Proof that $\frac{1}{2}(c_{n}-d_{n})\pi=1$ if $n$ is odd, for $f(z)=\csc(z)$ and $\{c_{n}\}$ and $\{d_{n}\}$ Laurent coefficients of $f$ Let $f(z)=\csc(z)$ and $\{c_{n}\}$ and $\{d_{n}\}$ Laurent coefficients of $f$ in $\{z\in \mathbb{C}:0<|z|<1\}$ and $\{z\in \mathbb{C}:1<|z|<2\}$ respectively. Proof that $\frac{1}{2}(c_{n}-d_{n})\pi=1$ if $n$ is odd. 
In addition, you find a similar expression when $n$ is even.
Hi! I have problems to solve this problem so my idea was obtner comprehensive performing Laurent coefficients, but these came out very hard and I could not find a difference or similarity between even and odd coefficients. Is needed for integration or is there an easier way?
 A: The function must be $f(z) = \csc (\pi z)$, actually. $\csc z$ has no singularities in $0 < \lvert z\rvert < \pi$, so the Laurent expansions of that in the two annuli $0 < \lvert z\rvert < 1$ and $1 < \lvert z\rvert < 2$ are identical (and are actually the Laurent expansion in $0 < \lvert z\rvert < \pi$).
Remember that the coefficients of the Laurent expansion of a function $g$ holomorphic on the annulus $r < \lvert z\rvert < R$ are given by
$$a_n = \frac{1}{2\pi i}\int_{\lvert z\rvert = \rho} \frac{g(z)}{z^{n+1}}\,dz$$
for an arbitrary $\rho \in (r,R)$.
So to compute $d_n - c_n$ here, we pick two arbitrary radii $0 < \rho_1 < 1 < \rho_2 < 2$ and compute
$$\begin{align}
d_n - c_n &= \frac{1}{2\pi i}\int_{\lvert z\rvert = \rho_2} \frac{\csc (\pi z)}{z^{n+1}}\,dz - \frac{1}{2\pi i}\int_{\lvert z\rvert = \rho_1} \frac{\csc (\pi z)}{z^{n+1}}\,dz\\
&= \frac{1}{2\pi i} \int_{\partial A} \frac{\csc (\pi z)}{z^{n+1}}\,dz,\tag{1}
\end{align}$$
where $A$ is the annulus $\rho_1 < \lvert z\rvert < \rho_2$.
By the residue theorem, $(1)$ is the sum of the residues of $\dfrac{\csc(\pi z)}{z^{n+1}}$ in $A$. The denominator $z^{n+1}$ is holomorphic and nonzero in $A$, so the residues come only from the poles of $\csc (\pi z)$ in $A$. $\csc (\pi z)$ has poles in $k \in \mathbb{Z}$ and is holomorphic everywhere else. Of these poles, the two poles in $1$ and $-1$ lie in $A$, thus
$$d_n - c_n = \operatorname{Res}\left(\frac{1}{z^{n+1}\sin (\pi z)}; 1\right) + \operatorname{Res}\left(\frac{1}{z^{n+1}\sin (\pi z)}; -1\right).$$
If $h$ has a simple zero in $\alpha$, and $g$ is holomorphic and nonzero in a neighbourhood of $\alpha$, the residue of $\dfrac{g}{h}$ in $\alpha$ is $\dfrac{g(\alpha)}{h'(\alpha)}$.
Now, the zeros of $\sin (\pi z)$ are simple, and $z^{-(n+1)}$ is holomorphic and nonzero except at the origin, hence
$$\operatorname{Res}\left(\frac{1}{z^{n+1}\sin (\pi z)}; k\right) = \frac{1}{k^{n+1}\pi\cos (\pi k)} = \frac{(-1)^k}{\pi k^{n+1}}$$
for $ \neq 0$. Inserting $k = \pm 1$ yields
$$\begin{align}
d_n - c_n &= \frac{-1}{\pi 1^{n+1}} + \frac{-1}{\pi(-1)^{n+1}} = \frac{-(1+(-1)^{n+1})}{\pi}\\
&= \begin{cases} -\frac{2}{\pi} &, n \text{ odd}\\ \quad 0 &, n \text{ even}. \end{cases}\tag{2}
\end{align}$$
Rearranging $(2)$ easily yields
$$\frac12(c_n-d_n)\pi = 1 \text{ for } n \equiv 1 \pmod{2}; \qquad \frac12(c_n-d_n)\pi = 0 \text{ for } n \equiv 0 \pmod{2}.$$
Note: That the even-indexed coefficients are the same for both annuli fell out of the general computation with no extra effort. It can, however, be seen entirely without computation: $\csc (\pi z)$ is an odd function, hence all even-indexed coefficients in any Laurent expansion with centre $0$ are always $0$.
