Complex isolated singularities

I require some help with identifying and classifying the isolated singularities of complex-valued functions.

• For example: $f(z)=\frac{(z^3+1)}{z^2(z+1)}$, here I understand that there is a pole of order 2 @ $z=0$, and a removable singularity @$z=1$, but this is really only from intuition, and would appreciate some clarification on how to formally demonstrate this.
• Another example: $f(z)=\frac{cos(z)}{z^2+1}+4z$, here the singularities are less obvious to me, and further I am particularly confused by $cos(z)$.

So far I am inclined to use a Laurent series, however defining an $R$ is unclear to me, as is how to simply for intricate functions as mentioned above. Thanks.

Note that we obtain these functions by dividing entire functions. The quotient rule tells us that we will get differentiability everywhere the functions are defined: i.e. where the denominators are $0$. That is, our functions will be analytic everywhere except where the denominators are $0$.
So, in the first case, this is everywhere except $z = 0$ and $z = -1$ (not $z = +1$ - careful!). In the second case, this is everywhere except $z = \pm i$.
To show that the singularity $z_0$ of a function $f$ is a pole and compute its order, you multiply $f(z)$ by some power of $z - z_0$ until the singularity becomes removable (note that this isn't always possible). So, in the case of the first function with $z = 0$, we multiply $$z f(z) = \frac{z^3 + 1}{z(z + 1)},$$ which is not removable at $0$ since the numerator tends to a non-zero number as $z \to 0$. If we try a higher power, $$z^2 f(z) = \frac{z^3 + 1}{z + 1},$$ on its domain, which is $z \neq 0, -1$. Note that the resulting function can be extended to an analytic function at $z = 0$. Thus, the singularity is removable at $z = 0$, and the pole of $f$ at $z = 0$ was order $2$.
For the second function, note that the singularities at $z = \pm i$ are not removable, since $$\cos(z) \to \frac{e^{-1} - e}{2} \neq 0$$ as $z \to i$, and similarly for $z \to -i$. However, multiplying $f$ by $z - i$ or by $z + i$ will make the singularities removable, hence the singularities are poles of order $1$.