欧拉公式 （Euler's formula）是复分析 领域的公式，它将三角函数 与复指数函数 关联起来，因其提出者莱昂哈德·欧拉而得名。欧拉公式提出，对任意实数
x
{\displaystyle x}
e
i
x
=
cos
x
+
i
sin
x
{\displaystyle \text{e}^{\text{i}x} = \cos x + \text{i} \sin x}
e
{\displaystyle \text{e}}
自然对数 的底数，
i
{\displaystyle \text{i}}
cos
{\displaystyle \cos}
sin
{\displaystyle \sin}
余弦 、正弦 对应的三角函数，参数
x
{\displaystyle x}
弧度 为单位。这一复数指数函数有时还写作
cis
x
{\displaystyle \operatorname{cis} x}
x
{\displaystyle x}
欧拉公式在数学、物理和工程领域应用广泛。物理学家理查德·费曼将欧拉公式称为：“我们的珍宝”和“数学中最非凡的公式”。
证明 [ ] 首先，在复数域上对
e
z
{\displaystyle \text{e}^z}
a
,
b
∈
R
,
z
=
a
+
i
b
∈
C
{\displaystyle a, b \in \R, z = a + \text{i} b \in \C}
e
z
:=
lim
n
→
∞
(
1
+
z
n
)
n
{\displaystyle \text{e}^z := \lim_{n \to \infty} \left( 1 + \dfrac{z}{n} \right)^n}
复数
w
=
u
+
i
v
=
r
(
cos
θ
+
i
sin
θ
)
{\displaystyle w = u + \text{i} v = r (\cos \theta + \text{i} \sin \theta)}
r
=
u
2
+
v
2
∈
R
,
θ
=
arctan
(
v
u
)
∈
R
.
{\displaystyle r = \sqrt{u^2 + v^2} \in \R, \theta = \arctan \left( \dfrac{v}{u} \right) \in \R.}
棣莫弗公式
w
n
=
(
u
+
i
v
)
n
=
r
n
(
cos
n
θ
+
i
sin
n
θ
)
,
{\displaystyle w^n = (u + \text{i} v)^n = r^n (\cos n \theta + \text{i} \sin n \theta),}
(
1
+
a
+
b
i
n
)
n
=
[
(
1
+
a
n
)
+
i
b
n
]
n
=
r
n
(
cos
θ
n
+
i
sin
θ
n
)
.
{\displaystyle \left( 1 + \dfrac{a + b \text{i}}{n} \right)^n = \left[ \left( 1 + \dfrac{a}{n} \right) + \text{i} \dfrac{b}{n} \right]^n = r_n (\cos \theta_n + \text{i} \sin \theta_n).}
n
>
|
a
|
{\displaystyle n > |a|}
r
n
=
[
(
1
+
a
n
)
2
+
(
b
n
)
2
]
n
2
,
θ
n
=
n
arctan
b
n
1
+
a
n
.
{\displaystyle r_n = \left[ \left( 1 + \dfrac{a}{n} \right)^2 + \left( \dfrac{b}{n} \right)^2 \right]^\frac{n}{2}, \theta_n = n \arctan \dfrac{\frac{b}{n}}{1 + \frac{a}{n}}.}
lim
n
→
∞
ln
r
n
=
lim
n
→
∞
[
n
2
ln
(
1
+
2
a
n
+
a
2
+
b
2
n
2
)
]
=
lim
n
→
∞
[
n
2
(
2
a
n
+
a
2
+
b
2
n
2
)
]
=
a
.
{\displaystyle
\begin{align} \lim_{n \to \infty} \ln r_n
& = \lim_{n \to \infty}\left[ \dfrac{n}{2} \ln \left( 1 + \dfrac{2a}{n} + \dfrac{a^2 + b^2}{n^2} \right) \right] \\
& = \lim_{n \to \infty}\left[ \dfrac{n}{2} \left( \dfrac{2a}{n} + \dfrac{a^2 + b^2}{n^2} \right) \right] \\
& = a.
\end{align}}
lim
n
→
∞
r
n
=
lim
n
→
∞
e
ln
r
n
=
e
a
.
{\displaystyle
\lim_{n \to \infty} r_n = \lim_{n \to \infty} \text{e}^{\ln r_n} = \text{e}^a.}
lim
n
→
∞
θ
n
=
lim
n
→
∞
(
n
arctan
b
n
1
+
a
n
)
=
lim
n
→
∞
(
n
b
n
1
+
a
n
)
=
b
.
{\displaystyle \begin{align}
\lim_{n \rightarrow\infty}\theta_n
& = \lim_{n\to \infty} \left( n \arctan \dfrac{\frac{b}{n}}{1 + \frac{a}{n}} \right) \\
& = \lim_{n\to \infty} \left( n \dfrac{\frac{b}{n}}{1 + \frac{a}{n}} \right) \\
& = b.
\end{align}}
lim
n
→
∞
(
1
+
a
+
b
i
n
)
n
=
e
a
(
cos
b
+
i
sin
b
)
{\displaystyle \lim_{n \to \infty} \left( 1 + \dfrac{a + b\text{i}}{n} \right)^n = \text{e}^a (\cos b + \text{i} \sin b)}
e
a
+
i
b
=
e
a
(
cos
b
+
i
sin
b
)
{\displaystyle \text{e}^{a + \text{i} b} = \text{e}^a (\cos b + \text{i} \sin b)}
a
=
0
{\displaystyle a = 0}
上下节 [ ] 
![{\displaystyle \left(1+{\dfrac {a+b{\text{i}}}{n}}\right)^{n}=\left[\left(1+{\dfrac {a}{n}}\right)+{\text{i}}{\dfrac {b}{n}}\right]^{n}=r_{n}(\cos \theta _{n}+{\text{i}}\sin \theta _{n}).}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/5fb6f4e0130df2a7feaa671ea9003025a4ab1aa9)
![{\displaystyle r_{n}=\left[\left(1+{\dfrac {a}{n}}\right)^{2}+\left({\dfrac {b}{n}}\right)^{2}\right]^{\frac {n}{2}},\theta _{n}=n\arctan {\dfrac {\frac {b}{n}}{1+{\frac {a}{n}}}}.}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/fca5bff9e4e88f1eb2f096eceeb6aaf2de1a8fd8)
![{\displaystyle {\begin{aligned}\lim _{n\to \infty }\ln r_{n}&=\lim _{n\to \infty }\left[{\dfrac {n}{2}}\ln \left(1+{\dfrac {2a}{n}}+{\dfrac {a^{2}+b^{2}}{n^{2}}}\right)\right]\\&=\lim _{n\to \infty }\left[{\dfrac {n}{2}}\left({\dfrac {2a}{n}}+{\dfrac {a^{2}+b^{2}}{n^{2}}}\right)\right]\\&=a.\end{aligned}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/33093d0f6a42b346ba6802cf7f6a8ad4c5445f44)
