I have the following integral: $$\oint_{\left | z \right | = 1}{\frac{z+\bar{z}}{\left ( 2z-i \right )^{2}}dz}$$ In the official answer, they calculate that using Cauchy's integral formula and the answer is: $$\frac{\pi}{2}i$$ I'm trying to solve this using the Residue theorem but I get different answer.
First, I wanted to find the singularities points, so: $$\frac{z+\bar{z}}{\left ( 2z-i \right )^{2}} = \frac{z}{z}\cdot \frac{z+\bar{z}}{\left ( 2z-i \right )^{2}} = \frac{z^{2}+\left | z \right |^{2}}{\left ( 2z-i \right )^{2}} = \frac{z^{2}+1}{z\left ( 2z-i \right )^{2}}$$ and now, I can see that my points are: $$z_{1}=0$$ $$z_{2}=\frac{i}{2}$$but I'm getting that:$$Res(f,0)=1$$ $$Res(f,\frac{i}{2})=\frac{5}{4}$$ what am I doing wrong?
thank to @WimC, I found My mistake:
$$Res(f,0)=-1$$ and therefore: $$\oint_{\left | z \right | = 1}{\frac{z+\bar{z}}{\left ( 2z-i \right )^{2}}dz} = 2\pi \cdot i\cdot ((-1)+\frac{5}{4})=\frac{\pi \cdot i}{2}$$ as expected.