Problem: ( a , b ) ( a , b ) ( a + b i ) 2002 = ( a + b i ) 2 0 0 2 = a − b i a − 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 ˉ = a − b i z = a + b i , z ˉ = a − b i ∣ z ∣ = a 2 + b 2 ∣ z ∣ = a 2 + b 2 ​ z 2002 = z ˉ z 2 0 0 2 = z ˉ
∣ z ∣ 2002 = ∣ z 2002 ∣ = ∣ z ˉ ∣ = ∣ z ∣ ∣ z ∣ 2 0 0 2 = ∣ ∣ ∣ ​ z 2 0 0 2 ∣ ∣ ∣ ​ = ∣ z ˉ ∣ = ∣ z ∣
from which it follows that
∣ z ∣ ( ∣ z ∣ 2001 − 1 ) = 0 ∣ z ∣ ( ∣ z ∣ 2 0 0 1 − 1 ) = 0
Hence ∣ z ∣ = 0 ∣ z ∣ = 0 ( a , b ) = ( 0 , 0 ) ( a , b ) = ( 0 , 0 ) ∣ z ∣ = 1 ∣ z ∣ = 1 ∣ z ∣ = 1 ∣ z ∣ = 1 z 2002 = z ˉ z 2 0 0 2 = z ˉ z 2003 = z ˉ ⋅ z = ∣ z ∣ 2 = 1 z 2 0 0 3 = z ˉ ⋅ z = ∣ z ∣ 2 = 1 z 2003 = 1 z 2 0 0 3 = 1 1 + 2003 = 2004 1 + 2 0 0 3 = 2 0 0 4 ​
The problems on this page are the property of the MAA's American Mathematics Competitions