I think that 0 is the only fixed point of the sin function in the complex unit disc.
In other words sinz=z. I'm not sure how to prove it.
Maybe you can use Rouche's theorem? It seems possible that
$\displaystyle |(\sin z - z)+z|< |z|$
If this is true, then you can conclude that $\displaystyle \sin z-z$ and $\displaystyle z$ have the same number of roots, which would be 1.
Update: This turns out to be false, for $\displaystyle z=i$.
You could also try to use the argument principle to measure the number of roots in the disk, but the function might be a little nasty to deal with. Putting $\displaystyle f(z)=\sin z-z$,
$\displaystyle \displaystyle \frac{1}{2\pi i}\oint_{\partial D(0,1)} \frac{f'(z)}{f(z)}\text{ d}z=\frac{1}{2\pi i}\oint_{\partial D(0,1)} \frac{\cos z-1}{\sin z-z}\text{ d}z$
gives the number of roots in the disk. If you can evaluate this, then you are done.
Maybe somebody has a cleaner way of doing this.