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


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Question Number 21235 by Tinkutara last updated on 17/Sep/17

$For any integer k, let α_{k} = \cos (\frac{k π}{7}) +$ $i \sin (\frac{k π}{7}), where i = \sqrt{- 1} . The value of$ $the expression \frac{\underset{k = 1}{\sum^{12}} ∣ α_{k + 1} - α_{k} ∣}{\underset{k = 1}{\sum^{3}} ∣ α_{4 k - 1} - α_{4 k - 2} ∣} is$
	Answered by sma3l2996 last updated on 17/Sep/17

	$α_{k} = e^{i \frac{k π}{7}}; α_{k + 1} = e^{i (k + 1) \frac{π}{7}}$ $α_{k + 1} - α_{k} = e^{i k \frac{π}{7}} (e^{i \frac{π}{7}} - 1)$ $∣ α_{k + 1} - α_{k} ∣=∣ e^{i k \frac{π}{7}} ∣ . ∣ e^{i \frac{π}{7}} - 1 ∣=∣ e^{i \frac{π}{7}} - 1 ∣$ $α_{4 k - 1} = e^{- i \frac{π}{7}} . e^{i 4 k \frac{π}{7}}; α_{4 k - 2} = e^{- 2 i \frac{π}{7}} . e^{4 i k \frac{π}{7}}$ $α_{4 k - 1} - α_{4 k - 2} = e^{4 i k \frac{π}{7}} (e^{- i \frac{π}{7}} - e^{- 2 i \frac{π}{7}}) = e^{i (4 k - 2) \frac{π}{7}} (e^{i \frac{π}{7}} - 1)$ $s o ∣ α_{4 k - 1} - α_{4 k - 2} ∣=∣ e^{i (4 k - 2) \frac{π}{7}} ∣ . ∣ e^{i \frac{π}{7}} - 1 ∣=∣ e^{i \frac{π}{7}} - 1 ∣$ $s o : \frac{\underset{k = 1}{\sum^{12}} ∣ α_{k + 1} - α_{k} ∣}{\underset{k = 1}{\sum^{3}} ∣ α_{4 k - 1} - α_{4 k - 2} ∣} = \frac{12 ∣ e^{i \frac{π}{7}} - 1 ∣}{3 ∣ e^{i \frac{π}{7}} - 1 ∣} = 4$
	Commented by Tinkutara last updated on 17/Sep/17

	$Thank you very much Sir!$
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