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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)
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