# 棣莫弗公式

（重定向自棣莫弗定理

${\displaystyle \left(\cos(x)+i\sin(x)\right)^{n}=\cos(nx)+i\sin(nx)}$

${\displaystyle \operatorname {cis} ^{n}(x)=\operatorname {cis} (nx)}$

## 證明

### 數學歸納法

#### 正整数情形

${\displaystyle P(n)=(\cos \theta +i\sin \theta )^{n}=\cos(n\theta )+i\sin(n\theta ),n\in \mathbb {N} }$

${\displaystyle n=k+1}$

${\displaystyle (\cos \theta +i\sin \theta )^{k+1}}$
${\displaystyle =(\cos \theta +i\sin \theta )^{k}\cdot (\cos \theta +i\sin \theta )}$
${\displaystyle =(\cos k\theta +i\sin k\theta )\cdot (\cos \theta +i\sin \theta )}$
${\displaystyle =(\cos k\theta \cdot \cos \theta +\cos k\theta \cdot i\sin \theta )+(i\sin k\theta \cdot \cos \theta +i\sin k\theta \cdot i\sin \theta )}$
${\displaystyle =[\cos k\theta \cdot \cos \theta -\sin k\theta \cdot \sin \theta ]+i[\cos k\theta \cdot \sin \theta +\sin k\theta \cdot \cos \theta ]}$
${\displaystyle =\cos(k+1)\theta +i\sin(k+1)\theta }$

#### 負整数情形

${\displaystyle (\cos(n\theta )+i\sin(n\theta ))\cdot (\cos(-n\theta )+i\sin(-n\theta ))=1}$ 即可證明。

## 用棣莫弗公式求根

${\displaystyle z=\cos \theta +i\sin \theta }$

${\displaystyle w^{n}=z}$ ，則 ${\displaystyle w}$  也可以表成：

${\displaystyle w=\cos \phi +i\sin \phi }$

${\displaystyle w^{n}=(\cos \phi +i\sin \phi )^{n}=\cos n\phi +i\sin n\phi =\cos \theta +i\sin \theta =z}$

${\displaystyle n\phi =\theta +2k\pi }$ （其中 ${\displaystyle k\in \mathbb {Z} }$

${\displaystyle \phi ={\dfrac {\theta +2k\pi }{n}}}$

${\displaystyle k}$ ${\displaystyle 0,1,\ldots ,n-1}$ ，我們得到 ${\displaystyle n}$  個不同的根：

${\displaystyle w=\cos({\dfrac {\theta +2k\pi }{n}})+i\sin({\dfrac {\theta +2k\pi }{n}}),k=0,1,\ldots ,n-1}$

## 參考文獻

1. ^ Lial, Margaret L.; Hornsby, John; Schneider, David I.; Callie J., Daniels. College Algebra and Trigonometry 4th. Boston: Pearson/Addison Wesley. 2008: 792. ISBN 9780321497444.
2. ^ 林琦焜. 棣美弗定理與 Euler 公式 (PDF). 中央研究院. 2006-12-22 [2017-06-18].