Can someone help me: Assuming that none of the points $0$, $1$ and $-1$ lies on $L$, calculate all possible values of the integral $$ \mathop{\int}\limits_{L}{\frac{dz}{{z}{\mathrm{(}}{z}^{2}\mathrm{{-}}{1}{\mathrm{)}}}}\text{,} $$
for various positions of $ L $ .
We have $z(z^2-1) = z(z+1)(z-1)$.
Calculating the residues, we have:
\begin{align} \text{Res}\left(\frac{1}{z(z+1)(z-1)},0\right)&=\lim\limits_{z\rightarrow0}\frac{1}{(z+1)(z-1)}=-1\\ \text{Res}\left(\frac{1}{z(z+1)(z-1)},1\right)&=\lim\limits_{z\rightarrow1}\frac{1}{z(z+1)}=\frac{1}{2}\\ \text{Res}\left(\frac{1}{z(z+1)(z-1)},-1\right)&=\lim\limits_{z\rightarrow-1}\frac{1}{z(z-1)}=\frac{1}{2} \end{align}
Now, by the Residue Theorem, we have
$$\oint_L\frac{dz}{z(z^2-1)}=2\pi i\sum_k\text{Res}\left(\frac{1}{z(z+1)(z-1)},z_k\right)$$
where $z_k$ are the zeros of the denominator of the function that are enclosed by L.
For example, if $L$ is a circle about the origin with radius $R>1$, we have
$$\oint_L\frac{dz}{z(z^2-1)}=2\pi i\left(-1+\frac{1}{2}+\frac{1}{2}\right)=0$$