We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+2
139
7
avatar+140 

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 \[\frac{2a_1 - 1}{a_1^2 - a_1 + 1} + \frac{2a_2 - 1}{a_2^2 - a_2 + 1} + \dots + \frac{2a_n - 1}{a_n^2 - a_n + 1}.\]

 Jun 12, 2019
edited by LeoIsTheBest  Jun 13, 2019
edited by LeoIsTheBest  Jun 13, 2019

Best Answer 

 #7
avatar
+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,

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

.
 Jun 13, 2019
edited by Guest  Jun 13, 2019
 #1
avatar
+1

Can anybody read this question?

 Jun 13, 2019
 #2
avatar+25 
-4

nope

ProffesorNobody  Jun 13, 2019
 #3
avatar+25 
-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
 #4
avatar+25 
-4

actually it doesn't help...

ProffesorNobody  Jun 13, 2019
 #5
avatar+25 
-4

ok wow i can read it now..... but im confused now

 Jun 13, 2019
 #6
avatar+140 
0

I fixed it now.

 Jun 13, 2019
 #7
avatar
+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

39 Online Users