if 1, $\alpha_1$, $\alpha_2$, $\alpha_3$, $\ldots$, $\alpha_{n-1}$ are nth roots of unity then...

Question

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?

Answer 1

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 $$\sum_i y_i=\frac{_nC_2}{_nC_1}.$$

Answer 2

The sum of the roots of $a_nx^{n}+a_{n-1}x^{n-1}+\cdots +a_0$ is $-\frac {a_{n-1}} {a_n}$.

Answer 3

Since you already obtained some clarification on the textbook's solution, I am introducing a different solution. You can also prove by noting that the roots of $x^n-1=0$ are $1,\xi,\xi^2,\ldots,\xi^{n-1}$, where $$\xi=e^{\frac{2\pi i}{n}}.$$ Therefore, we may take $a_k$ to be $\xi^k$ for $k=1,2,\ldots,n-1$. Now, $$\frac{1}{1-a_k}+\frac{1}{1-a_{n-k}}=\frac{1}{1-\xi^k}+\frac{1}{1-\xi^{n-k}}=\frac{1}{1-\xi^k}+\frac{1}{1-\frac{\xi^n}{\xi^k}}.$$ Since $\xi^n=1$, we obtain $$\frac{1}{1-a_k}+\frac{1}{1-a_{n-k}}=\frac{1}{1-\xi^k}+\frac{1}{1-\frac{1}{\xi^k}}=\frac{1}{1-\xi^k}+\frac{\xi^k}{\xi^k-1}=1.$$ Therefore, $$2\sum_{k=1}^{n-1}\frac{1}{1-a_k}=\sum_{k=1}^{n-1}\left(\frac{1}{1-a_k}+\frac{1}{1-a_{n-k}}\right)=\sum_{k=1}^{n-1}1=n-1,$$ so $$\sum_{k=1}^{n-1}\frac{1}{1-a_k}=\frac{n-1}{2}.$$

Answer 4

I know that this problem is tagged alebra-precalculus, but it can be solved quite easily with some knowledge of derivatives as well.

For a polynomial $P(x)$ with roots $r_i$, it follows that:

$$ P(x) = \prod_{cyc} (x-r_i) $$

$$ \ln \ P(x) = \sum_{cyc} \ln(x-r_i) $$

Then differentiate:

$$ \frac{P'(x)}{P(x)} = \sum_{cyc} \frac{1}{x-r_i} $$

Now consider the polynomial $P(x) = \frac{x^n-1}{x-1} = \sum_{r=0}^{n-1} x^r$, which has for its roots the complex $n$th roots of unity, denoted here by $\omega_r$.

From the previous result it follows that:

$$\frac{\sum_{r=0}^{n-1} r\cdot x^{r-1}}{\sum_{r=0}^{n-1} x^r} = \sum_{cyc} \frac{1}{x-\omega_r}$$

Now put $x=1$:

$$\frac{\frac{(n-1)\cdot n}{2}}{n} = \sum_{cyc} \frac{1}{1-\omega_r}$$

$$\sum_{cyc} \frac{1}{1-\omega_r} = \boxed {\frac{n-1}{2}}$$