Is it true that:
$\oint _{C(5i+1,8\sqrt3)} \frac {z}{sh(z)} dz = \oint _{C(i,\sqrt{10})} \frac {z^2}{(1-cos2z)^4}dz = \oint _{C(\pi + i,4)} \frac {z}{tan(z)} dz = 0$
The problem is that i lost my time finding poles and applying the Residue theorem in order to find the integral of $\oint _{C(1,\sqrt \pi)} \frac {z^4-1}{z^6-1} dz$ which was after all equal to $0$ !
Is the integral of any even function on any contour equal to zero since the term $a_{-1}$ in the Laurent series expansion of any even function equal to $0$ ?