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Question Number 21235 by Tinkutara last updated on 17/Sep/17
For any integer k, let α_k = cos (((kπ)/7)) + i sin (((kπ)/7)), where i = (√(−1)). The value of the expression ((Σ_(k=1) ^(12) ∣α_(k+1) − α_k ∣)/(Σ_(k=1) ^3 ∣α_(4k−1) − α_(4k−2) ∣)) is
$$\mathrm{For}\:\mathrm{any}\:\mathrm{integer}\:{k},\:\mathrm{let}\:\alpha_{{k}} \:=\:\mathrm{cos}\:\left(\frac{{k}\pi}{\mathrm{7}}\right)\:+ \\ $$$${i}\:\mathrm{sin}\:\left(\frac{{k}\pi}{\mathrm{7}}\right),\:\mathrm{where}\:{i}\:=\:\sqrt{−\mathrm{1}}.\:\mathrm{The}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{expression}\:\frac{\underset{{k}=\mathrm{1}} {\overset{\mathrm{12}} {\sum}}\mid\alpha_{{k}+\mathrm{1}} \:−\:\alpha_{{k}} \mid}{\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\mid\alpha_{\mathrm{4}{k}−\mathrm{1}} \:−\:\alpha_{\mathrm{4}{k}−\mathrm{2}} \mid}\:\mathrm{is} \\ $$
Answered by sma3l2996 last updated on 17/Sep/17
α_k =e^(i((kπ)/7)) ; α_(k+1) =e^(i(k+1)(π/7)) α_(k+1) −α_k =e^(ik(π/7)) (e^(i(π/7)) −1) ∣α_(k+1) −α_k ∣=∣e^(ik(π/7)) ∣.∣e^(i(π/7)) −1∣=∣e^(i(π/7)) −1∣ α_(4k−1) =e^(−i(π/7)) .e^(i4k(π/7)) ; α_(4k−2) =e^(−2i(π/7)) .e^(4ik(π/7)) α_(4k−1) −α_(4k−2) =e^(4ik(π/7)) (e^(−i(π/7)) −e^(−2i(π/7)) )=e^(i(4k−2)(π/7)) (e^(i(π/7)) −1) so ∣α_(4k−1) −α_(4k−2) ∣=∣e^(i(4k−2)(π/7)) ∣.∣e^(i(π/7)) −1∣=∣e^(i(π/7)) −1∣ so : ((Σ_(k=1) ^(12) ∣α_(k+1) −α_k ∣)/(Σ_(k=1) ^3 ∣α_(4k−1) −α_(4k−2) ∣))=((12∣e^(i(π/7)) −1∣)/(3∣e^(i(π/7)) −1∣))=4
$$\alpha_{{k}} ={e}^{{i}\frac{{k}\pi}{\mathrm{7}}} ;\:\alpha_{{k}+\mathrm{1}} ={e}^{{i}\left({k}+\mathrm{1}\right)\frac{\pi}{\mathrm{7}}} \\ $$$$\alpha_{{k}+\mathrm{1}} −\alpha_{{k}} ={e}^{{ik}\frac{\pi}{\mathrm{7}}} \left({e}^{{i}\frac{\pi}{\mathrm{7}}} −\mathrm{1}\right) \\ $$$$\mid\alpha_{{k}+\mathrm{1}} −\alpha_{{k}} \mid=\mid{e}^{{ik}\frac{\pi}{\mathrm{7}}} \mid.\mid{e}^{{i}\frac{\pi}{\mathrm{7}}} −\mathrm{1}\mid=\mid{e}^{{i}\frac{\pi}{\mathrm{7}}} −\mathrm{1}\mid \\ $$$$\alpha_{\mathrm{4}{k}−\mathrm{1}} ={e}^{−{i}\frac{\pi}{\mathrm{7}}} .{e}^{{i}\mathrm{4}{k}\frac{\pi}{\mathrm{7}}} \:\:\:;\:\:\alpha_{\mathrm{4}{k}−\mathrm{2}} ={e}^{−\mathrm{2}{i}\frac{\pi}{\mathrm{7}}} .{e}^{\mathrm{4}{ik}\frac{\pi}{\mathrm{7}}} \\ $$$$\alpha_{\mathrm{4}{k}−\mathrm{1}} −\alpha_{\mathrm{4}{k}−\mathrm{2}} ={e}^{\mathrm{4}{ik}\frac{\pi}{\mathrm{7}}} \left({e}^{−{i}\frac{\pi}{\mathrm{7}}} −{e}^{−\mathrm{2}{i}\frac{\pi}{\mathrm{7}}} \right)={e}^{{i}\left(\mathrm{4}{k}−\mathrm{2}\right)\frac{\pi}{\mathrm{7}}} \left({e}^{{i}\frac{\pi}{\mathrm{7}}} −\mathrm{1}\right) \\ $$$${so}\:\:\mid\alpha_{\mathrm{4}{k}−\mathrm{1}} −\alpha_{\mathrm{4}{k}−\mathrm{2}} \mid=\mid{e}^{{i}\left(\mathrm{4}{k}−\mathrm{2}\right)\frac{\pi}{\mathrm{7}}} \mid.\mid{e}^{{i}\frac{\pi}{\mathrm{7}}} −\mathrm{1}\mid=\mid{e}^{{i}\frac{\pi}{\mathrm{7}}} −\mathrm{1}\mid \\ $$$${so}\:\::\:\frac{\underset{{k}=\mathrm{1}} {\overset{\mathrm{12}} {\sum}}\mid\alpha_{{k}+\mathrm{1}} −\alpha_{{k}} \mid}{\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\mid\alpha_{\mathrm{4}{k}−\mathrm{1}} −\alpha_{\mathrm{4}{k}−\mathrm{2}} \mid}=\frac{\mathrm{12}\mid{e}^{{i}\frac{\pi}{\mathrm{7}}} −\mathrm{1}\mid}{\mathrm{3}\mid{e}^{{i}\frac{\pi}{\mathrm{7}}} −\mathrm{1}\mid}=\mathrm{4} \\ $$
Commented by Tinkutara last updated on 17/Sep/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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