Let $\phi\colon[0,1]\longrightarrow\mathbb C$ be the curve $\phi(t)=\exp(2\pi it)$ where $0\leqslant t\leqslant1$. Find, giving justification, the values of contour integral $$\int \frac{dz}{4z^2-1}$$ over $\phi$.
I have resolved the function into partial fractions but I am unable to decide how to take the limits.
There are no limits to take here. By the residue theorem, your integral is equal to$$2\pi i\left(\operatorname{res}_{z=\frac12}\frac1{4z^2-1}+\operatorname{res}_{z=-\frac12}\frac1{4z^2-1}\right).$$Besides,$$\operatorname{res}_{z=\frac12}\frac1{4z^2-1}=\frac14\text{ and }\operatorname{res}_{z=-\frac12}\frac1{4z^2-1}=-\frac14.$$So, your integral is equal to $0$.