Contradiction in sin(z) = 2 proof There is something I can't understand in the proof for $\sin(z)=2$.
So, if we know that $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$,
we can end up with the pretty quadratic $(e^{iz})^2-4ie^{iz}-1=0$,
   and if we let $x = e^{iz}$ we get roots $x_{1,2}=i(2\pm\sqrt3)$.  

That's great, as we can solve for $z.$ However, doesn't $e^{iz}=i(2\pm\sqrt3)$ imply that $\sin(z)=2\pm\sqrt3$ ?  

This is confusing.  

EDIT:
If $e^{iz}=i(2\pm\sqrt3)$, then $\cos(z)=0$  and $\sin(z)=2\pm\sqrt3$
If $e^{\imath z} = \imath(2+\sqrt{3}) \implies  e^{-y}\cos(x) = 0$, thus $ x = \frac{\pi}{2} \imath e^{-y}\sin(x) = \imath(2 \pm \sqrt 3 )$ thus $y = -\ln( 2 \pm \sqrt 3 )$
If $e^{iz}=i(2\pm\sqrt{3})$ then 
$$
e^{-iz}=\frac1{e^{iz}}=\frac1{i(2\pm\sqrt{3})}=\frac{i(2\mp\sqrt{3})}{i(2\pm\sqrt{3})i(2\mp\sqrt{3})}=\frac{i(2\mp\sqrt{3})}{-(2^2-\sqrt{3}^2)}=-i(2\mp\sqrt{3})
$$
Now you can deduce
\begin{align*}
\sin(z)=\frac1{2i}\left(e^{iz}-e^{-iz}\right)=\frac1{2i}\left(i(2\pm\sqrt{3})+i(2\mp\sqrt{3})\right)=\frac1{2i}\left(4i\right)=2.
\end{align*}
and
\begin{align*}
\cos(z)=\frac1{2}\left(e^{iz}-e^{-iz}\right)=\frac1{2}\left(i(2\pm\sqrt{3})-i(2\mp\sqrt{3})\right)=\frac1{2}\left(\pm2\sqrt{3}i\right)=\pm\sqrt{3}i.
\end{align*}

If $a,b,c,d\in\mathbb R$ then you can say
$$
a+bi=c+di \Leftrightarrow a-c=(b-d)i
$$
Hence $a-c\in\mathbb R$ while $(d-b)i\in i\mathbb R$ and you can deduce $a=c$ and $b=d$. But this works just if and only if $a,b,c,d\in\mathbb R$. And I think you thought
$$
0+i(2\pm\sqrt{3})=e^{iz}=\cos(z)+\sin(z)i
$$
implies $\cos(z)=0$ and $\sin(z)=2\pm\sqrt{3}$. But this conclusion is just valid if $\cos(z),\sin(z)\in\mathbb R$. As you can see $\cos(z)=\pm\sqrt{3}i\notin\mathbb R$.

Further, you can check
$$
e^{iz}=\cos(z)+\sin(z)i=\pm\sqrt{3}i+2i=(2\pm\sqrt{3})i.
$$