IF A,B,C ARE COMPLEX NUMBERS SUCH THAT MODA=1 MODB=ROOT2,AND MODC=ROOT3,ALSO MOD A+B+C=2 THEN MOD BC+2CA+3AB=?.ANSWER MUST BE IN ONE FIGURE.

 ANS IS 9. PLZ  APPROVE THE ANSWER.

Hi Sagar,

consider |bc+2ca+3ab|^2 = (bc+2ca+3ab)*{(bc+2ca+3ab) whole bar}

                                     = (multiply and add) ==> |bc|^2 + 4|ca|^2 + 9|ab|^2 + 6*{ (a(bar)b)+(b(bar)a)+..... cyclic terms}

                                     = 6 + 12 + 18 + 6*{...........}   -------------- (1)

consider |a+b+c|^2 = (do the same thing)

                             = |a|^2 + |b|^2 + |c|^2 + {...........}

==> {..........} = -2

Substitute in (1)  ==>  |req quantity|^2 = 24

Hence answer = Root(24) = 4*Root(6)