Verification of Complex Integral $\int \frac{e^{i\pi z}}{2z^2-5z+2}dz$ What is the value of the greatest integer less than or equal to the value of integral 
$$\int_C\frac{e^{i\pi z}}{2z^2-5z+2}dz$$
where $C$ be the curve $\cos t+i\sin t, t\in[0,2\pi]$. 
I think the value is $4$.I used the residue theorem to evaluate the integral. The answer key shows the answer as $2$. The value exactly I obtained was $\frac{4\pi}{3}$. Is there any mistake that I committed? Any hints? Thanks beforehand.
 A: On where the mistake is:
$$\frac{1}{2z^2-5z+2}=\frac{-\frac23}{2z-1}+\frac{\frac13}{z-2}$$
The catch? For the residue at $\frac12$, we'll need to divide numerator and denominator by $2$ to get $\frac{-1/3}{z-\frac12}$. After all, we're comparing to $\frac1{z-r}$; we need $z-\frac12$ instead of $2z-1$ in the denominator.
This is not the full story, of course; the actual problem also has that $e^{i\pi z}$ factor multiplying the residue by $i$. It does look like where the doubling to $\frac{4\pi}{3}$ instead of $\frac{2\pi}{3}$ came from.
A: The value of this integral is equal to$$2\pi i\operatorname{res}_{z=\frac12}\frac{e^{i\pi z}}{2z^2-5z+2}=\frac{2\pi}3\approx2.0944.$$So, yes, the answer is $2$.
A: There is only 1 residue to consider in the unit circle as $2z^2-5z+2 =(2z-1)(z-2)$.
So you get
$$\int \frac{e^{i\pi z}}{2z^2-5z+2}dz = 2\pi i Res_{z=\frac{1}{2}}\frac{e^{i\pi z}}{2z^2-5z+2}$$ $$=2\pi i Res_{z=\frac{1}{2}}\frac{e^{i\pi z}}{2(z-\frac{1}{2})(z-2)} =\pi i\frac{e^{\frac{\pi}{2}i}}{\frac{1}{2}-2} = \frac{2}{3}\pi$$
