# 歐拉公式

${\displaystyle e^{ix}=\cos x+i\sin x}$

## 形式

${\displaystyle e^{ix}=\cos x+i\sin x}$

## cis函數

${\displaystyle \operatorname {cis} \theta =\cos \theta +i\sin \theta }$
${\displaystyle \operatorname {cis} \theta =e^{i\theta }}$

${\displaystyle \theta }$值為複數時，cis函數仍然是有效的，所以有些人可利用cis函數將歐拉公式推廣到更複雜的版本。[1]

## 證明

${\displaystyle e^{x}=1+x+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+\cdots }$
${\displaystyle \cos x=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+\cdots }$
${\displaystyle \sin x=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\cdots }$
${\displaystyle x=iz\,}$代入${\displaystyle e^{x}\,}$可得：
{\displaystyle {\begin{aligned}e^{iz}&=1+iz+{\frac {(iz)^{2}}{2!}}+{\frac {(iz)^{3}}{3!}}+{\frac {(iz)^{4}}{4!}}+{\frac {(iz)^{5}}{5!}}+{\frac {(iz)^{6}}{6!}}+{\frac {(iz)^{7}}{7!}}+{\frac {(iz)^{8}}{8!}}+\cdots \\&=1+iz-{\frac {z^{2}}{2!}}-{\frac {iz^{3}}{3!}}+{\frac {z^{4}}{4!}}+{\frac {iz^{5}}{5!}}-{\frac {z^{6}}{6!}}-{\frac {iz^{7}}{7!}}+{\frac {z^{8}}{8!}}+\cdots \\&=\left(1-{\frac {z^{2}}{2!}}+{\frac {z^{4}}{4!}}-{\frac {z^{6}}{6!}}+{\frac {z^{8}}{8!}}-\cdots \right)+i\left(z-{\frac {z^{3}}{3!}}+{\frac {z^{5}}{5!}}-{\frac {z^{7}}{7!}}+\cdots \right)\\&=\cos z+i\sin z\end{aligned}}}

${\displaystyle f(x)\,}$之導數為：

{\displaystyle {\begin{aligned}f'(x)&={\frac {(-\sin x+i\cos x)\cdot e^{ix}-(\cos x+i\sin x)\cdot i\cdot e^{ix}}{(e^{ix})^{2}}}\\&={\frac {-\sin x\cdot e^{ix}-i^{2}\sin x\cdot e^{ix}}{(e^{ix})^{2}}}\\&={\frac {-\sin x\cdot e^{ix}+\sin x\cdot e^{ix}}{(e^{ix})^{2}}}\\&=0\end{aligned}}}

${\displaystyle [a,b]\in I}$${\displaystyle c\in (a,b)}$

${\displaystyle f'(c)={\frac {f(b)-f(a)}{b-a}}.}$拉格朗日中值定理
${\displaystyle \because f'(x)=0}$
${\displaystyle \therefore f'(c)=0}$
${\displaystyle f(a)=f(b)}$

${\displaystyle f(x)=f(0)}$
${\displaystyle {\frac {\cos x+i\sin x}{e^{ix}}}={\frac {\cos 0+i\sin 0}{e^{0}}}=1}$

${\displaystyle e^{ix}=\cos x+i\sin x}$

${\displaystyle {\frac {d}{dx}}e^{ix}=ie^{ix}=iy}$
{\displaystyle {\begin{aligned}{\frac {d}{dx}}(\cos x+i\sin x)&=-\sin x+i\cos x\\&=i(i\sin x+\cos x)\\&=iy\end{aligned}}}
${\displaystyle e^{i0}=e^{0}=1}$
${\displaystyle \cos 0+i\sin 0=1+i(0)=1}$

${\displaystyle x=0}$時，原函數的值相等，所以以上兩個函數相等。

${\displaystyle e^{ix}=\cos x+i\sin x}$

## 證明和角公式

{\displaystyle {\begin{aligned}e^{i\alpha }\times e^{i\beta }&=e^{i\alpha +i\beta }=e^{i(\alpha +\beta )}\\&=\cos(\alpha +\beta )+i\sin(\alpha +\beta )\\&=(\cos \alpha \times \cos \beta +i\sin \alpha \times i\sin \beta )+(i\sin \alpha \times \cos \beta +\cos \alpha \times i\sin \beta )\\&=(\cos \alpha \cos \beta -\sin \alpha \sin \beta )+i(\sin \alpha \cos \beta +\cos \alpha \sin \beta )\\\end{aligned}}}

${\displaystyle \cos(\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta }$
${\displaystyle \sin(\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta }$

## 在複變分析的應用

${\displaystyle z=x+iy=|z|(\cos \phi +i\sin \phi )=|z|e^{i\phi }\,}$
${\displaystyle {\bar {z}}=x-iy=|z|(\cos \phi -i\sin \phi )=|z|e^{-i\phi }\,}$

${\displaystyle x=\mathrm {Re} \{z\}\,}$為實部
${\displaystyle y=\mathrm {Im} \{z\}\,}$為虛部
${\displaystyle |z|={\sqrt {x^{2}+y^{2}}}}$z
${\displaystyle \phi =\mathrm {atan2} {(y,x)}}$，其中${\displaystyle \mathrm {atan2} {(y,x)}={\begin{cases}\arctan \left({\frac {y}{x}}\right)&\qquad x>0\\\pi +\arctan \left({\frac {y}{x}}\right)&\qquad y\geq 0,x<0\\-\pi +\arctan \left({\frac {y}{x}}\right)&\qquad y<0,x<0\\{\frac {\pi }{2}}&\qquad y>0,x=0\\-{\frac {\pi }{2}}&\qquad y<0,x=0\\{\text{undefined}}&\qquad y=0,x=0\end{cases}}}$

## 參考資料

1. ^ Moskowitz, Martin A. A Course in Complex Analysis in One Variable. World Scientific Publishing Co. 2002: 7. ISBN 981-02-4780-X.