Calculation of $\int_{\partial B(\frac 1 2, \frac 3 4)} \frac{dz}{z(z^2-1)}$
So far, it is shown that $$ \int_{\partial B(\frac 1 2, \frac 3 4)} \frac{dz}{z(z^2-1)}= -\int_{\partial B(\frac 1 2, \frac 3 4)} \frac{dz}{z}+\frac{1}{2}\int_{\partial B(\frac 1 2, \frac 3 4)} \frac{dz}{z-1}-\frac{1}{2}\int_{\partial B(\frac 1 2, \frac 3 4)} \frac{dz}{z+1} $$ Next, using the fact that $$ \int_{\partial B(a,r)}\frac{dz}{z-z_0}=\left\{ \begin{array}{cc} 2\pi i & \text{if} & z_0\in B(a,r), \\ 0 & \text{if} & z_0\not\in \overline{B}(a,r), \\ \end{array} \right. $$ we obtain that $$ \int_{\partial B(\frac 1 2, \frac 3 4)} \frac{dz}{z(z^2-1)}= -\pi i. $$