Compute all the possible values for the integral $$\int_C \frac{dz}{z(z^2-1)}$$ for different positions of contour C.
Note: C doesn't pass through the points $0,1,-1$.
The answer is:
If contour C contains point $0$ in it's interior and does not contains $1$ nor $-1$, then we have $l=-2\pi i$.
If it contains only one of the points $1$ or $-1$ and does not contain $0$, then $l=\pi i$.
Therefore the integral can only take 5 values $(-2\pi i,-\pi i,0,\pi i,2\pi i )$.
Can someone explain me the whole answers with details please?? Because honestly I don't understand it.
The integrand has poles at $0,\pm 1$ and no other singularities.
If no poles lie within the contour, the value of the integral must be zero (the function is analytic on the interior of the contour). All three poles are simple, so the residues at each of them are $$ r_0 = \text{Res}_{z = 0}\frac{1}{z(z^2-1)} = \lim_{z \rightarrow 0}z\frac{1}{z(z^2-1)} = -1, $$ $$ r_1 = \text{Res}_{z = 1}\frac{1}{z(z^2-1)} = \lim_{z \rightarrow 1}(z-1)\frac{1}{z(z^2-1)} = \frac{1}{2}, $$ $$ r_{-1} = \text{Res}_{z = -1}\frac{1}{z(z^2-1)} = \lim_{z \rightarrow -1}(z+1)\frac{1}{z(z^2-1)} = \frac{1}{2}. $$ The residue theorem says $$ \int_C f(z) dz = 2\pi i\sum_{\text{$p$, poles of $f$}}\text{Res}_{z = p}f(z), $$
so if either $\pm 1$ but no other pole is contained within $C$ then $2 \pi i$ times the residue is $\pi i$.
If only zero is contained then $2 \pi i$ times the residue is $-2\pi i$.
If zero and either $\pm 1$ is contained then $2 \pi i$ times the sum of residues is $-\pi i$.
If both $\pm 1$ but not zero are contained, then $2 \pi i$ times the sum of residues is $2\pi i$.
If all three poles are contained then then $2 \pi i$ times the sum of residues is $0$.
These are the only possible cases.