Simple Residue calculation

Question

$$\int_{\gamma(0;2)}\frac{e^{i\pi z/2}}{z^2-1} \, dz$$ Using the residue calculus i got $$-2\pi$$But the answer is $$=i$$ I must be wrong at this, but shouldn't the answer have $\pi$ at least since the integral already requires $2\pi i \cdot\text{residue }$?

Answer 1

I agree with Gid Gut's commentary : we have $$\frac{e^{i\frac{\pi}{2}z}}{z^{2}-1}=\frac{e^{i\frac{\pi}{2}z}}{\left(z+1\right)\left(z-1\right)} $$ thus there are two simple poles at $z=\pm1 $. Then by the residue theorem, we get $$\oint_{\gamma\left(0,2\right)}\frac{e^{i\frac{\pi}{2}z}}{z^{2}-1}dz=2\pi i\left(\frac{e^{i\frac{\pi}{2}}}{2}-\frac{e^{-i\frac{\pi}{2}}}{2}\right)=2\pi i\left(i\sin\frac{\pi}{2}\right)=-2\pi $$