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

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?

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

• my knowledge is clearly lacking here. Let me start with a very basic example to dumb it down for me: $x^2+2x -15 = (x+5)(x-3)$ -5 and 3 are the roots of this polynomial. Is there a general theorem that you can tell me to apply so I can get from your first $P(z)$ to your second $P(z)$? How did you get the prefix coefficient of $_nC_1$? Is this because it's the highest common factor of all the terms? – Bucephalus Nov 7 '18 at 10:26
• No, it's just the coefficient of the highest degree term. For example if you have $2x^2+4x-30$, the roots are $3$ and $-5$ so the polynomial is $2(x+5)(x-3)$. – Arnaud D. Nov 7 '18 at 10:27
• oh yeah so the other ones are understood to be included in there. I see, let me look at this more. – Bucephalus Nov 7 '18 at 10:28
• I cannot get from your second equation to your third equation. I'm reading here about the Fundamental Theorem of Algebra and The Factor Theorem, and I think what you're doing here is something to do with their interaction but I cannot see it. – Bucephalus Nov 7 '18 at 10:47
• I see this rule: if P(x) is a polynomial of degree n and r is a zero of P(x), then P(x) can be written as $P(x) = (x-r)Q(x)$ where Q(x) is a polynomial of degreen n-1. I think you're using this, but I still cannot see it. @ArnaudD – Bucephalus Nov 7 '18 at 10:49

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 $$\sum_{k=1}^{n-1}\frac{1}{1-a_k}=\frac{n-1}{2}.$$

• this is interesting. I will look at this when I get home from work. Thankyou – Bucephalus Nov 7 '18 at 21:56

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

• Thankyou too @KaviRamaMuthy – Bucephalus Nov 7 '18 at 12:04