PUMaC 2022 · 代数（A 组） · 第 4 题

PUMaC 2022 — Algebra (Division A) — Problem 4

The set C of all complex numbers z satisfying ( z + 1) = az for some a ∈ [ − 10 , 3] is the union of two curves intersecting at a single point in the complex plane. If the sum of the lengths of these two curves is ℓ , find ⌊ ℓ ⌋ .

解析

The set C of all complex numbers z satisfying ( z + 1) = az for some a ∈ [ − 10 , 3] is the union of two curves intersecting at a single point in the complex plane. If the sum of the lengths of these two curves is ℓ , find ⌊ ℓ ⌋ . Proposed by Julian Shah Answer: 16 2 We want solutions to z + (2 − a ) z + 1 = 0. The discriminant is non-negative when a ∈ ( −∞ , 0] ∪ [4 , ∞ ), so for our purposes, a ≤ 0. When the discriminant is non-negative, it can 2 be seen that the solutions lie between the solutions to x + (2 − ( − 10)) z + 1; this interval has √ length 2 35. The remaining values of a are in (0 , 3]. The solutions when a ∈ (0 , 3] are non-real, so they must be conjugates, and they are reciprocals, so it follows that they lie on the unit circle. − (2 − a ) 1 Furthermore, they’re real part is equal to , which ranges from − 1 to ; thus, the 2 2 1 solution set here is the portion of the unit circle with real part less than , which comprises 2 4 π two thirds of the unit circle. Thus, the length of this region is . 3 √ 4 π The desired length is then the sum of the lengths of these two regions, which is 2 35 + . 3 √ 4 π Rewriting, this is 140 + , which has floor 16. 3