欧拉公式

$e^{i\pi }+1=0$$e^{i\pi}+1=0$

推导

已知：

$e^z=\underset{n\to \mathrm{\infty }}{lim}(1+\frac{z}{n}{)}^n$$e^z=\lim\limits_{n \to \infty} (1+\frac{z}{n})^n$

推导：

令$z=a+bi$$z=a+bi$

$e^{a+bi}=\underset{n\to \mathrm{\infty }}{lim}(1+\frac{a+bi}{n}{)}^n$$e^{a+bi}=\lim\limits_{n \to \infty}(1+\frac{a+bi}{n})^n$

$e^{a+bi}=\underset{n\to \mathrm{\infty }}{lim}[(1+\frac{a}{n})+\frac{b}{n}i{]}^n$$e^{a+bi}=\lim\limits_{n \to \infty}[(1+\frac{a}{n})+\frac{b}{n}i]^n$

$\underset{n\to \mathrm{\infty }}{lim}[(1+\frac{a}{n})+\frac{b}{n}i{]}^n$$\lim\limits_{n \to \infty}[(1+\frac{a}{n})+\frac{b}{n} i]^n$

$=\underset{n\to \mathrm{\infty }}{lim}{r}^n[cos(n\theta )+isin(n\theta )]$$=\lim\limits_{n \to \infty}r^n[cos(n \theta)+i sin(n \theta)]$ (棣莫弗公式）

则有$r=\underset{n\to \mathrm{\infty }}{lim}\sqrt{(1+\frac{a}{n}{)}^2+(\frac{b}{n}{)}^2}$$r=\lim\limits_{n \to \infty}\sqrt{(1+\frac{a}{n})^2+(\frac{b}{n})^2}$

$tan\theta =\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{b}{n}}{1+\frac{a}{n}}$$tan\theta=\lim\limits_{n \to \infty}\cfrac{\cfrac{b}{n}}{1+\cfrac{a}{n}}$

$\theta =arctan\frac{\frac{b}{n}}{1+\frac{a}{n}}$$\theta=arctan\cfrac{\cfrac{b}{n}}{1+\cfrac{a}{n}}$

$ln{r}^n=\underset{n\to \mathrm{\infty }}{lim}\frac{n}{2}ln(1+\frac{a^2+b^2}{n^2}+\frac{2a}{n})$$lnr^n=\lim\limits_{n \to \infty}\frac{n}{2}ln(1+\frac{a^2+b^2}{n^2}+\frac{2 a}{n})$

又有

$x\to 0,ln(x+1)\to x,x\to arctanx$${x \to 0},{ln (x+1) \to x},{x \to arctanx}$

则有

$ln{r}^n=\underset{n\to \mathrm{\infty }}{lim}(a+\frac{a^2+b^2}{2n})$$lnr^n=\lim\limits_{n \to \infty}(a+\frac{a^2+b^2}{2 n})$

$r^n=e^a$$r^n=e^a$

$n\theta =\underset{n\to \mathrm{\infty }}{lim}\frac{b}{1+\frac{a}{n}}$$n\theta=\lim\limits_{n \to \infty}\cfrac{b}{1+\cfrac{a}{n}}$

$n\theta =b$$n\theta=b$

带入原式，即得：

$e^{a+bi}=e^a(cosb+isinb)$$e^{a+bi}=e^a(cosb+i sinb)$

将$b$$b$换为$\theta$$\theta$

$e^{i\theta }=cos\theta +isin\theta$$e^{i\theta}=cos\theta+i sin\theta$

当$\theta =\pi$$\theta=\pi$时

$e^{i\pi }+1=0.$$e^{i\pi}+1=0.$

棣莫弗公式：

$z^n=r^n[cos(n\theta )+isin(n\theta )]$$z^n=r^n[cos(n\theta)+i sin(n\theta)]$

简证：

$z=r(cos\theta +isin\theta )$$z=r(cos\theta+isin\theta)$

$z^2=r^2(cos\theta +isin\theta )(cos\theta +isin\theta )$$z^2=r^2(cos\theta+isin\theta)(cos\theta+isin\theta)$

$=r^2(co{s}^2\theta -si{n}^2\theta +2isin\theta cos\theta )$$=r^2(cos^2\theta-sin^2\theta+2isin\theta cos\theta)$

$=r^2(cos2\theta +isin2\theta )$$=r^2(cos2\theta+isin2\theta)$

若$z^n=r^n[cos(n\theta )+isin(n\theta )]$$z^n=r^n[cos(n\theta)+i sin(n\theta)]$

则$z^{n+1}=r^{n+1}[cos(n\theta )+isin(n\theta )](cos\theta +isin\theta )$$z^{n+1}=r^{n+1}[cos(n\theta)+i sin(n\theta)](cos\theta+i sin\theta)$

$=r^{n+1}[cos(n\theta )cos\theta -sin(n\theta )sin\theta +icos(n\theta )sin\theta +isin(n\theta )cos\theta ]$$=r^{n+1}[cos(n\theta)cos\theta-sin(n\theta)sin\theta+i cos(n\theta)sin\theta+i sin(n\theta)cos\theta]$

$=r^{n+1}[cos(n+1)\theta +isin(n+1)\theta ]$$=r^{n+1}[cos(n+1)\theta+isin(n+1)\theta]$

故$z^n=r^n[cos(n\theta )+isin(n\theta )]$$z^n=r^n[cos(n\theta)+i sin(n\theta)]$成立.