$$z^{3}\cos(\frac{1}{z+2.})d\ z=2$$
$d=\frac{2}{z^{4}\cos(\frac{1}{z+2})}$
$z\neq 0\text{ and }z\neq -2\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }z=-\frac{4\pi n_{1}}{2\pi n_{1}+\pi }+\frac{2\left(1-\pi \right)}{2\pi n_{1}+\pi }$
