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 Complex Number Puzzle (Posted on 2015-11-08)
Each of A, B and C is a complex number such that:

A+B+C = 2, and:
A2 + B2 + C2 = 3, and:
A*B*C = 4

Evaluate:
(A*B + C - 1)-1 + (B*C + A - 1)-1 + (C*A + B - 1)-1

 Submitted by K Sengupta No Rating Solution: (Hide) A+B+C = 2, ----(i) A2 + B2 + C2 = 3....(ii) A*B*C = 4 ....(iii) Squaring (i) and subtracting (ii) from the result, we get: 2(A*B+B*C+C*A) =1, so that: A*B+B*C+C*A =1/2. Accordingly, A, B and C are roots of the equation: x3 - 2x2+x/2 - 4= 0 .....(iv) Now, A*B + C – 1 = 4/C+C-1, and so: (A*B + C - l)-1 = (4/C+C-1)-1 = C/(C2 - C +4) .....(v) Similarly, (B*C + A - l)-1 = (4/A+A-1)-1 = A/(A2 - A +4) ...(vi) (C*A + B - l)-1 = (4/B+B-1)-1 = B/(B2 - B +4) ...(vii) From (iv): C3 - 2C2+C/2 - 4= 0 or, (C2 - C +4)(C-1) = 9C/2 or, C/(C2 - C +4) = 2(C-1)/9 Similarly, A/(A2 - A +4) = 2(A-1)/9 and, B/(B2 - B+4) = 2(B-1)/9 Accordingly from (v), (vi) and (vii): (A*B+C-1)-1 + (B*C+A-1)-1 + (C*A+B-1)-1 = 2(C-1)/9 + 2(A-1)/9 + 2(B-1)/9 = 2(A+B+C-3)/9 = 2*(2-3)/9 = -2/9

 Subject Author Date A more confident answer Jer 2015-11-09 14:11:51 Maybe I got it (possible spoiler) Jer 2015-11-08 11:26:25

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