I am supposed to do the integral $$ \int_{\gamma_2} \frac{\sin(z)}{z+\frac{i}{2}} dz$$ where $\gamma_2:[-\pi, 3\pi] \rightarrow \mathbb{C}$ , $\gamma_2(t)=\exp(it)$ for $ t\in [-\pi,\pi]$, $\gamma_2(t)=(1+t-\pi)\exp(it)$ for $t\in [\pi,2\pi)$ and $\gamma_2(t)=(1+3\pi-t) \exp(it)$ for $t\in[2\pi,3\pi]$.
My idea was to say that this is equal to $2 \cdot 2\pi i \sin(-\frac{i}{2})$. Since we have two loops and the rest is cauchy's integral formula, is this correct?
Yes, indeed, Lipschitz: Your conclusions are correct. Nice work.