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