1anyone for this one
let x,y and z be complex nos such that
x+y+z=2
x^{2}+y^{2}+z^{2}=3
xyz=4
evaluate \frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{zx+y-1}
not my doubt[1]
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5 Answers
1interestingly sum of all imaginary parts is 0 in all conditions
since product of all terms is real so i think 1 term will have no imaginary part
and the other two terms will have imaginary parts equal in magnitude and opposite in sign
and since the sum of the squares has no imaginary part too, then the real part of the two nos will be equal in magnitude and in sign......
1[7][3][6].....@ nkj well its written that x y z are complex nos.....but believe me the soln has nothing much to do with all that u hav written and the fact that they r complex nos.........
1using equation 1
\sum_{cyclic}^{ }{ \frac{1}{1-(x+y)+xy}}
\sum_{cyclic}^{ }{ \frac{(1-z)}{1-x)(1-y)(1-z)}}
{ \frac{(1-z)+(1-x)+(1-y)}{1-x)(1-y)(1-z)}}
{ \frac{(3-\sum{x})}{(1-\sum{x}+\sum{xy}-xyz)}}
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