Problem: z z w w
z + 20 i w = 5 + i w + 12 i z = − 4 + 10 i z + w 2 0 i w + z 1 2 i = 5 + i = − 4 + 1 0 i
Find the smallest possible value of ∣ z w ∣ 2 ∣ z w ∣ 2
Solution:
Multiply corresponding sides of the given two equations to yield
( z + 20 i w ) ( w + 12 i z ) = ( 5 + i ) ( − 4 + 10 i ) , and z w + 32 i − 240 z w = − 30 + 46 i . ( z + w 2 0 i ) ( w + z 1 2 i ) z w + 3 2 i − z w 2 4 0 = ( 5 + i ) ( − 4 + 1 0 i ) , and = − 3 0 + 4 6 i .
Letting v = z w v = z w v 2 − ( − 30 + 14 i ) v − 240 = 0 v 2 − ( − 3 0 + 1 4 i ) v − 2 4 0 = 0
v = − 30 + 14 i ± ( 30 − 14 i ) 2 + 960 2 = − 15 + 7 i ± ( 15 − 7 i ) 2 + 240 = − 15 + 7 i ± 416 − 210 i v = 2 − 3 0 + 1 4 i ± ( 3 0 − 1 4 i ) 2 + 9 6 0 = − 1 5 + 7 i ± ( 1 5 − 7 i ) 2 + 2 4 0 = − 1 5 + 7 i ± 4 1 6 − 2 1 0 i
Letting 416 − 210 i = ( a + b i ) 2 4 1 6 − 2 1 0 i = ( a + b i ) 2 a b = − 105 a b = − 1 0 5 a 2 − b 2 = 416 a 2 − b 2 = 4 1 6 ( a , b ) = ( 21 , − 5 ) ( a , b ) = ( 2 1 , − 5 ) ( − 21 , 5 ) ( − 2 1 , 5 ) v = − 15 + 7 i ± ( 21 − 5 i ) v = − 1 5 + 7 i ± ( 2 1 − 5 i ) v = 6 + 2 i v = 6 + 2 i − 36 + 12 i − 3 6 + 1 2 i ∣ z w ∣ 2 ∣ z w ∣ 2 6 2 + 2 2 = 40 6 2 + 2 2 = 4 0 w = 2 + 4 i w = 2 + 4 i z = 1 − i z = 1 − i
The problems on this page are the property of the MAA's American Mathematics Competitions