( ) ( ) ( ) ( )Z I+ = Z 1 1 I− − ⇔ + =Z I 2 Z 1 1− TA CÓ
2015)
( ) ( ) ( ) ( )
z i+ = z 1 1 i− − ⇔ + =z i 2 z 1 1−Ta có:
z x yi;x;y R = + ∈
. Thay vào (1) ta có:
Đặt:
x yi i− + = 2 x 1 yi− +( )
2
( )
2
( ) (
2
)
2
2
2
⇔ + − = − + ⇔ − + + =x y 1 2 x 1 y x 2 y 1 4