棣莫弗公式

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

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

證明

（证明的思路是用数学归纳法证明正整数的情形，并推广到负整数。）

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

（1）当${\displaystyle n=0}$ 时，显然成立。

（2）當${\displaystyle n=1}$ 时：

（3）當${\displaystyle n>1}$ 时：

${\displaystyle n=k+1}$ 时：

{\displaystyle {\begin{aligned}(\cos \theta +i\sin \theta )^{k+1}&=(\cos \theta +i\sin \theta )^{k}\cdot (\cos \theta +i\sin \theta )\\&=(\cos k\theta +i\sin k\theta )\cdot (\cos \theta +i\sin \theta )\\&=(\cos k\theta \cdot \cos \theta +i\sin k\theta \cdot i\sin \theta )+(\cos k\theta \cdot i\sin \theta +i\sin k\theta \cdot \cos \theta )\\&=(\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 )\\&\ {\overset {1}{=}}\cos(k\theta +\theta )+i\sin(k\theta +\theta )\\&\ =\cos[(k+1)\theta ]+i\sin[(k+1)\theta ]\\\end{aligned}}}

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

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：

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

參考資料

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]. （原始内容存档 (PDF)于2021-01-19）.