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

Rearranging       x¹⁰  + (13x - 1)¹⁰ = 0

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

So the ten 10^th roots, R_n, are given by

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

Thus     1/R_n = 13 – exp(i*n*pi/10)

Taking conjugates (shown in bold):

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

So it is clear that 1/R_n = 1/R_-n so that

(R_n*R_n)^-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