（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。）
令P ( n ) = ( cos θ + i sin θ ) n = cos ( n θ ) + i sin ( n θ ) , n ∈ N {\displaystyle P(n)=(\cos \theta +i\sin \theta )^{n}=\cos(n\theta )+i\sin(n\theta ),n\in \mathbb {N} }
（1）当n = 0 {\displaystyle n=0}
（2）当n = 1 {\displaystyle n=1}
左式 = ( cos θ + i sin θ ) 1 = cos θ + i sin θ = cos ( 1 ⋅ θ ) + i sin ( 1 ⋅ θ ) = {\displaystyle =(\cos \theta +i\sin \theta )^{1}=\cos \theta +i\sin \theta =\cos(1\cdot \theta )+i\sin(1\cdot \theta )=}
因此，P ( 1 ) {\displaystyle P(1)}
（3）当n > 1 {\displaystyle n>1}
假设P ( k ) {\displaystyle P(k)} ( cos θ + i sin θ ) k = cos ( k θ ) + i sin ( k θ ) {\displaystyle (\cos \theta +i\sin \theta )^{k}=\cos(k\theta )+i\sin(k\theta )}
当n = k + 1 {\displaystyle n=k+1}
( cos θ + i sin θ ) k + 1 = ( cos θ + i sin θ ) k ⋅ ( cos θ + i sin θ ) = ( cos k θ + i sin k θ ) ⋅ ( cos θ + i sin θ ) = ( cos k θ ⋅ cos θ + i sin k θ ⋅ i sin θ ) + ( cos k θ ⋅ i sin θ + i sin k θ ⋅ cos θ ) = ( cos k θ ⋅ cos θ − sin k θ ⋅ sin θ ) + i ( cos k θ ⋅ sin θ + sin k θ ⋅ cos θ ) = 1 cos ( k θ + θ ) + i sin ( k θ + θ ) = cos [ ( k + 1 ) θ ] + i sin [ ( k + 1 ) θ ] {\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}}}
等号1处使用和角公式 。
因此，P ( k + 1 ) {\displaystyle P(k+1)}
综上所述，根据数学归纳法，∀ n ∈ N {\displaystyle \forall n\in \mathbb {N} } P ( n ) {\displaystyle P(n)}
另外，由恒等式：
( cos ( n θ ) + i sin ( n θ ) ) ⋅ ( cos ( − n θ ) + i sin ( − n θ ) ) = 1 {\displaystyle (\cos(n\theta )+i\sin(n\theta ))\cdot (\cos(-n\theta )+i\sin(-n\theta ))=1} 可知，公式对于负整数情况也成立。
证毕。
最简单的方法是应用欧拉公式 [2]
由于e i x = cos x + i sin x {\displaystyle e^{ix}=\cos x+i\sin x\,}
所以( cos x + i sin x ) n = ( e i x ) n = e i n x = e i ( n x ) = cos ( n x ) + i sin ( n x ) {\displaystyle {\color {Green}(\cos x+i\sin x)^{n}}=(e^{ix})^{n}=e^{inx}=e^{i(nx)}={\color {Green}\cos(nx)+i\sin(nx)}} 用棣莫弗公式求根
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此定理可用来求单位复数的 n {\displaystyle n} | z | = 1 {\displaystyle |z|=1}
z = cos θ + i sin θ {\displaystyle z=\cos \theta +i\sin \theta } 若 w n = z {\displaystyle w^{n}=z} w {\displaystyle w}
w = cos ϕ + i sin ϕ {\displaystyle w=\cos \phi +i\sin \phi } 按照棣莫弗公式：
w n = ( cos ϕ + i sin ϕ ) n = cos n ϕ + i sin n ϕ = cos θ + i sin θ = z {\displaystyle w^{n}=(\cos \phi +i\sin \phi )^{n}=\cos n\phi +i\sin n\phi =\cos \theta +i\sin \theta =z} 于是得到
n ϕ = θ + 2 k π {\displaystyle n\phi =\theta +2k\pi } k ∈ Z {\displaystyle k\in \mathbb {Z} } 也就是：
ϕ = θ + 2 k π n {\displaystyle \phi ={\dfrac {\theta +2k\pi }{n}}} 当 k {\displaystyle k} 0 , 1 , … , n − 1 {\displaystyle 0,1,\ldots ,n-1} n {\displaystyle n}
w = cos ( θ + 2 k π n ) + i sin ( θ + 2 k π n ) , k = 0 , 1 , … , n − 1 {\displaystyle w=\cos({\dfrac {\theta +2k\pi }{n}})+i\sin({\dfrac {\theta +2k\pi }{n}}),k=0,1,\ldots ,n-1} 参考资料
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^ Lial, Margaret L.; Hornsby, John; Schneider, David I.; Callie J., Daniels. College Algebra and Trigonometry 4th. Boston: Pearson/Addison Wesley. 2008: 792. ISBN 9780321497444 ^ 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）.
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/21413d9f5c297b9f261a27ad0830a9db892c3257)
