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# System prepended metadata

title: 複數的n次方根

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### 複數的n次方根
複數$z^n = r (\cos\theta + i\sin\theta)\;, r > 0,\;0\leq\theta \leq 2\pi$ 解為
$$
z_{k} = \sqrt[n]{r}\;(\cos\frac{\theta + 2k\pi}{n} + i\sin\frac{\theta +2k\pi}{n}),\; k= 0,1,2,\dots,n-1
$$

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$Proof:$

$$
|z^n| = |r (\cos\theta + i\sin\theta)| \Rightarrow \; |z|^n = r,\;|z| = \sqrt[n]{ r }
$$
$$
令z = \sqrt[n]{ r }(\cos \phi + i\sin \phi)
$$
$$
又z^n = r (\sin\theta + i\sin\theta)\;
$$
$$
\therefore [\sqrt[n]{ r }(\cos \phi + i\sin \phi)]^n = r(\cos\theta +i\sin\theta)
$$

由 de Moivre's Theorem: 
$$\Rightarrow [\sqrt[n]{ r }(\cos \phi + i\sin \phi)]^n =r(\cos n\phi + i\sin n\phi) = r(\cos\theta +i\sin\theta)
$$
根據複數相等的意義，有
$$
n\phi = \theta + 2k\pi\;\Rightarrow \phi = \frac{\theta+ 2k\pi}{n}
$$
所以
$$
z = \sqrt[n]{r}\;(\cos\frac{\theta + 2k\pi}{n} + i\sin\frac{\theta +2k\pi}{n}),\; k= 0,1,2,\dots,n-1
$$

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### 1的n次方根
$z^n = 1$ 的解為
$$
z_{k} = \text{cos} \frac{2k\pi}{n} + i\text{ sin} \frac{2k\pi}{n}, k= 0,1,2,\dots,n-1
$$
當代入$k= 0,1,2,\dots,n-1$
$$
z_{0}=\text{cos}\text{0} + i\text{ sin}\text{0}  = 1 
$$
$$
z_{1} = \omega = \text{cos}\frac{2\pi}{n} + i\text{sin}\frac{2\pi}{n}
$$
$$
z_{2} = \omega^2 = \text{cos} \frac{4\pi}{n} + i\text{sin} \frac{4\pi}{n}$$
$$
\vdots
$$
$$
z_{n-1} = \omega^{n-1} =\cos\frac{{2}{(n-1)}\pi}{n} + i\sin \frac{{2}{(n-1)}\pi}{n}
$$

#### 性質
1.  $\omega^n =1$
2. $\omega^{n-1}+\omega^{n-2}+\dots+\omega+1 = 0$
3. $$(\alpha - \omega)(\alpha - \omega^2) \dots (\alpha - \omega^{n-1}) = \alpha^{n-1} + \alpha^{n-2} + \dots + \alpha + 1 = \frac{1(1-\alpha^n)}{1-\alpha}$$
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#### $Proof:$
1. 
$$
z_{n} = \omega^n = \text{cos}{2\pi} + i\text{ sin}{2\pi} = 1
$$
2. $x^n -1$ 有下列因式分解

$$
\left\{
    \begin{matrix}
        x^n -1 &=& (x-1)(x-\omega)(x-\omega^2)\dots(x-\omega^{n-1})\\
         &=& (x-1)(x^{n-1}+x^{n-2}+\dots+x+1)
    \end{matrix}
\right.
$$
所以
$$
(x-1)(x-\omega)(x-\omega^2)\dots(x-\omega^{n-1}) = (x-1)(x^{n-1}+x^{n-2}+\dots+x+1)
$$
$$
(x-\omega)(x-\omega^2)\dots(x-\omega^{n-1}) = (x^{n-1}+x^{n-2}+\dots+x+1)
$$
將$\omega代入x$，可得到
$$
\omega^{n-1}+\omega^{n-2}+\dots+\omega+1 = 0
$$
3. 將$\alpha代入x$，可得到
$$(\alpha - \omega)(\alpha - \omega^2) \dots (\alpha - \omega^{n-1}) = \alpha^{n-1} + \alpha^{n-2} + \dots + \alpha + 1 = \frac{1(1-\alpha^n)}{1-\alpha}$$