Someone please confirm that what i'm doing below is correct, -Thanks.
$$sin(1-z^{-1}) \tag{1}$$
$sin(1-z^{-1})=\frac{z-1}{z}-\frac{(z-1)^3}{3!z^3}+\frac{(z-1)^5}{5!z^5}$
-has only an essential singularity at $z=0$ because of the recurring $z$ in the denominator that makes it singular.
$$\frac{tanz}{z} \tag{2}$$
$\frac{tanz}{z}=1+..$
-has only a removable singularity at $z=0$ since the pole disappears after we take the series expansion.
$$\frac{cosz}{z^2+1}+4z \tag{3}$$
$\frac{cosz}{z^2+1}+4z = 4z+\{1-0.5z^2+0.25z^3+..\}\{1-z^2+z^4+..\}$
-has only a removable singularity at $z=\pm i$
$$\frac{sin3z}{z^2}+\frac{3}{z} \tag{4}$$
$\frac{sin3z}{z^2}+\frac{3}{z} = \frac{1}{z}\{6-\frac{(3z)^2}{3!}+..\}$
-has only a simple pole at $z = 0$ since $\frac{1}{z}$ of order one can be factored out
$$\frac{1+e^z}{z(e^z-1)} \tag{5}$$
$\frac{1+e^z}{z(e^z-1)} = \frac{1}{z^2}\{2+z+..\}\{1-(\frac{z}{2!}+..)+..\}$
-has a pole of order 2 at $z = 0$ and a simple pole at $z=2k\pi i$