Let f(x)=x^2004+2x^2003+3x^2002+...+2005. Let z=cis(pi/1003). Find f(z)f(z^2)f(z^3)f(z^4)...f(z^2005).

Let f(x)=x^2004+2x^2003+3x^2002+...+2005. Let z=cis(pi/1003). Find f(z)f(z^2)f(z^3)f(z^4)...f(z^2005).

AIME question

f(x) = x^2004 + 2x^2003 + 3x^2002 + ... + 2005
.....= Σ(n = 0 to 2004) (n+1) x^n
.....= (d/dx) Σ(n = 0 to 2004) x^(n+1)
.....= (d/dx) x(x^2005 - 1)/(x - 1)
.....= [(2006 x^2005 - 1) (x - 1) - (x^2006 - x)] / (x - 1)^2
.....= (2005 x^2006 - 2006 x^2005 + 1) / (x - 1)^2

Letting x = z^k for any k yields
f(z^k) = (2005 (z^k)^2006 - 2006 (z^k)^2005 + 1) / (z^k - 1)^2
........= (2005 * 1 - 2006 (z^k)^(-1) + 1) / (z^k - 1)^2, since z^2006 = 1.
........= (2006 - 2006 z^(-k)) / (z^k - 1)^2
........= 2006 z^(-k) (z^k - 1) / (z^k - 1)^2
........= 2006 z^(-k) / (z^k - 1).

Hence, f(z) f(z^2) ... f(z^2005)
= Π(k = 1 to 2005) f(z^k)
= Π(k = 1 to 2005) 2006 [z^(-k) / (z^k - 1)]
= 2006^2005 * Π(k = 1 to 2005) z^(-k) / [Π(k = 1 to 2005) (z^k - 1)]
= 2006^2005 * z^(-(1 + 2 + ... + 2005)) / [Π(k = 1 to 2005) (z^k - 1)]

However,
Π(k = 1 to 2005) z^(-k)
= z^(-(1 + 2 + ... + 2005))
= z^(-2005 * 2006/2)
= (z^(-2005))^1003
= z^1003
= -1, since z is a primitive 2006th root of unity and 2006 is even.

Π(k = 1 to 2005) (z^k - 1)
= Π(k = 1 to 2005) -(1 - z^k)
= (-1)^2005 * Π(k = 1 to 2005) (1 - z^k)
= -1 * 2006, since Π(k = 1 to n-1) (1 - cis(2πk/n)) = n
= -2006.

Therefore, f(z) f(z^2) ... f(z^2005)
= 2006^2005 * z^(-(1 + 2 + ... + 2005)) / [Π(k = 1 to 2005) (z^k - 1)]
= 2006^2005 * -1 / (-2006)
= 2006^2004.

I hope this helps!