find the isolated singular points of $f(z)$ 
find the isolated singular points of $f(z)$, and what the type of this point ?
1) $f(z)={1\over {z^2{(z-3)}^3}}$
2) $f(z)={e^{2z}\over {(z-1)^2}}$

my try:
1) the isolated singular points are 0 and 3, but what about there types ?
2)
the isolated singular point is 1
and for the Type :
$e^{2z}=1+2z+{{(2z)}^2\over 2!}+{{(2z)}^3\over 3!}+...$
so 
$f(z)={1\over {(z-1)}^2} (1+2z+{{(2z)}^2\over 2!}+{{(2z)}^3\over 3!}+...)$
but what about $(z-1)^2$ ?
 A: I hope I remember everything correctly from a Complex Analysis course I took and I hope I'm not making mistakes here. I understand this answer may not be the best, but I'm trying to help anyway. :) Corrections from the community are very welcome.
First question: You can write the Taylor series of the function around the singularity like $$\sum_{n = 0}^{N}a_n(z-z_0)^n +\sum_{m=1}^{M}b_m(z-z_0)^{-m}$$ If the series has infinitly many $b_m \neq 0$, then the singularity is essisential. If $b_m = 0$ for all $m$, then the singularity is isolated. If only finitely there are only finitely many $b_m \neq 0$, then the singularity is a pole of order $m$ with $m$ the highest integer for which $b_m \neq 0$
Second question:
Let's say $f(z) = \frac{p(z)}{q(z)}$ with $p(z) = e^{2z}$ and $q(z) = (z-1)^2 = z^2 - 2z + 1$ and call the singularity $z_0$.
The singularity is a pole. ($\lim_{z\to z_0} (z-z_0)^2f(z)$ is neither $0$ nor $\infty$.)
If $p(z_0) \neq 0$ then the order of the pole is the lowest integer $n$ for which $$q^{(n)}(z_0) \neq 0$$
Let's see:
$$q^{(1)}(z) = 2z - 2$$
$q^{(1)}(1) = 0$, so we have to differentiate more:
$$q^{(2)}(z) = 2$$
Obviously $q^{(2)}(1) = 2$ (since it equals $2$ for any $z$), so the singularity is a pole of order $2$.
