Problem: z 12 = 64 z 1 2 = 6 4
Answer Choices:
A. 2 2 4 4 2 + 2 3 2 β + 2 3 β 2 2 + 6 2 2 β + 6 β ( 1 + 3 ) + ( 1 + 3 ) i ( 1 + 3 β ) + ( 1 + 3 β ) i Solution:
The principal root of the equation z 12 = 64 z 1 2 = 6 4
z = 6 4 1 12 β
( cos β‘ Ο 6 + i sin β‘ Ο 6 ) = 2 β
( cos β‘ Ο 6 + i sin β‘ Ο 6 ) z = 6 4 1 2 1 β β
( cos 6 Ο β + i sin 6 Ο β ) = 2 β β
( cos 6 Ο β + i sin 6 Ο β )
The 12 roots lie in the complex plane on the circle of radius 2 2 β 0 , Β± Ο 6 0 , Β± 6 Ο β Β± Ο 3 Β± 3 Ο β
2 + 2 2 β
cos β‘ Ο 6 + 2 2 β
cos β‘ Ο 3 = 2 β
( 1 + 3 + 1 ) = 2 2 + 6 2 β + 2 2 β β
cos 6 Ο β + 2 2 β β
cos 3 Ο β = 2 β β
( 1 + 3 β + 1 ) = 2 2 β + 6 β β
The problems on this page are the property of the MAA's American Mathematics Competitions