What is the value of the following contour integral? The contour is a circle with radius $0.5$ around $z=i$ point: $|z-i|<\frac{1}{2}$
$$\oint_C\frac{dz}{2-\sin z}$$
I think it is $0$ because the fuction doesn't have any singularities in $|z-i|<\frac{1}{2}$. It was a question on a multichoice exam yesterday and I chose $0$, but I suspect it may have some tricky point in it.
You have to show that $\sin z$ is never equal to $2$ inside the contour. So suppose $\sin(x+iy) = \sin x \cosh y + i\cos x \sinh y$ is equal to $2$. Then we must have
$$\sin x \cosh y = 2$$ $$\cos x \sinh y = 0$$
For the second equation to hold, we must have $\cos x = 0$ or $\sinh y = 0$. But $|x| \le \frac12$ inside the contour, so $\cos x$ must be non-zero. And $y \ge \frac12$ inside the contour, so $\sinh y$ must be non-zero.
Hence $\sin z - 2$ is non-zero inside the contour, as you say.