z + ω 1 n
- · · · + = 0 . z − ω z − ω 1 n Prove that | z | = 1. Proposed by: Karthik Venkata Vedula Solution 1: We show that no solutions z not on the unit circle can exist. First, we eliminate | z | > 1. z + ω j Claim 1. For all j and | z | > 1, the real part of is positive. z − ω j Proof. We use geometry. Note that ω and − ω are antipodes on the unit circle. Since z lies outside the j j z + ω j unit circle, it follows that ∠ ω z ( − ω ) is acute. But this means the complex number lies strictly j j z − ω j in the first or fourth quadrant of the complex plane and thus has positive real part, as desired. P n z + ω j It is then clear that whenever | z | > 1, the sum has positive real part and thus cannot be j =1 z − ω j z + ω j
- The case where | z | < 1 is analogous, except ∠ ω z ( − ω ) is obtuse instead, so has negative real j j z − ω j part for all j . Therefore, all solutions z to the original equation must satisfy | z | = 1. Solution 2: We show more generally that for any positive integers k , a , . . . , a , and distinct ω on 1 k j the unit circle, the equation k X z + ω i a = 0 j z − ω j j =1 has k distinct solutions on the unit circle. The original problem then follows upon consolidating duplicate ω ’s. Without loss of generality, assume that ω , . . . , ω are in this order going clockwise j 1 k around the unit circle. Claim 2. There is a solution on the (clockwise) arc from ω to ω for all j (where ω = ω ). j j +1 k +1 1 ◦ Proof. First, ω and − ω are antipodes on the unit circle, so if z is on the unit circle, ∠ ω z ( − ω ) = 90 j j j j z + ω j This means is purely imaginary. Now consider the imaginary part of the left hand side of the z − ω j equation, which is a real and continuous function on the arc strictly between ω and ω for each j . j j +1 In particular, as z approaches ω from the clockwise direction, this function approaches ∞ . On the j +1 other hand, as z approaches ω from the counterclockwise direction, this function approaches −∞ . By j the Intermediate Value Theorem, there must be a solution on this arc, as desired. It follows that there are at least k solutions on the unit circle. But the equation is equivalent to a polynomial of degree k . Hence, there are exactly k solutions, all of which lie on the unit circle. Remark. The coefficients a ’s are introduced to handle the case where some of ω , . . . , ω are equal. j 1 n Another way to get around this case is to utilize the fact that roots of polynomials are continuous, so we can take the limit where several ω ’s approach each other. j z − 1 Remark. The M¨ obius transformation z 7 → i sends the unit circle to a real line. One can rephrase z +1 both solutions as working on the real line instead of the unit circle.