HMMT 二月 2013 · 代数 · 第 9 题

英文解析

9. Let z be a non-real complex number with z = 1. Compute
   ∑22
   .1
   k 2 k
   1 + z + zk =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 23 rd 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 arbitrarycomplex number a :
   2222
   ∑
   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 )
   Now, returning to our orignal sum, let ω 6 = 1 satisfy ω = 1. Then 3
   22 22
   ∑ ∑
   1 1 1 1 = −
   k 2 k 2 k 2 k
   1 + z + z ω − ω ω − z ω − zk =0 k =0
   ( )
   22 22
   ∑ ∑
   1 1 1 = −
   2 k 2 kω − ω ω − z ω − zk =0 k =0
   ( )
   22 44
   1 23 ω 23 ω = −
   2 23 46
   ω − ω ω − 1 ω − 1
   ( )
   23 ω ω2 = −
   2 2
   ω − ω ω − 1 ω − 1
   2 2
   23 ( ω − ω ) − ( ω − ω )
   2 2=
   ω − ω 2 − ω − ω = .46 3