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Question

From the equation

$x^{21} - 1 = 0 $

Define 20 complex roots as $\omega_{1},\omega_{2}, \omega_{3},...,\omega_{20}$

What is the value of $(1-\omega_{1})(1-\omega_{2})(1-\omega_{3})...(1-\omega_{20})$?

We can rewrite the equation as

$x^{21}- 1 = (x-1)(x-\omega_{1})(x-\omega_{2})...(x-\omega_{20})$

Divide both sides by $x-1$

$ \dfrac{x^{21}- 1}{x- 1} =(x-\omega_{1})(x-\omega_{2})...(x-\omega_{20})$

On the other hand, we can also do the long division:

$x^{21}-1 = (x-1)(x^{20}+x^{19}+x^{18}+...+x+1)$

So substitute this:

$\dfrac{(x-1)(x^{20}+x^{19}+x^{18}+...+x+1)}{x- 1} =(x-\omega_{1})(x-\omega_{2})...(x-\omega_{20})$

$x^{20}+x^{19}+x^{18}+...+x+1 =(x-\omega_{1})(x-\omega_{2})...(x-\omega_{20})$

Plug in 1, we have

$(1-\omega_{1})(1-\omega_{2})(1-\omega_{3})...(1-\omega_{20}) = 21$.

But I'm not too sure if this process is entirely justified. Because ultimately the original equation is:

$ \dfrac{x^{21}- 1}{x- 1} =(x-\omega_{1})(x-\omega_{2})...(x-\omega_{20})$

And the LHS is NOT defined when $x = 1$, or rather take one more step backward we have:

$x^{21}- 1 = (x-1)(x-\omega_{1})(x-\omega_{2})...(x-\omega_{20})$

And if we were to divide both sides by $(x-1)$ we can only do this when $ x \neq 1$, because otherwise the process is undefined.

How can I resolve this?

VladeKR
  • 433

2 Answers2

1

Hint:

You may consider $$\lim_{x\to1}\frac{x^{21}-1}{x-1},$$ which exists.

user
  • 26,272
  • forgive me but isn't the limit of a function different from the value of a function. Why can we invoke a limit in situations like this? – VladeKR Dec 19 '20 at 21:15
  • @VladeKR The limit of a function can be of course equal to the value of the function. Consider for example two functions $f(x),g(x)$ equal on the whole domain $x\in\mathbb R$ except for the point $x_0$. Then obviously $\lim_{x\to x_0}[f(x)-g(x)]=0\implies \lim_{x\to x_0}f(x)=\lim_{x\to x_0}g(x)$. – user Dec 20 '20 at 12:57
  • I understand the limit of a function can be equal to the value of the function, but in our example, the value of the function is undefined. – VladeKR Dec 20 '20 at 15:40
  • @VladeKR Therefore I recommend you to use the limit instead of the undefined value of the function. – user Dec 20 '20 at 19:33
1

The roots of $f(z):=z^{21}-1$ form the cyclic $\Bbb C^*$ subgroup $\langle\omega\rangle$ of size $21$ where $\omega=\cos(\frac{2\pi}{21})+i\sin(\frac{2\pi}{21})$. $$f(z)=z^{21}-1=\prod_{j=1}^{21}(z-\omega^j)$$ $$f'(z)=21z^{20}=\cdots$$ $$(\text{Product Rule})\;\;\cdots=\prod_{k\neq 21}(z-\omega^k)\cdot\frac{d}{dz}(z-1)+(z-1)\frac{d}{dz}\prod_{k\neq 21}(z-\omega^k)$$ $$\therefore\;\; f'(1)=\prod_{k\neq 21}(1-\omega^k)=21$$

Oliver Kayende
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