I’m trying to calculate
$$ I = \oint \frac{2+ 3\mathrm{sin}(\pi z)}{z(z-1)^2} $$ on a closed contour of radius $3$ centered on $O(0,0)$.
I’ve written this equation substituting $\mathrm{sin}(\pi z)$ by $ \frac{e^{i\pi z} - e^{-i \pi z}}{2i}$.
The equation becomes :
$$ I = 1/2i \oint \frac{4i + 3 e^{i \pi z} -3 e^{-i \pi z} }{z(z-1)^2} $$
Using Residue theorem, I got the sum of the residues $-6i \pi$, dividing by $2i$, it will equal $-3\pi$
Edit: and multiplying it by $2 \pi i$ of the residue formula I get $I =-6 i \pi ^2$.
The thing is that the professor gave us the answer which is : $ I =-6i\pi$. What’s my fault?
Hints: By substituting $\sin(\pi z)$ the integration becomes $$ I = \frac 1 {2i} \oint \frac{4i + 3 e^{i \pi \color{red}z} -3 e^{-i \pi \color{red}z} }{z(z-1)^2}dz. $$ The mistake is in red color.
Simply, $$Res(z=0)=\lim_{z\to0}\frac{2+3\sin{\pi z}}{(z-1)^2}=2$$ and $$Res(z=1)=\lim_{z\to1}\frac{d}{dz}\Big(\frac{2+3\sin{\pi z}}{z}\Big)=-3\pi-2.$$ Thus $$I=2\pi i (2-3\pi-2)=-6\pi^2 i.$$ So your answer is correct.