JEE-Advanced 2015 (21/40)

Algebra Level 3

∑ k = 1 12 ∣ α k + 1 − α k ∣ ∑ k = 1 3 ∣ α 4 k − 1 − α 4 k − 2 ∣ \dfrac{\displaystyle \sum_{k=1}^{12} |\alpha_{k+1}-\alpha_k|}{\displaystyle \sum_{k=1}^3 |\alpha_{4k-1}-\alpha_{4k-2}|}

For integer k k , let α k = cos ⁡ ( k π 7 ) + i sin ⁡ ( k π 7 ) \alpha_k=\cos \left( \frac{k \pi}{7} \right)+i \sin \left( \frac{k \pi}{7} \right) , where i = − 1 i=\sqrt{-1} . Evaluate the value of the expression above.


The answer is 4.

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

Prakhar Bindal
Nov 26, 2015

Using geometrical interpretation of complex number kills this problem completely.

consider the 14th roots of unity. all of them will lie on the unit circle in argand plane centred at origin .

now we know that modulus of (a-b) signifies distance between complex numbers a and b in argand plane .

also we know that the roots of unity constitute the vertices of a regular polygon . both the numerator and

denominator in the problem represent the length of sides of that regular polygon . hence all the terms in

numerator and denominator are equal . so the sum evaluated to 12/3 = 4

Q.E.D :)

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