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| Complex Number Puzzle (Posted on 2015-11-08) |
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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
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Submitted by K Sengupta
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Solution:
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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
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