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$$e^{i\theta }=\cos \theta +\sin \theta i,i=\sqrt{-1}$$
证明：
$$\begin{gathered} f\left(\theta \right)=\frac{e^{i\theta }}{\cos \theta +\sin \theta i} \\ {f}^{\prime }\left(\theta \right)=\frac{i{e}^{i\theta }\left(\cos \theta +\sin \theta i\right)-e^{i\theta }\left(-\sin \theta +\cos \theta i\right)}{{\left(\cos \theta +\sin \theta i\right)}^2}=0 \\ \Rightarrow f\left(\theta \right)=\mathrm{cons}\tan \mathrm{t} \\ f\left(\theta \right)=f\left(0\right)=\frac{e^{i0}}{\cos 0+\sin 0i}=1\Rightarrow \frac{e^{i\theta }}{\cos \theta +\sin \theta i}=1 \\ \Rightarrow e^{i\theta }=\cos \theta +\sin \theta i \end{gathered}$$

求解：$\sin x=2$
令：$\sin z=2=c,z\in \mathbb{C}$
$\{\begin{array}{l}{e}^{iz}=\cos z+\sin zi\\ e^{i\left(-z\right)}=\cos z-\sin zi\end{array}\Rightarrow e^{iz}-e^{-iz}=2\sin zi$
$\therefore \sin z=\frac{e^{iz}-e^{-iz}}{2i}=c\Rightarrow \frac{e^{ai-b}-e^{b-ai}}{2i}=\frac{e^{ai}{e}^{-b}-e^b{e}^{-ai}}{2i}=c$
且有：$\{\begin{array}{l}{e}^{ia}=\cos a+\sin ai\\ e^{i\left(-a\right)}=\cos a-\sin ai\end{array}$
$$\begin{gathered} \Rightarrow \frac{e^{-b}\left(\cos a+\sin ai\right)-e^b\left(\cos a-\sin ai\right)}{2i}=\frac{\left(e^{-b}-e^b\right)\cos a-\left(e^{-b}+e^b\right)\sin ai}{2i}=c \\ \Rightarrow \frac{1}{2}\left(e^b-e^{-b}\right)\cos ai+\frac{1}{2}\left(e^{-b}+e^b\right)\sin a=c=c+0i \end{gathered}$$
$\Rightarrow \{\begin{array}{l}\frac{1}{2}\left(e^{-b}+e^b\right)\sin a=c\\ \frac{1}{2}\left(e^b-e^{-b}\right)\cos a=0\end{array}$

• 当 $b=0$ 时，$\sin a=c$ 不成立（所设$a,b\in \mathbb{R}$）
• 当 $\cos a=0$ 时，$\frac{1}{2}\left(e^{-b}+e^b\right)=\pm c\Rightarrow 1+e^{2b}\pm 2c{e}^b=0$
  设$u=e^b>0$ ，则有：$u=\pm c\pm \sqrt{c^2-1}$
  $\therefore b=\ln \left(c\pm \sqrt{c^2-1}\right)$
  $\Rightarrow z=\frac{\pi }{2}+2k\pi +\ln \left(c\pm \sqrt{c^2-1}\right)i=\frac{\pi }{2}+2k\pi +\ln \left(2\pm \sqrt{3}\right)i$