# Quod Erat Demonstrandum

## 2008/01/04

### [AL][PM] Primitive mth root of unity (1 的 m 次本原根)

Filed under: HKALE,Pure Mathematics,Teaching — johnmayhk @ 5:55 下午
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(a) 求 $\lim_{x\to 1}\frac{x^{27} - 1}{x - 1}$

(b) 證明 $\sin\frac{\pi}{27}\sin\frac{2\pi}{27}\sin\frac{3\pi}{27}\times\dots\times \sin\frac{13\pi}{27} = \frac{3\sqrt{3}}{2^{13}}$

(c) A primitive $m^{th}$ root of unity is a root $\alpha$ of the equation $x^m = 1$ such that $\alpha^h \neq 1$ for any factor $h$ of $m$ where $1 \leq h < m$. Expand $\prod_{i=1}^{n}(x - \alpha_{i})$ where $\alpha_1, \alpha_2, \dots \alpha_{n}$ are all primitive $27^{th}$ roots of unity.

$\omega_0 = 1$
$\omega_1 = \omega$ (where $\omega = \cos(\frac{2\pi}{6}) + i\sin(\frac{2\pi}{6})$)
$\omega_2 = \omega^2$
$\omega_3 = \omega^3$
$\omega_4 = \omega^4$
$\omega_5 = \omega^5$

$\omega_1, \omega_5$ 是 1 的 6 次本原根（primitive $6^{th}$ roots of unity）。而 $1, \omega_2, \omega_3, \omega_4$ 就不是 1 的 6 次本原根

$\omega^m = 1$ and
$\omega^h \neq 1$ for any positive integer $h$ less than $m$

$\omega^m = 1$ and
$\omega^h \neq 1$ for any factor $h$ of $m$ ($1 \leq h < m$)

$z^2 = 1$
$z^3 = 1$

$z^3 = 1$
$z^9 = 1$

$x^{27} - 1 = (x^9)^3 - 1^3 = (x^9 - 1)(x^{18} + x^9 + 1)$

$x^{18} + x^9 + 1 = 0$ 的根，所以

$\prod_{i=1}^{18}(x - \alpha_{i}) = x^{18} + x^9 + 1$

SBA 時間：對於正整數 m，試找出 1 的 m 次本原根之數目。（答 $\phi(m)$，其中 $\phi$ 是歐拉函數 Euler’s totient function

## 2 則迴響 »

1. (a)和(c)都能解出，但不知道為甚麼解不出(b), 可以解釋下嗎?

迴響 由 NT — 2010/08/11 @ 4:55 下午 | 回覆

• It is easy to have

$x^{27} - 1 \equiv (x - 1)\displaystyle \prod_{k = 1}^{13}(x^2 - 2x\cos\frac{2k\pi}{27} + 1)$

then

$\frac{x^{27} - 1}{x - 1} = \displaystyle \prod_{k = 1}^{13}(x^2 - 2x\cos\frac{2k\pi}{27} + 1)$ for $x \ne 1$

now taking $x \rightarrow 1$, and by (a), result follows.

迴響 由 johnmayhk — 2010/08/14 @ 1:23 下午 | 回覆