Problem: ( a , b ) ( a , b ) ( a + b i ) 2002 = ( a + b i ) 2 0 0 2 = a β b i a β b i
Answer Choices:
A. 1001 1 0 0 1 1002 1 0 0 2 2001 2 0 0 1 2002 2 0 0 2 2004 2 0 0 4 Solution:
Let z = a + b i , z Λ = a β b i z = a + b i , z Λ = a β b i β£ z β£ = a 2 + b 2 β£ z β£ = a 2 + b 2 β z 2002 = z Λ z 2 0 0 2 = z Λ
β£ z β£ 2002 = β£ z 2002 β£ = β£ z Λ β£ = β£ z β£ β£ z β£ 2 0 0 2 = β£ β£ β£ β z 2 0 0 2 β£ β£ β£ β = β£ z Λ β£ = β£ z β£
from which it follows that
β£ z β£ ( β£ z β£ 2001 β 1 ) = 0 β£ z β£ ( β£ z β£ 2 0 0 1 β 1 ) = 0
Hence β£ z β£ = 0 β£ z β£ = 0 ( a , b ) = ( 0 , 0 ) ( a , b ) = ( 0 , 0 ) β£ z β£ = 1 β£ z β£ = 1 β£ z β£ = 1 β£ z β£ = 1 z 2002 = z Λ z 2 0 0 2 = z Λ z 2003 = z Λ β
z = β£ z β£ 2 = 1 z 2 0 0 3 = z Λ β
z = β£ z β£ 2 = 1 z 2003 = 1 z 2 0 0 3 = 1 1 + 2003 = 2004 1 + 2 0 0 3 = 2 0 0 4 β
The problems on this page are the property of the MAA's American Mathematics Competitions