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Complex Conclusion II (Posted on 2014-11-08) Difficulty: 3 of 5
Given that the equation: x10 + (13x - 1)10 = 0 has ten complex roots:
R1, R1, R2, R2, R3, R3, R4, R4, R5 and R5
where the bolded form represents complex conjugation.

Determine the value of:

(R1*R1)-1+( R2*R2) -1+(R3*R3) -1+(R4*R4) -1+( R5*R5) -1

No Solution Yet Submitted by K Sengupta    
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Solution Solution | Comment 1 of 2
Rearranging       x10  + (13x - 1)10 = 0

gives                 (13 – 1/x)10 = -1 = exp(i*pi)

So the ten 10th roots, Rn, are given by

13 – 1/Rn = exp(i*n*pi/10) for n = -9, -7, -5,..  …5, 7, 9.

Thus     1/Rn = 13 – exp(i*n*pi/10)

Taking conjugates (shown in bold):

            1/Rn = 13 – exp(-i*n*pi/10)

So it is clear that 1/Rn = 1/R-n so that

(Rn*Rn)-1   = (13 – exp(i*n*pi/10))(13 – exp(-i*n*pi/10))

                = (169 + 1 – (exp(i*n*pi/10) + exp(-i*n*pi/10)))

                = 170 – 2*cos(n*pi/10)

and summing over n = 1, 3, 5, 7, 9 gives: required sum

= 850 – 2(cos(.1pi)+cos(.3pi)+cos(.5pi)+cos(.7pi)+cos(.9pi))

Since cos(.1pi) = -cos(.9pi), cos(.3pi) = -cos(.7pi) & cos(.5pi)=0

the sum reduces to 850.



  Posted by Harry on 2014-11-10 11:40:56
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