How to find residues of $\frac{1+z}{1-\sin z}$

Question

I have to compute an integral: $$\int_C \frac{1+z}{1-\sin z}$$ where $C$ is the circle of radius $8$. I would do this way: the function is holomorphic in $\mathbb{C}-\{\frac{\pi}{2}+k\pi\}$, $C$ is homologous to $0$ in $\mathbb{C}$ and the singularities don't intercept the curve so I can use the residue theorem, so: $$\int_C \frac{1+z}{1-\sin z}=2\pi i( Res_{\frac{\pi}{2}}(f)+Res_{\frac{\pi}{2}+2\pi}(f)+Res_{\frac{\pi}{2}-2\pi}(f))$$ where $$\frac{\pi}{2},\frac{\pi}{2}+2\pi,\frac{\pi}{2}-2\pi$$ are the only singularities inside the circle of radius $8$. Now my problem is how to compute the residues: I did this: $$\frac{1+z}{1-\sin z}=\frac{1+z}{1-\cos(z-\frac{\pi}{2})}\frac{(z-\frac{\pi}{2})^2}{(z-\frac{\pi}{2})^2}=[\frac{1+z}{1-\cos(z-\frac{\pi}{2})}(z-\frac{\pi}{2})^2]\frac{1}{(z-\frac{\pi}{2})^2}$$ now,say $g$ the function in the square brackets. $g$ has a removable singularity at $\frac{\pi}{2}$ and hence the residues of $f$ in $\frac{\pi}{2}$ is $g'(\frac{\pi}{2})$. The problem is that $g'$ is not defined in $\frac{\pi}{2}$. Is there a more elegant way to proceed?

Answer 1

Let's try the residue at $z = k\pi/2$ where $k \equiv 1 \mod 4$. If $w = z - k \pi/2$, near $w=0$ we have $$\eqalign{\sin(z) &= \cos(w) = 1 - \frac{1}{2} w^2 + O(w^4)\cr \dfrac{1}{1 - \sin(z)} &= \frac{2}{w^2} + O(w^0)\cr \dfrac{1+z}{1-\sin(z)} &= (1 + k\pi/2 + w)\left(\frac{2}{w^2}+O(w^0)\right)\cr &= \frac{(2 + k \pi)}{w^2} + \frac{2}{w} + O(w^0) } $$ So the residue is the coefficient of $w^{-1}$, namely $2$.