Complex integral with $1-\sin z$ as denominator I want to evaluate $$\int_{|z|=8}\dfrac{1+z}{1-\sin z}dz$$ using the residue theorem. But I'm not sure what the residues are. In $|z|=8$, for $z$ real, we have $\sin z=1$ for $z=-3\pi/2,\pi/2,5\pi/2$. But what about $z$ complex? I only know how to write $\sin$ as a power series, but then it's hard to determine when it equals $1$.
 A: The equation $\sin z=1$ only has real solutions, i.e.
$$
z_n=\frac\pi2+2n\pi,\quad n \mathbb{Z}.
$$
Since 
$$
(1-\sin z)'|_{z_n}=-\cos z_n=0,
$$
it follows that each $z_n$ is a pole of order $2$ for the function $z \mapsto f(z)=(1+z)/(1-\sin z)$.
For every $n$ we have
\begin{eqnarray}
\text{Res}(f,z_n)&=&\lim_{z\to z_n}\frac{d}{dz}\left[(z-z_n)^2f(z)\right]=\lim_{\xi\to 0}\frac{d}{d\xi}\left[\xi^2f(z_n+\xi)\right]\\
&=&\lim_{\xi\to 0}\frac{d}{d\xi}\left[\frac{(1+z_n+\xi)\xi^2}{1-\cos\xi}\right]=\lim_{\xi\to 0}\frac{d}{d\xi}\left(\frac{1+z_n+\xi}{\frac12-\frac{\xi^2}{4!}+o(\xi^3)}\right)\\
&=&\frac12\lim_{\xi\to 0}\frac{1-\frac{\xi^2}{48}+o(\xi^3)+(1+z_n+\xi)(\frac{\xi}{24}+o(\xi^2))}{(1-\frac{\xi^2}{48})^2}=\frac12.
\end{eqnarray}
Since the only singularities inside the open disk $|z|<8$ are $z_{-1},z_0,z_1$, we get
$$
\int_{|z|=8}f(z)\,dz=2i\pi\sum_{n=-1}^1\text{Res}(f,z_n)=3\pi i.
$$


To see that the equation
  $$
\sin z=1
$$
  only has real solutions, set $u=e^{iz}$, and since
  $$
\sin z=\frac{e^{iz}-e^{-iz}}{2i}=\frac{u-u^{-1}}{2i}=\frac{u^2-1}{2iu},
$$
  we have
  $$
\sin z=1 \iff \frac{u^2-1}{2iu}=1 \iff u^2-2iu-1=0 \iff (u-i)^2=0 \iff u-i=0,
$$
  i.e. $u=e^{iz}=i=e^{i(\frac\pi2+2n\pi)}$, and so 
  $$
z \in \left\{\frac\pi2+2n\pi: \ n \in \mathbb{N}\right\}.
$$

A: $$\sin (x+iy) = \frac{1}{2i}\left(e^{i\pi (x+iy)}-e^{-i\pi (x+iy)}\right)$$
The absolute value of $e^{i\pi(x+iy)}$ is $e^{-y\pi}$, and of $e^{-i\pi(x+iy)}$ is $e^{\pi y}$ so they have different absolute values if $y\ne 0$.  So $\sin (x+iy)$ can only be zero if $y=0$.
