If and are the roots of the equation 2z2-3z-2i=0, where i= -1, then 16 ( 19+ 19+ 11+ 11 15+ 15 ) ( 19+ 19+ 11+ 11 15+ 15 ) is equal to
- 441
- 312
- 409
- 398
Answer
(1) 441
Solution
2z2-3z-2i=0 z- iz= 32 for both ,
Square: z2- 1z2= 94+2i; square again: z4+ 1z4= ( 94+2i )2+2= 4916+9i
19+ 11= 15 ( 4+ 1 4 ), same for
the fraction = 4916+9i
16 4916 9=441
Answer: (1)