HMMT 二月 2013 · 代数 · 第 9 题

HMMT February 2013 — Algebra — Problem 9

Let z be a non-real complex number with z = 1. Compute 22 ∑ 1 . k 2 k 1 + z + z k =0

解析

Let z be a non-real complex number with z = 1. Compute 22 ∑ 1 . k 2 k 1 + z + z k =0 Answer: 46/3 First solution: Note that 22 22 22 22 7 k 24 k ∑ ∑ ∑ ∑ ∑ 1 1 1 − z 1 1 − ( z ) 1 3 kℓ = + = + = + z . k 2 k 3 k 3 k 1 + z + z 3 1 − z 3 1 − z 3 k =0 k =1 k =1 k =1 ℓ =0 3 and 23 are prime, so every non-zero residue modulo 23 appears in an exponent in the last sum exactly 7 times, and the summand 1 appears 22 times. Because the sum of the 23rd roots of unity is zero, our 1 46 answer is + (22 − 7) = . 3 3 Second solution: For an alternate approach, we first prove the following identity for an arbitrary complex number a : 22 22 ∑ 1 23 a = . k 23 a − z a − 1 k =0 23 2 22 To see this, let f ( x ) = x − 1 = ( x − 1)( x − z )( x − z ) . . . ( x − z ). Note that the sum in question is ′ f ( a ) merely , from which the identity follows. f ( a ) 3 Now, returning to our orignal sum, let ω 6 = 1 satisfy ω = 1. Then 22 22 ∑ ∑ 1 1 1 1 = − k 2 k 2 k 2 k 1 + z + z ω − ω ω − z ω − z k =0 k =0 ( ) 22 22 ∑ ∑ 1 1 1 = − 2 k 2 k ω − ω ω − z ω − z k =0 k =0 ( ) 22 44 1 23 ω 23 ω = − 2 23 46 ω − ω ω − 1 ω − 1 ( ) 2 23 ω ω = − 2 2 ω − ω ω − 1 ω − 1 2 2 23 ( ω − ω ) − ( ω − ω ) = 2 2 ω − ω 2 − ω − ω 46 = . 3