$$\begin{eqnarray}
\sin\left(z\right)&=&2\;\cdots\;z\in\mathbb{C}
\end{eqnarray}$$
$$\begin{eqnarray} \sin\left(z\right)&=&\frac{ e^{iz}-e^{-iz} }{2i} \\&=&\frac{ X-X^{-1} }{2i}\;\cdots\;X=e^{iz} \\\frac{ X-X^{-1} }{2i}&=&2 \\X-X^{-1}&=&2\cdot2i \\X-X^{-1}&=&4i \\X-4i-X^{-1}&=&0 \\X\cdot\left(X-4i-X^{-1}\right)&=&X\cdot0 \\X^2-4iX-1&=&0 \\X&=&\frac{-(-4i)\pm\sqrt{(-4i)^2-4\cdot1\cdot(-1)}}{2\cdot1} \\&=&\frac{4i\pm\sqrt{-12}}{2} \\&=&\frac{4i\pm2\sqrt{3}i}{2} \\&=&2i \pm \sqrt{3}i \\X=e^{iz}&=&2i \pm \sqrt{3}i \\&=&\left(2\pm \sqrt{3}\right)i \\iz&=&\log{\left(\left(2\pm \sqrt{3}\right)i\right)}+2\pi i n \;\cdots\; n \in \mathbb{Z} \\z&=&\frac{1}{i}\left\{\log{\left(\left(2\pm \sqrt{3}\right)i\right)}+2\pi i n\right\} \\&=&\frac{i}{i}\frac{1}{i}\log{\left(\left(2\pm \sqrt{3}\right)i\right)}+\frac{1}{\cancel{i}}2\pi \cancel{i} n \\&=&\frac{i}{-1}\log{\left(\left(2\pm \sqrt{3}\right)i\right)}+2\pi n \\&=&-i\log{\left(\left(2\pm \sqrt{3}\right)i\right)}+2\pi n \end{eqnarray}$$
$$\begin{eqnarray} \sin\left(z\right)&=&\frac{ e^{iz}-e^{-iz} }{2i} \\&=&\frac{ X-X^{-1} }{2i}\;\cdots\;X=e^{iz} \\\frac{ X-X^{-1} }{2i}&=&2 \\X-X^{-1}&=&2\cdot2i \\X-X^{-1}&=&4i \\X-4i-X^{-1}&=&0 \\X\cdot\left(X-4i-X^{-1}\right)&=&X\cdot0 \\X^2-4iX-1&=&0 \\X&=&\frac{-(-4i)\pm\sqrt{(-4i)^2-4\cdot1\cdot(-1)}}{2\cdot1} \\&=&\frac{4i\pm\sqrt{-12}}{2} \\&=&\frac{4i\pm2\sqrt{3}i}{2} \\&=&2i \pm \sqrt{3}i \\X=e^{iz}&=&2i \pm \sqrt{3}i \\&=&\left(2\pm \sqrt{3}\right)i \\iz&=&\log{\left(\left(2\pm \sqrt{3}\right)i\right)}+2\pi i n \;\cdots\; n \in \mathbb{Z} \\z&=&\frac{1}{i}\left\{\log{\left(\left(2\pm \sqrt{3}\right)i\right)}+2\pi i n\right\} \\&=&\frac{i}{i}\frac{1}{i}\log{\left(\left(2\pm \sqrt{3}\right)i\right)}+\frac{1}{\cancel{i}}2\pi \cancel{i} n \\&=&\frac{i}{-1}\log{\left(\left(2\pm \sqrt{3}\right)i\right)}+2\pi n \\&=&-i\log{\left(\left(2\pm \sqrt{3}\right)i\right)}+2\pi n \end{eqnarray}$$
$$\begin{eqnarray}
\frac{
e^{iz}-e^{-iz}
}{2i}
&=&
\frac{
e^{iz}-e^{i(-z)}
}{2i}
\\&=&
\frac{
\left(\cos{\left(z\right)}+i\sin{\left(z\right)}\right)
-\left(\cos{\left(-z\right)}+i\sin{\left(-z\right)}\right)
}{2i}
\\&=&
\frac{
\left(\cos{\left(z\right)}+i\sin{\left(z\right)}\right)
-\left(\cos{\left(z\right)}-i\sin{\left(z\right)}\right)
}{2i}
\\&=&
\frac{
\cancel{\cos{\left(z\right)}}+i\sin{\left(z\right)}
\cancel{-\cos{\left(z\right)}}+i\sin{\left(z\right)}
}{2i}
\\&=&
\frac{
\cancel{2i}\sin{\left(z\right)}
}{\cancel{2i}}
\\&=&\sin{\left(z\right)}
\end{eqnarray}$$
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