$\int _{C^{+}(0,3)} \frac {dz}{2-\sin z}$ 
$$\int _{C^{+}(0,3)} \frac {dz}{2-\sin z},$$ 
  $z$ is complex. I have no idea how to solve $2-\sin z$.

I will be really grateful for any help
 A: $\displaystyle{\sin z= \sin x+iy= \sin x \cos iy + \cos x \sin iy= \sin x \frac{e^{i*iy}+e^{-i*iy}}{2}}+\cos x \frac{e^{i*iy}-e^{i*iy}}{2i}=$
$\displaystyle \sin x \frac{e^y+e^{-y}}{2}+ \cos x \frac{e^{-y}-e^y}{2i}=\sin x \frac{e^y+e^{-y}}{2}- i\cos x \frac{e^{-y}-e^y}{2}$
I've used formula $\sin (\alpha + \beta) =\sin \alpha \cos \beta + \cos \alpha \sin \beta$, you can check that it's true also for complex $\alpha,\beta$ and formulas $\sin t=\frac{e^{it}-e^{it}}{2i}$ and $\cos t=\frac{e^{it}+e^{-it}}{2}$. Now you can solve $sin z=2$ this way:
$\displaystyle \sin x \frac{e^y+e^{-y}}{2}=2$
$\displaystyle \cos x \frac{e^{-y}-e^y}{2}=0$
From second equation $\displaystyle \cos x=0$ or $\displaystyle \frac{e^{-y}-e^y}{2}=0$. Note that if $\displaystyle \frac{e^{-y}-e^y}{2}=0$ then $y=0$, so $\displaystyle \frac{e^y+e^{-y}}{2}=1$ and by first $\sin x=2$ and it's contradiction because $x \in \mathbb{R}$.
For $\displaystyle \cos x=0$ you have $\sin x=1$ or $\sin x=-1$. In second case $\displaystyle \frac{e^y+e^{-y}}{2}=-2$, it's contradiction because $e^a >0$ for real $a$. In first case you have $\displaystyle \frac{e^y+e^{-y}}{2}=2$, it's equal $e^{2y}+1=e^{y}$, if you substitute $t=e^y$ you have quadratic equation $t^2+1=4t$ and there are two solution $t>0$, see Wolfram. And finally $\cos x=0$ and $\sin x=1$ when $x=2k \pi+\frac{\pi}{2}$, where $k \in \mathbb{Z}$, so $z=2k \pi+\frac{\pi}{2}+i\log{(2+\sqrt{3})}$ or $z=2k \pi+\frac{\pi}{2}+i\log{(2-\sqrt{3})}$
