HMMT 二月 2011 · ALGGEO 赛 · 第 17 题

HMMT February 2011 — ALGGEO Round — Problem 17

Let z = cos + i sin , and let 2011 2011 2008 · 2009 2009 · 2010 2008 2007 2006 P ( x ) = x + 3 x + 6 x + . . . x + 2 2 2 3 2010 for all complex numbers x . Evaluate P ( z ) P ( z ) P ( z ) . . . P ( z ).

解析

Let z = cos + i sin , and let 2011 2011 2008 · 2009 2009 · 2010 2008 2007 2006 P ( x ) = x + 3 x + 6 x + . . . x + 2 2 2 3 2010 for all complex numbers x . Evaluate P ( z ) P ( z ) P ( z ) . . . P ( z ). 2009 2011 2011 Answer: 2011 · (1005 − 1004 ) Multiply P ( x ) by x − 1 to get 2009 · 2010 2009 2008 P ( x )( x − 1) = x + 2 x + . . . + 2009 x − , 2 or, 2009 2008 P ( x )( x − 1) + 2010 · 1005 = x + 2 x + . . . + 2009 x + 2010 . Multiplying by x − 1 once again: Algebra & Geometry Individual Test 2010 · 2011 2010 2009 ( x − 1)( P ( x )( x − 1) + ) = x + x + . . . + x − 2010 , 2 2010 2009 = ( x + x + . . . + x + 1) − 2011 . Hence, 2010 2009 ( x + x + . . . + x + 1) − 2011 − 2011 · 1005 x − 1 P ( x ) = x − 1 2010 2009 2 2010 Note that x + x + . . . + x +1 has z, z , . . . , z as roots, so they vanish at those points. Plugging those 2010 powers of z into the last equation, and multiplying them together, we obtain 1004 2010 ( − 2011) · 1005 · ( x − ) ∏ i 1005 P ( z ) = . 2 ( x − 1) i =1 2 2010 2010 2009 Note that ( x − z )( x − z ) . . . ( x − z ) = x + x + . . . + 1. Using this, the product turns out to 2009 2011 2011 be 2011 · (1005 − 1004 ).