2
$\begingroup$

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!

$\endgroup$
15
  • 1
    $\begingroup$ 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.) $\endgroup$
    – nejimban
    Commented Nov 19, 2022 at 16:10
  • 1
    $\begingroup$ 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. $\endgroup$
    – nejimban
    Commented Nov 19, 2022 at 16:21
  • 1
    $\begingroup$ 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. $\endgroup$
    – nejimban
    Commented Nov 19, 2022 at 16:26
  • 1
    $\begingroup$ 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. $\endgroup$
    – nejimban
    Commented Nov 19, 2022 at 16:33
  • 1
    $\begingroup$ 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 $z=1$). $\endgroup$
    – nejimban
    Commented Nov 19, 2022 at 16:36

1 Answer 1

Reset to default
4
$\begingroup$

Hint:

Observe that,

$$\begin{align}\lim_{z\to 1}\frac {z^n-1}{z-1}&=\lim_{z\to 1}P_n(z)\\ &=P_n(z)\mid_{z=1}\\ &=P_n(1).\end{align}$$

$\endgroup$
15
  • $\begingroup$ Can you do that? $P_n$ is not defined at 1 $\endgroup$
    – nonuser
    Commented Nov 19, 2022 at 16:14
  • $\begingroup$ 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? $\endgroup$
    – Rajdeep Sindhu
    Commented Nov 19, 2022 at 16:15
  • 1
    $\begingroup$ @nonuser Why can't you define $P_n(z):=(z-\alpha_1)\cdots(z-\alpha_{n-1})$ for $z=1$? It is a polynomial! $\endgroup$
    – nejimban
    Commented Nov 19, 2022 at 16:17
  • $\begingroup$ @RajdeepSindhu Think about $\frac xx ≠1$, when $x=0$. But, their limit is equal, at $x=0$, right? $\endgroup$
    – lone student
    Commented Nov 19, 2022 at 16:22
  • $\begingroup$ @lonestudent Yes, but saying $\left.\dfrac xx\right|_{x=0}=1$ is still wrong, right? $\endgroup$
    – Rajdeep Sindhu
    Commented Nov 19, 2022 at 16:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .