Laurent series of $\frac{e^{iz}}{z^2+p^2}$, $ p>0$. I need help finding the main part of the laurent series of $f(z)=\frac{e^{iz}}{z^2+p^2}$ in $ip,-ip$ since these are the two poles of $f$. Due to the orders of the poles are 1 I just have to find $a_{-1}(z-ip)^{-1}$ and $b_{-1}(z+ip)^{-1}$. 
I want to use $2ipf(z)=\frac{e^{(z-ip)i-p}}{z-ip}-\frac{e^{(z-ip)i-p}}{z+ip}$ since this is a given hint. 
Any help appreciated.
 A: $$f(z)=f_1(z)-f_2(z):=\frac{e^{(z-ip)i-p}}{2ip(z-ip)}-\frac{e^{(z-ip)i-p}}{2ip(z+ip)}$$
Since
$$e^{iy}=1+\sum_{k=1}^\infty \frac{i^k y^k}{k!}$$
We have
$$\frac{e^{iy}}{y}=\frac{1}{y}+\sum_{j=0}^\infty \frac{i^{j+1} y^j}{(j+1)!}$$
$$f_1 (z)=\frac{e^{-p}}{2ip}\left(\frac{1}{(z-ip)}+\sum_{j=0}^\infty \frac{i^{j+1}(z-ip)^{j}}{(j+1)!}\right):=\sum_{k=-1}^\infty B_k (z-ip)^k$$
$$B_{-1}=\frac{e^{-p}}{2ip}, B_j=\frac{e^{-p}}{2ip}\frac{i^{j+1}}{(j+1)!}(j\ge 0)$$
$$(z+ip)^{-1}=(z-ip+2ip)^{-1}=(2ip)^{-1}\left(1+\frac{z-ip}{2ip}\right)^{-1}=(2ip)^{-1}\sum_{k=0}^\infty(-1)^k \left(\frac{z-ip}{2ip}\right)^k$$
Therefore:
$$e^{(z-ip)i}(z+ip)^{-1}=\sum_{j=0}^\infty \frac{i^j (z-ip)^j}{j!}\sum_{k=0}^\infty(-1)^k\frac{(z-ip)^k}{(2ip)^k}=\sum_{l=0}^\infty\sum_{k=0}^l \frac{i^{l} }{(2p)^{k}(l-k)!} (z-ip)^l:=\sum_{l=0}^\infty C_l (z-ip)^l$$
And $C_{-1}=0$.
$$f_2(z):=\frac{e^{(z-ip)i-p}}{2ip(z+ip)}=\frac{e^{-p}}{2ip}\frac{e^{(z-ip)i}}{(z+ip)}=\frac{e^{-p}}{2ip}\sum_{l=0}^\infty C_l (z-ip)^l$$.
Thus we obtained the Laurent series for $f(z)$ near $z=ip$.
$$f(z)=\sum_{l=0}^\infty a_l (z-ip)^l, a_l=B_l+B_{-1}C_l$$
$$a_{-1}=B_{-1}=\frac{e^{-p}}{2ip}$$
Similarly we can obtain:
$$f(z)=\sum_{l=0}^\infty b_l (z+ip)^l$$
$$b_{-1}=-\frac{e^{p}}{2ip}$$
