I want to integrate $\displaystyle \int_\gamma \frac{z}{(z-1)(z-2)}dz$ along the curve $\gamma(t)=2+\frac{1}{2}e^{it}$, with $t\in [0,2\pi]$
Can this be done by mean of residues? The curve is not closed, so I think it might be a problem. The fact that exercise was in a chapter calleed "Cauchy integral formula and residues" suggest that this theorems should be used to solve the integral
On
The curve $\gamma$ described parametrically by $z=2+\frac12e^{it}$, $t\in [0,2\pi]$ is a circle centered at $z=2$ with radius $1/2$. That curve encloses the pole of $f(z)=\frac{z}{(z-1)(z-2)}$ at $z=2$, but not at $z=1$.
From the residue theorem, we have
$$\begin{align} \oint_\gamma f(z)\,dz&=\int_{|z-2|=\frac12} \frac{z}{(z-1)(z-2)}\,dz\\\\ &=2\pi i\text{Res}\left(\frac{z}{(z-1)(z-2)}, z=2\right)\\\\ &=2\pi i \lim_{z\to 2}(z-2)\frac{z}{(z-1)(z-2)}\\\\ &=4\pi i \end{align}$$
The curve is closed. What made you think it is not?
You can compute is using Cauchy's integral formula:\begin{align}\int_\gamma\frac z{(z-1)(z-2)}\,\mathrm dz&=\int_\gamma\frac{\frac z{z-1}}{z-2}\,\mathrm dz\\&=2\pi i\frac2{2-1}\\&=4\pi i.\end{align}