First slide

De-moivre's theorem

Question

If $ω = \cos \frac{π}{n} + i \sin \frac{π}{n},$ then value of $1 + ω + ω^{2} + . . . . + ω^{n - 1}$ is

Moderate

A

$1 + i c o t (\frac{π}{2 n})$
B

$1 + i \tan (\frac{π}{n})$
C

$1 + i$
D

none of these

Solution

We have
$1 + ω + ω^{2} + \dots + ω^{n - 1} = \frac{1 - ω^{n}}{1 - ω} = \frac{2}{1 - ω}$
as $ω^{n} = \cos (\frac{n π}{n}) + i \sin (\frac{n π}{n}) = \cos π + i \sin π = - 1$
Now,
$\begin{array}{l} \frac{2}{1 - ω} = \frac{2 (1 - \bar{ω})}{(1 - ω) (1 - \bar{ω})} = \frac{2 (1 - \bar{ω})}{1 - (ω + \bar{ω}) + ω \bar{ω}} \\ = \frac{2 (1 - \bar{ω})}{2 - 2 Re (ω)} [∵ ω \bar{ω} = | ω |^{2} = 1] \\ = \frac{1 - Re (ω) + i Im (ω)}{1 - Re (ω)} = 1 + i \frac{Im (ω)}{1 - Re (ω)} \\ = 1 + i \frac{\sin (\frac{π}{n})}{1 - \cos (\frac{π}{n})} = 1 + i \frac{2 \sin (\frac{π}{2 n}) \cos (\frac{π}{2 n})}{2 \sin^{2} (\frac{π}{2 n})} \\ = 1 + i \cot (π / 2 n) \end{array}$

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