If $Z_{r}, r = 1, 2, 3, . . . . . 2 m, m ϵ N$ are the roots of the equation $Z^{2 m} + Z^{2 m 1} + Z^{2 m 2} + . . . . . . . + Z + 1 = 0$ then find $2 m \sum r = 1 \frac{1}{Z_{r} - 1} =$

-2m

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-m

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-(2m+1)

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-(m+1)

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Solution

The correct option is B -m
Let $z^{2 m} + z^{2 m 1} 1... + z + 1 = (z - z_{1}) (z - z_{2}) . . . . . (z - z_{2 m})$
Taking log on both the sides & differentiating w.r.t. $z$
$\frac{2 m z^{2 m - 1} + (2 m - 1) z^{2 m - 2} + . . . . . + 2 z + 1}{z^{2 m} + z^{2 m - 1} + . . . . . . + z^{2} + z + 1}$ $= \frac{1}{z - z_{1}} + \frac{1}{z - z_{2}} + . . . . . . + \frac{1}{z - z_{2 m}}$

$\Rightarrow \frac{1 + 2 + 3 + . . . . . + 2 m}{(2 m + 1)} (p u t z = 1)$ $= \frac{1}{1 - z_{1}} + \frac{1}{1 - z_{2}} + . . . . . + \frac{1}{1 - z_{2 m}}$

$\Rightarrow 2 m \sum r = 1 \frac{1}{Z_{r} - 1} = - [\frac{2 m (2 m + 1)}{2 (2 m + 1)}] = - m$

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