Complex Sum.

Algebra Level 3

1 a 1 + ω + 1 a 2 + ω + 1 a 3 + ω + ⋯ + 1 a n + ω = − i \dfrac{1}{a_1+\omega} + \dfrac{1}{a_2+\omega} + \frac 1{a_3+\omega} + \cdots + \dfrac{1}{a_n+\omega}=-i

Given that a 1 , a 2 , ⋯ , a n ∈ R a_1,a_2,\cdots,a_n \in \mathbb{R} satisfy the equation above, evaluate ∑ r = 1 n 2 a r − 1 a r 2 − a r + 1 \displaystyle \sum_{r=1}^n \dfrac{2a_r-1}{a_r^2-a_r+1} .

Notations:

  • i = − 1 i = \sqrt{-1} denotes the imaginary unit
  • ω \omega denotes the cube root of unity.
n n 0 0 i i ω \omega 1 1

Vilakshan Gupta by Vilakshan Gupta , 19, India

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2 solutions

Brian Moehring
Mar 12, 2019

First, note that if a a is real, then Real part of 1 a + ω = Real part of a + ω 2 a 2 − a + 1 = a − 1 2 a 2 − a + 1 \text{Real part of } \frac{1}{a+\omega} = \text{ Real part of } \frac{a+\omega^2}{a^2-a+1} = \frac{a-\tfrac{1}{2}}{a^2-a+1}

In particular, if we just consider the real part of the given sum ∑ r = 1 n 1 a r + ω = − i \sum_{r=1}^n \frac{1}{a_r+\omega} = -i then we find that 0 = Real part of − i = Real part of ∑ r = 1 n 1 a r + ω = ∑ r = 1 n ( Real part of 1 a r + ω ) = ∑ r = 1 n a r − 1 2 a r 2 − a r + 1 \begin{aligned} 0 &= \text{ Real part of } -i \\ &= \text{ Real part of } \sum_{r=1}^n \frac{1}{a_r+\omega} \\ &= \sum_{r=1}^n \left(\text{Real part of } \frac{1}{a_r+\omega}\right) \\ &= \sum_{r=1}^n \frac{a_r-\tfrac{1}{2}}{a_r^2-a_r+1} \end{aligned} so now just by multiplying both sides by two, we have ∑ r = 1 n 2 a r − 1 a r 2 − a r + 1 = 0. \sum_{r=1}^n \frac{2a_r-1}{a_r^2-a_r+1} = \boxed{0.}

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Nice solution.

Also read my solution.

Vilakshan Gupta - 2 years, 3 months ago
Vilakshan Gupta
Mar 13, 2019

Note that a 2 − a + 1 = ( a + ω ) ( a + ω 2 ) a^2-a+1=(a+\omega)(a+\omega^2) .

Also note that 2 a − 1 = ( a + ω ) + ( a + ω 2 ) 2a-1=(a+\omega)+(a+\omega^2)

So the given sum can be written as ∑ r = 1 n 2 a r − 1 a r 2 − a r + 1 = ∑ r = 1 n ( a r + ω ) + ( a r + ω 2 ) ( a r + ω ) ( a r + ω 2 ) = ∑ r = 1 n 1 a r + ω + ∑ r = 1 n 1 a + ω 2 = i − i = 0 \begin{aligned} \sum_{r=1}^n \dfrac{2a_r-1}{a_r^2-a_r+1}=\sum_{r=1}^n \dfrac{(a_r+\omega)+(a_r+\omega^2)}{(a_r+\omega)(a_r+\omega^2)} & =\sum_{r=1}^n \dfrac{1}{a_r+\omega} + \sum_{r=1}^n \dfrac{1}{a+\omega^2} & = i - i & =\boxed{0} \end{aligned}

The latter sum can be found by taking conjugates on both sides of the original equation as all a r ∈ R a_r \in \mathbb{R} we can write ∑ r = 1 n 1 a + ω 2 = − i \displaystyle \sum_{r=1}^n \dfrac{1}{a+\omega^2}=-i .

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