Problem: z 4 + 4 z 3 i − 6 z 2 − 4 z i − i = 0 z 4 + 4 z 3 i − 6 z 2 − 4 z i − i = 0
Answer Choices:
A. 2 5 8 2 8 5 2 3 4 2 4 3 2 2 2 5 4 2 4 5 2 3 2 2 2 3 Solution:
Adding 1 + i 1 + i
1 + i = ( z 4 + 4 z 3 i − 6 z 2 − 4 z i − i ) + 1 + i = z 4 + 4 z 3 i − 6 z 2 − 4 z i + 1 = ( z + i ) 4 1 + i = ( z 4 + 4 z 3 i − 6 z 2 − 4 z i − i ) + 1 + i = z 4 + 4 z 3 i − 6 z 2 − 4 z i + 1 = ( z + i ) 4
Let w = z + i = r ( cos θ + i sin θ ) w = z + i = r ( cos θ + i sin θ )
i + 1 = 2 ( cos π 4 + i sin π 4 ) i + 1 = 2 ( cos 4 π + i sin 4 π )
the solutions of w 4 = 1 + i w 4 = 1 + i
r 4 = 2 and θ = 1 4 ( π 4 + 2 k π ) = π 16 + π 2 k r 4 = 2 and θ = 4 1 ( 4 π + 2 k π ) = 1 6 π + 2 π k
for k = 0 , 1 , 2 k = 0 , 1 , 2
w k = 2 1 / 8 ( cos ( π 16 + π 2 k ) + i sin ( π 16 + π 2 k ) ) for k = 0 , 1 , 2 , or 3 w k = 2 1 / 8 ( cos ( 1 6 π + 2 π k ) + i sin ( 1 6 π + 2 π k ) ) for k = 0 , 1 , 2 , or 3
and the four solutions for z = w − i z = w − i
z k = 2 1 / 8 ( cos ( π 16 + π 2 k ) + i sin ( π 16 + π 2 k ) ) − i for k = 0 , 1 , 2 , or 3 z k = 2 1 / 8 ( cos ( 1 6 π + 2 π k ) + i sin ( 1 6 π + 2 π k ) ) − i for k = 0 , 1 , 2 , or 3
Note that w 0 , w 1 , w 2 w 0 , w 1 , w 2 w 3 w 3 2 1 / 8 2 1 / 8 ( 0 , 0 ) ( 0 , 0 ) z 0 , z 1 , z 2 z 0 , z 1 , z 2 z 3 z 3 2 1 / 8 2 1 / 8 ( 0 , − 1 ) ( 0 , − 1 ) z 0 , z 1 , z 2 z 0 , z 1 , z 2 z 3 z 3 2 1 / 8 2 = 2 5 / 8 2 1 / 8 2 = 2 5 / 8 ( 2 5 / 8 ) 2 = 2 5 / 4 ( 2 5 / 8 ) 2 = 2 5 / 4
OR
The Binomial Theorem gives
( z + i ) 4 = z 4 + 4 z 3 i − 6 z 2 − 4 z i + 1 = ( z 4 + 4 z 3 i − 6 z 2 − 4 z i − i ) + 1 + i = 1 + i ( z + i ) 4 = z 4 + 4 z 3 i − 6 z 2 − 4 z i + 1 = ( z 4 + 4 z 3 i − 6 z 2 − 4 z i − i ) + 1 + i = 1 + i
Let a a a 4 = 1 + i a 4 = 1 + i w = z + i w = z + i w 4 = a 4 w 4 = a 4 w w a , i a , − a a , i a , − a − i a − i a 2 ∣ a ∣ = 2 2 8 2 ∣ a ∣ = 2 8 2 w = z + i w = z + i ( 2 2 8 ) 2 / 2 = 2 2 4 = 2 5 / 4 ( 2 8 2 ) 2 / 2 = 2 4 2 = 2 5 / 4
The problems on this page are the property of the MAA's American Mathematics Competitions