+0

# Complex Numbers

+2
644
7
+141

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

#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,

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

Jun 13, 2019
edited by Guest  Jun 13, 2019

#1
+1

Jun 13, 2019
#2
+23
-4

nope

ProffesorNobody  Jun 13, 2019
#3
+23
-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
+23
-4

actually it doesn't help...

ProffesorNobody  Jun 13, 2019
#5
+23
-4

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

Jun 13, 2019
#6
+141
0

I fixed it now.

Jun 13, 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