Problem:
The equation
x10+(13xβ1)10=0
has 10 complex roots r1β,r1ββ,r2β,r2ββ,r3β,r3ββ,r4β,r4ββ,r5β,r5ββ, where the bar denotes complex conjugation. Find the value of
r1βr1ββ1β+r2βr2ββ1β+r3βr3ββ1β+r4βr4ββ1β+r5βr5ββ1β
Solution:
Let p(x)=x10+(13xβ1)10. If r is a zero of p(x), then
β1=(r13rβ1β)10=(r1ββ13)10
Thus
(r1ββ13)(rΛ1ββ13)=1
so that
(r1β1ββ13)(r1ββ1ββ13)+β―+(r5β1ββ13)(r5ββ1ββ13)=5
Expanding and rearranging, we find
(r1βr1ββ1β+β―+r5βr5ββ1β)β13(r1β1β+r1ββ1β+β―+r5β1β+r5ββ1β)+5β
169=5
Note that 1/r1β,1/r1ββ,β¦,1/r5β,1/r5ββ are the zeros of
x10p(x1β)=x10β130x9+β―
so
r1β1β+r1β1β+β―+r5β1β+r5ββ1β=130
Therefore,
r1βr1ββ1β+β―+r5βr5ββ1β=13β
130β5β
169+5=850β
OR
Let p(x)=x10+(13xβ1)10. If p(r)=0, then
(13βr1β)10=β1=cos180β+isin180β
It follows that
r1β=13β(cosΞΈ+isinΞΈ)
where ΞΈ is an odd multiple of 18β. Hence
rrΛ1β=(13β(cosΞΈ+isinΞΈ))(13β(cosΞΈβisinΞΈ))=170β26cosΞΈ
Letting ΞΈ take on the values 18β,54β,90β,126β,162β, we obtain all of the desired products. Thus
r1βr1ββ1β+β―+r5βr5ββ1β=5β
170β26(cos18β+cos54β+cos90β+cos126β+cos162β)
Applying the identity cosΞΈ+cos(180ββΞΈ)=0, we see the sum is 5β
170=850β.
The problems on this page are the property of the MAA's American Mathematics Competitions