For-any-integer-k-let-k-cos-kpi-7-i-sin-kpi-7-where-i-1-The-value-of-the-expression-k-1-12-k-1-k-k-1-3-4k-1-4k-2-is-

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!