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JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Complex Numbers

For any integer

k

, let

α_{k} = cos (\frac{k π}{7}) + i sin (\frac{k π}{7})

, where

i = - 1

. The value of the expression

\frac{\sum _{k = 1}^{12} ∣ α _{k + 1} - α _{k} ∣}{\sum _{k = 1}^{3} ∣ α _{4 k - 1} - α _{4 k - 2} ∣}

is

Answer: 4

Solution

Recognize that

α_{k} = e^{i \frac{2 k π}{14}}

in exponential form. The numerator becomes

\sum_{k = 1}^{12} e^{i \frac{2 ( k + 1 ) π}{14}} - e^{i \frac{2 k π}{14}} = \sum_{k = 1}^{12} e^{i \frac{2 π}{14}} - 1 = 12 e^{i \frac{π}{7}} - 1

. Similarly, the denominator is

\sum_{k = 1}^{3} e^{i \frac{2 π}{14}} - 1 = 3 e^{i \frac{π}{7}} - 1

. Therefore, the ratio is

\frac{12}{3} = 4

.

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Integer answer

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