# If $1,\alpha_1,\alpha_2,..,\alpha_{n-1}$ denote the $n^{th}$ roots of unity, then what is $\prod_{k=1}^{n-1}(1-\alpha_k)$

Here's what we did in class :

$$f(z)=z^n-1$$ has all of the $$n$$ $$n^{th}$$ roots of $$1$$ as it's zeroes. Let's say that $$\alpha_1=e^{\iota(2\pi/n)}$$ and $$\alpha_j=(\alpha_1)^j$$ for $$j\in[1,n-1]$$. So $$1,\alpha_1,\alpha_2,...,\alpha_{n-1}$$ are the said $$n$$ roots. This implies that :

$$z^n-1=(z-1)(z-\alpha_1)(z-\alpha_2)...(z-\alpha_{n-1})$$ $$\implies \dfrac{z^n-1}{z-1}=(z-\alpha_1)(z-\alpha_2)...(z-\alpha_{n-1})\overset{\text{def}}{=}P_n(z),\text{where }z\neq1$$

And as $$\dfrac{z^n-1}{z-1}$$ denotes the sum of the first $$n$$ terms of a geometric progression with the first term as $$1$$ and the common ratio as $$z$$, we can say that $$\dfrac{z^n-1}{z-1}=1+z+z^2+...+z^{n-1}$$, again for $$z\neq 1$$

From this, it can be said that :

$$P_n(z)=1+z+z^2+...+z^{n-1},\text{for }z\neq1~~~~~~~~~~~~~~...(1)$$

What our teacher did next was write $$(1-\alpha_1)(1-\alpha_2)...(1-\alpha_{n-1})$$ as $$P_n(1)$$ and equate it to $$1+1+1^2+...+1^{n-1}=n$$, from relation $$(1)$$, even though it forbids $$z$$ being $$1$$, claiming that $$(1)$$ was an identity and was applicable to $$\Bbb C$$.

I'm really confused about this. We derived it with the assumption that $$z\neq1$$ and I don't see why it shouldn't be taken into account.

Thank you!

• And if you do not want to consider limits: your teacher showed that the polynomial $P_n(z):=(z-\alpha_1)\cdots(z-\alpha_{n-1})$ coincides with the other polynomial $Q_n(z):=1+z+z^2+\cdots+z^{n-1}$ on $\mathbb C\setminus\{1\}$. That is more than enough for the two polynomials to be equal. (You would only need $P_n(z)$ and $Q_n(z)$ to coincide at $n$ distinct points.) Commented Nov 19, 2022 at 16:10
• If $P$ and $Q$ are two polynomials of degree at most $n$  and $P(a_i)=Q(a_i)$ for distinct $a_0,\ldots,a_n\in\Bbb C$, then $P\equiv Q$. That comes from the fundamental fact that a polynomial of degree at most $n$ cannot have more than $n$ roots. Commented Nov 19, 2022 at 16:21
• I (and I guess your teacher also) defined $P_n(z):=(z-\alpha_1)\cdots(z-\alpha_{n-1})$ for all $z\in\Bbb C$. But I agree that $P_n(z)=\frac{z^n-1}{z-1}$ for $z\neq1$, of course. Commented Nov 19, 2022 at 16:26
• We don't care how the two polynomials are called. They are polynomials and you have derived that they coincide for all $z\neq1$. That is enough to conclude. Commented Nov 19, 2022 at 16:33
• I wrote why like three times? If you prefer, consider the difference of both sides (so $(z-\alpha_1)\cdots(z-\alpha_{n-1})-(1+z+z^2+\cdots+z^{n-1})$). It is still a polynomial, and it is equal to $0$ at infinitely many points (for all $z\in\Bbb C\setminus\{1\}$), so it is identically zero (thus zero also at $z=1$). Commented Nov 19, 2022 at 16:36

\begin{align}\lim_{z\to 1}\frac {z^n-1}{z-1}&=\lim_{z\to 1}P_n(z)\\ &=P_n(z)\mid_{z=1}\\ &=P_n(1).\end{align}
• Can you do that? $P_n$ is not defined at 1 Commented Nov 19, 2022 at 16:14
• Right, I see how the limit shows that the value of $P_n(1^+)$ and $P_n(1^-)$ is in fact $1$ but that still doesn't change that $P_n(1)$ is undefined, right? Commented Nov 19, 2022 at 16:15
• @nonuser Why can't you define $P_n(z):=(z-\alpha_1)\cdots(z-\alpha_{n-1})$ for $z=1$? It is a polynomial! Commented Nov 19, 2022 at 16:17
• @RajdeepSindhu Think about $\frac xx ≠1$, when $x=0$. But, their limit is equal, at $x=0$, right? Commented Nov 19, 2022 at 16:22
• @lonestudent Yes, but saying $\left.\dfrac xx\right|_{x=0}=1$ is still wrong, right? Commented Nov 19, 2022 at 16:23