Let \[\omega\] be a nonreal root of \[z^3 = 1.\] Let \[a_1, a_2, \dots, a_n\] be real numbers such that \[\frac{1}{a_1 + \omega} + \frac{1}{a_2 + \omega} + \dots + \frac{1}{a_n + \omega} = 2 + 5i.\]Compute \[\frac{2a_1 - 1}{a_1^2 - a_1 + 1} + \frac{2a_2 - 1}{a_2^2 - a_2 + 1} + \dots + \frac{2a_n - 1}{a_n^2 - a_n + 1}.\]

LeoIsTheBest Jun 12, 2019

#7**+1 **

\(\displaystyle \omega =-\frac{1}{2} \pm i\frac{\sqrt{3}}{2}.\)

Consider a general term on the lhs,

\(\displaystyle \frac{1}{a_{k}+\omega}=\frac{1}{(a_{k}-1/2) \pm i \sqrt{3}/2}=\frac{1}{(a_{k}-1/2) \pm i \sqrt{3}/2}.\frac{(a_{k}-1/2) \mp i \sqrt{3}/2}{(a_{k}-1/2) \pm i \sqrt{3}/2}\)

\(\displaystyle = \frac{(a_{k}-1/2) \mp i \sqrt{3}/2}{(a_{k}-1/2)^{2}+3/4}=\frac{(a_{k}-1/2) \mp i \sqrt{3}/2}{a_{k}^{2}-a_{k}+1}.\)

Substitute and equate reals,

\(\displaystyle \frac{a_{1}-1/2}{a_{1}^{2}-a_{1}+1}+\frac{a_{2}-1/2}{a_{2}^{2}-a_{2}+1}+ \dots + \frac{a_{n}-1/2}{a_{n}^{2}-a_{n}+1}=2,\)

and multiplying that by 2,

\(\displaystyle \frac{2a_{1}-1}{a_{1}^{2}-a_{1}+1}+\frac{2a_{2}-1}{a_{2}^{2}-a_{2}+1}+ \dots + \frac{2a_{n}-1}{a_{n}^{2}-a_{n}+1}=4.\)

.Guest Jun 13, 2019

edited by
Guest
Jun 13, 2019

#1

#3**-4 **

maybe this helps:

\(Let \omega be a nonreal root of z^3 = 1. Let a_1, a_2, \dots, a_n be real numbers such that \[\frac{1}{a_1 + \omega} + \frac{1}{a_2 + \omega} + \dots + \frac{1}{a_n + \omega} = 2 + 5i.\]Compute \[\frac{2a_1 - 1}{a_1^2 - a_1 + 1} + \frac{2a_2 - 1}{a_2^2 - a_2 + 1} + \dots + \frac{2a_n - 1}{a_n^2 - a_n + 1}.\]\)

ProffesorNobody
Jun 13, 2019

#7**+1 **

Best Answer

\(\displaystyle \omega =-\frac{1}{2} \pm i\frac{\sqrt{3}}{2}.\)

Consider a general term on the lhs,

\(\displaystyle \frac{1}{a_{k}+\omega}=\frac{1}{(a_{k}-1/2) \pm i \sqrt{3}/2}=\frac{1}{(a_{k}-1/2) \pm i \sqrt{3}/2}.\frac{(a_{k}-1/2) \mp i \sqrt{3}/2}{(a_{k}-1/2) \pm i \sqrt{3}/2}\)

\(\displaystyle = \frac{(a_{k}-1/2) \mp i \sqrt{3}/2}{(a_{k}-1/2)^{2}+3/4}=\frac{(a_{k}-1/2) \mp i \sqrt{3}/2}{a_{k}^{2}-a_{k}+1}.\)

Substitute and equate reals,

\(\displaystyle \frac{a_{1}-1/2}{a_{1}^{2}-a_{1}+1}+\frac{a_{2}-1/2}{a_{2}^{2}-a_{2}+1}+ \dots + \frac{a_{n}-1/2}{a_{n}^{2}-a_{n}+1}=2,\)

and multiplying that by 2,

\(\displaystyle \frac{2a_{1}-1}{a_{1}^{2}-a_{1}+1}+\frac{2a_{2}-1}{a_{2}^{2}-a_{2}+1}+ \dots + \frac{2a_{n}-1}{a_{n}^{2}-a_{n}+1}=4.\)

Guest Jun 13, 2019

edited by
Guest
Jun 13, 2019