First slide

De-moivre's theorem

Question

If $1, ω, . . . . . ω^{n - 1}$ are the nth roots of unity, then value of $\frac{1}{2 - ω} + \frac{1}{2 - ω^{2}} + . . . + \frac{1}{2 - ω^{n - 1}}$ equals

Moderate

A

$\frac{1}{2^{n} - 1}$
B

$\frac{n (2^{n} - 1)}{2^{n} + 1}$
C

$\frac{(n - 2) 2^{n - 1}}{2^{n} - 1}$
D

none of these

Solution

We know that
$\frac{1}{x - 1} + \frac{1}{x - ω} + \frac{1}{x - ω^{2}} + \dots + \frac{1}{x - ω^{n - 1}} = \frac{n (x^{n - 1})}{x^{n} - 1}$
Putting $x = 2,$ we get
$\frac{1}{2 - ω} + \frac{1}{2 - ω^{2}} + \dots + \frac{1}{2 - ω^{n - 1}} = \frac{n (2^{n - 1})}{2^{n} - 1}$

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