12 把 E=E_y mcosleftomega t-k x+phirightoverrightarrowa_y+E_x msinleftomega t-k x+phirightoverrightarrowa_x改写成复数形式

将 $E_{ym}\cos(\omega t-kx+\phi)$ 和 $E_{xm}\sin(\omega t-kx+\phi)$ 分别表示为复数形式：

$E_{ym}\cos(\omega t-kx+\phi)=\frac{E_{ym}}{2}\left(e^{i(\omega t-kx+\phi)}+e^{-i(\omega t-kx+\phi)}\right)$

$E_{xm}\sin(\omega t-kx+\phi)=\frac{E_{xm}}{2i}\left(e^{i(\omega t-kx+\phi)}-e^{-i(\omega t-kx+\phi)}\right)$

将两个复数相加：

$E=E_{ym}\cos(\omega t-kx+\phi)\hat{y}+E_{xm}\sin(\omega t-kx+\phi)\hat{x}$

$= \frac{E_{ym}}{2}\left(e^{i(\omega t-kx+\phi)}+e^{-i(\omega t-kx+\phi)}\right)\hat{y} + \frac{E_{xm}}{2i}\left(e^{i(\omega t-kx+\phi)}-e^{-i(\omega t-kx+\phi)}\right)\hat{x}$

$= \frac{E_{ym}}{2}(e^{i\phi}\hat{y}+e^{-i\phi}\hat{y})e^{i(\omega t-kx)} + \frac{E_{xm}}{2i}(e^{i\phi}\hat{x}-e^{-i\phi}\hat{x})e^{i(\omega t-kx)}$

$= \frac{E_{ym}}{2}(2i\sin\phi\hat{y})e^{i(\omega t-kx)} + \frac{E_{xm}}{2i}(2i\cos\phi\hat{x})e^{i(\omega t-kx)}$

$= iE_{ym}\sin\phi\hat{y}e^{i(\omega t-kx)} + iE_{xm}\cos\phi\hat{x}e^{i(\omega t-kx)}$

将 $e^{i(\omega t-kx)}$ 提出来：

$E=e^{i(\omega t-kx)}\left(iE_{ym}\sin\phi\hat{y}+iE_{xm}\cos\phi\hat{x}\right)$

令 $\alpha = \phi + \frac{\pi}{2}$，则可以进一步写成：

$E=e^{i(\omega t-kx)}\left(E_{ym}e^{i\alpha}\hat{y}+E_{xm}e^{-i\alpha}\hat{x}\right)