This might be a silly question but why does $sin(z)/z$ have a singularity at $z=0$?
Because if you expand $sin(z)$ out you can get rid of the z term from the denominator.
Similarly, why does $sin(z)/z^3$ have pole or order 2 at z=0?!
Any help would be appreciated!
On
Strictly speaking the expression $\frac{\sin z}{z}$ does not have a value when $z=0$, because that would requite dividing by zero.
This missing point in the domain is the only missing point in a neighborhood of it, so by definition it is a singularity.
Your reasoning about the series expansion shows that it is a removable singularity, however.
We have $\lim_{z \to 0} \sin(z)/z=1$, hence, by Riemann, the function $f(z)=\frac{\sin z}{z}$ has a removable singularity at $z=0$. Therefore the function $g$, defined by
$g(z)=f(z)$, if $z \ne 0$ and $g(0)=1$
is entire.
Since $\frac{\sin z}{z^3}=\frac{g(z)}{z^2}$ and $g(0) \ne 0$, $\frac{\sin z}{z^3}$ has a pole of order $2$ at $z=0$.