Problem:
The equation
x10+(13x−1)10=0
has 10 complex roots r1,r1,r2,r2,r3,r3,r4,r4,r5,r5, where the bar denotes complex conjugation. Find the value of
r1r11+r2r21+r3r31+r4r41+r5r51
Solution:
Let p(x)=x10+(13x−1)10. If r is a zero of p(x), then
−1=(r13r−1)10=(r1−13)10
Thus
(r1−13)(rˉ1−13)=1
so that
(r11−13)(r11−13)+⋯+(r51−13)(r51−13)=5
Expanding and rearranging, we find
(r1r11+⋯+r5r51)−13(r11+r11+⋯+r51+r51)+5⋅169=5
Note that 1/r1,1/r1,…,1/r5,1/r5 are the zeros of
x10p(x1)=x10−130x9+⋯
so
r11+r11+⋯+r51+r51=130
Therefore,
r1r11+⋯+r5r51=13⋅130−5⋅169+5=850
OR
Let p(x)=x10+(13x−1)10. If p(r)=0, then
(13−r1)10=−1=cos180∘+isin180∘
It follows that
r1=13−(cosθ+isinθ)
where θ is an odd multiple of 18∘. Hence
rrˉ1=(13−(cosθ+isinθ))(13−(cosθ−isinθ))=170−26cosθ
Letting θ take on the values 18∘,54∘,90∘,126∘,162∘, we obtain all of the desired products. Thus
r1r11+⋯+r5r51=5⋅170−26(cos18∘+cos54∘+cos90∘+cos126∘+cos162∘)
Applying the identity cosθ+cos(180∘−θ)=0, we see the sum is 5⋅170=850.
The problems on this page are the property of the MAA's American Mathematics Competitions