if 1, $\alpha_1$, $\alpha_2$, $\alpha_3$, $\ldots$, $\alpha_{n-1}$ are nth roots of unity then
$$\frac{1}{1-\alpha_1} + \frac{1}{1-\alpha_2} + \frac{1}{1-\alpha_3}+\ldots+\frac{1}{1-\alpha_n} = ?$$
Now this has the solution too, but I do not understand the last step of the solution so here is the solution from the book.
$1,\, \alpha_1, \, \alpha_2,\, \ldots, \, \alpha_n$ are the $n^\text{th}$ of unity. These are the roots of $x^n-1=0$
Let $y=\frac{1}{1-\alpha}$ where $\alpha = \alpha_1, \, \alpha_2, \, \alpha_3, \, \ldots, \, \alpha_n$
$$1 – \alpha = \frac{1}{y} \Rightarrow \alpha = \frac{y-1}{y}.$$
But $\alpha$ is a root of $x^n-1=0 \therefore \alpha^n=1 \Rightarrow (y-1)^n = y^n$
$$\Rightarrow y^n – _nC_1y^{n-1} + _nC_2y^{n-2}-\ldots+(-1)^n = y^n \\
\Rightarrow – _nC_1y^{n-1} + _nC_2y^{n-2}-\ldots+(-1)^n = 0.$$
Sum of roots
$$\frac{1}{1-\alpha_1}+\frac{1}{1-\alpha_2}+\ldots+\frac{1}{1-\alpha_{n-1}} = \frac{_nC_2}{_nC_1} = \frac{n-1}{2}$$
So this last part from "Sum of roots" I do not understand. I cannot see how this last shape relates to this binomial theorem notation. Can anyone help?
Best Answer
Consider the polynomial $$P(z)=-{_nC_1} z^{n-1}+{_nC_2}z^{n-2}-\cdots +(-1)^n.\tag{1}\label{1}$$ The answer shows that the numbers $y_i=\frac{1}{1-\alpha_i}$ are all roots of $P$, so they must be all the roots. Thus we must have $$P(z)=-{_nC_1}(z-y_1)\cdots(z-y_{n-1})\tag{2}\label{2}$$ Then comparing the coefficients of $z^{n-2}$ in $P$ from \eqref{1} and \eqref{2}, we find $${_nC_2}={_nC_1}\left(\sum_i y_i\right)$$ and thus $$\sum_i y_i=\frac{_nC_2}{_nC_1}.$$