HMMT 二月 2011 · ALGCALC 赛 · 第 18 题

英文解析

18. 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 − ,
    or,2
    2009 2008
    P ( x )( x − 1) + 2010 · 1005 = x + 2 x + . . . + 2009 x + 2010 .
    Multiplying by x − 1 once again:
    2010 · 2011
    2010 2009
    ( x − 1)( P ( x )( x − 1) + ) = x + x + . . . + x − 2010 ,
    2010 20092 = ( x + x + . . . + x + 1) − 2011 .
    Algebra & Calculus Individual Test
    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. Pluggingthose 2010 powers of z into the last equation, and multiplying them together, we obtain
    1004
    2010
    ( − 2011) · 1005 · ( x − )
    ∏
    i 1005
    P ( z ) = .
    ( x − 1)2
    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 ).