Just another geometric series

Algebra Level pending

For w = e i ( π / n ) w=e^{i(\pi/n)} , if the value of 1 + w + w 2 + … + w n − 1 1+w+w^2+\ldots +w^{n-1} equals to 1 + i cot ⁡ ( π k n ) , 1+i \cot \left( \frac\pi{kn} \right), find k k .


The answer is 2.

Sanchayan Dutta by Sanchayan Dutta , 23, India

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1 solution

Tom Engelsman
May 10, 2021

Taking the geometric series:

Σ k = 1 n w k − 1 = Σ k = 1 n ( e i π / n ) k − 1 = 1 − ( e i π / n ⋅ n ) 1 − e i π / n = 1 − ( − 1 ) 1 − e i π / n = 2 1 − e i π / n \Large \Sigma_{k=1}^{n} w^{k-1} = \Sigma_{k=1}^{n} (e^{i\pi/n})^{k-1} = \frac{1-(e^{i\pi/n \cdot n})}{1-e^{i\pi/n}} = \frac{1-(-1)}{1-e^{i\pi/n}} = \frac{2}{1-e^{i\pi/n}} ;

or 2 [ 1 − cos ⁡ ( π / n ) ] − i sin ⁡ ( π / n ) ⋅ [ 1 − cos ⁡ ( π / n ) ] + i sin ⁡ ( π / n ) [ 1 − cos ⁡ ( π / n ) ] + i sin ⁡ ( π / n ) = 2 [ [ 1 − cos ⁡ ( π / n ) ] + i sin ⁡ ( π / n ) ] [ 1 − cos ⁡ ( π / n ) ] 2 + sin ⁡ 2 ( π / n ) ; \Large \frac{2}{[1-\cos(\pi/n)] - i \sin(\pi/n)} \cdot \frac{[1-\cos(\pi/n)] + i \sin(\pi/n)}{[1-\cos(\pi/n)] + i \sin(\pi/n)} = \frac{2[[1-\cos(\pi/n)] + i \sin(\pi/n)]}{[1-\cos(\pi/n)]^2 + \sin^{2}(\pi/n)};

or 2 [ [ 1 − cos ⁡ ( π / n ) ] + i sin ⁡ ( π / n ) ] 1 − 2 cos ⁡ ( π / n ) + cos ⁡ 2 ( π / n ) + sin ⁡ 2 ( π / n ) \Large \frac{2[[1-\cos(\pi/n)] + i \sin(\pi/n)]}{1-2\cos(\pi/n) + \cos^{2}(\pi/n) + \sin^{2}(\pi/n)} ;

or [ 1 − cos ⁡ ( π / n ) ] + i sin ⁡ ( π / n ) 1 − cos ⁡ ( π / n ) ; \Large \frac{[1-\cos(\pi/n)] + i \sin(\pi/n)}{1-\cos(\pi/n) };

or 1 + i sin ⁡ ( π / n ) 1 − cos ⁡ ( π / n ) ; \large 1 + i \frac{\sin(\pi/n)}{1-\cos(\pi/n)};

or 1 + i cot ⁡ ( π 2 n ) . \large 1 + i \cot(\frac{\pi}{2n}).

Hence, k = 2 . \boxed{k=2}.

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