Find this sum out!

Level 2

Let z i ( i = 0 , 1 , 2 , 3 , 4 ) z_{i}( i= 0,1,2,3,4) denote the roots of the equation ( z + 1 ) 5 + z 5 = 0 (z+1)^{5} + z^{5} = 0 , where z ϵ C z \:\epsilon\: \mathbb{C} . Then find

2 ∑ i = 0 4 ∣ R e ( z i ) ∣ 2\sum_{i=0}^{4}\left | Re(z_{i}) \right |

PS: Don't cheat using wolfram alpha or such sites to get the roots of the equation.


The answer is 5.

Agni Purani by Agni Purani , 18, India

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1 solution

Mark Hennings
Feb 3, 2019

We are solving the equation ( z + 1 z ) 5 = − 1 \left(\frac{z+1}{z}\right)^5 \; = \; -1 and so z j + 1 z j = − e 2 π i j 5 z j = − 1 e 2 π i j 5 + 1 = − 1 e π i j 5 × 2 cos ⁡ π j 5 = − 1 2 cos ⁡ π j 5 e − π i j 5 = − 1 2 + 1 2 i tan ⁡ π j 5 0 ≤ j ≤ 4 \begin{aligned} \frac{z_j+1}{z_j} & = \; -e^{\frac{2\pi ij}{5}} \\ z_j & = \; -\frac{1}{e^{\frac{2\pi ij}{5}} + 1} \; = \; -\frac{1}{e^{\frac{\pi ij}{5}} \times 2\cos\frac{\pi j}{5}} & \\ & = \; -\frac{1}{2\cos\tfrac{\pi j}{5}}e^{-\frac{\pi ij}{5}} \; = \; -\tfrac12 + \tfrac12i\tan\tfrac{\pi j}{5} & 0 \le j \le 4 \end{aligned} Thus R e z j = − 1 2 \mathrm{Re}\,z_j = - \tfrac12 for 0 ≤ j ≤ 4 0 \le j \le 4 , and hence 2 ∑ j = 0 4 ∣ R e z j ∣ = 2 × 5 2 = 5 2\sum_{j=0}^4 \big|\mathrm{Re}\,z_j\big| \; = \; 2 \times \tfrac52 \; = \; \boxed{5}

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