Question

If $Z_r$, $r = 1, 2, 3, \dots, 2m$, $m \in N$ are the roots of the equation

$Z^{2m} + Z^{2m-1} + Z^{2m-2} + \dots + Z + 1 = 0$ then prove that $\sum_{i=1}^{2m} \frac{1}{Z_i - 1} = -m$

Answer 1

The statement $\sum_{i=1}^{2m} \frac{1}{Z_i - 1} = -m$ is proven to be true.

Answer 2

The given polynomial $P(Z) = Z^{2m} + Z^{2m-1} + \dots + Z + 1$ can be written as $P(Z) = \frac{Z^{2m+1}-1}{Z-1}$. The roots $Z_i$ are the $(2m+1)$-th roots of unity, excluding $Z=1$. Let $w_i = Z_i - 1$. These $w_i$ are the roots of the polynomial $Q(w) = P(w+1) = 0$. $Q(w) = \frac{(w+1)^{2m+1}-1}{w}$. Expanding $(w+1)^{2m+1}$ and dividing by $w$, we get: $Q(w) = w^{2m} + \binom{2m+1}{2}w^{2m-1} + \dots + \binom{2m+1}{2}w + \binom{2m+1}{1}$. For a polynomial $a_n x^n + \dots + a_1 x + a_0$, the sum of the reciprocals of its roots is $-\frac{a_1}{a_0}$. For $Q(w)$, $a_1 = \binom{2m+1}{2}$ and $a_0 = \binom{2m+1}{1} = 2m+1$. The sum of the reciprocals of the roots $w_i$ is $\sum_{i=1}^{2m} \frac{1}{w_i} = -\frac{\binom{2m+1}{2}}{2m+1} = -\frac{\frac{(2m+1)(2m)}{2}}{2m+1} = -m$. Since $w_i = Z_i - 1$, we have $\sum_{i=1}^{2m} \frac{1}{Z_i - 1} = -m$.