Evaluate $\int_{|z|=5} f(z)\ dz\ ;\ \ f(z)=\frac{z\sin \pi z}{z^2+2z+5}$
Here's my approach (please tell me if I'm correct):
By solving the denominator, simples poles of f(z) come out to be $z_1=-1+2\iota,\ z_2=-1-2\iota$.
Then, using $f(z)=\frac{\phi_1(z)}{z-z_1}=\frac{z\sin \pi z/(z-z_2)}{z-z_1}$ and $f(z)=\frac{\phi_2(z)}{z-z_2}=\frac{z\sin \pi z/(z-z_1)}{z-z_2}$,
$Res_{z=z_1}f(z)=\phi_1(z_1)=\frac{-1+2\iota}{4\iota}\sin(-\pi+2\pi\iota)$ and $Res_{z=z_2}f(z)=\phi_2(z_2)=\frac{1+2\iota}{4\iota}\sin(-\pi-2\pi\iota)$
Then after using Residue theorem and hyperbolic identities,
$\int_{|z|=5} f(z) dz=2\pi\iota(R_1+R_2)$
$=\frac{\pi}{2}[sin(-\pi - 2\pi\iota)-sin(-\pi + 2\pi\iota)+2\iota (sin(-\pi - 2\pi\iota)+sin(-\pi + 2\pi\iota))]$
$=\frac{\pi}{2}(2\cos\pi\ \sin(-2\pi\iota)+0)=\iota\pi \sinh 2\pi$