Finding residues of such functions in complex analysis I have to find the residues of these functions at all isolated singular points including $\infty$:
$f(z) = {z^{2n}\over (1+z)^n}, n\in \mathbb Z.$
$f(z) = e^{{z+1}\over z}. $
$f(z) = z^n sin{1\over z}, n\in \mathbb Z$
$ f(z) = {1\over sin(1/z)}.$
$f(z) = {\sqrt{z}\over sin{\sqrt{z}}}$
I do not know how to start questions like this, maybe if you show me one, I can do the rest of them. Or any hint or help on related topics in order to solve these would also be very helpful. 
 A: I'll demonstrate the first one.
First of all, you need to start of by finding the singularities of the given function. For example, this means that the function :
$$f(z) = {z^{2n}\over (1+z)^n}, \space n\in \mathbb Z$$
has singularities at the points where the denominator equals zero. More specifically : 
$$(1+z)^n = 0 \Leftrightarrow z=-1$$
Since the polynomial expansion of $(1+z)^n$ gives you $z$ raised at $n$ as a maximum power, this means that the pole $z=-1$ is a pole of order $n$. You'll need to find the residue : 
$$\text{Res}(f(z),-1)=\text{Res}\bigg({z^{2n}\over (1+z)^n}\bigg)$$
For higher order poles, there's the limit formula that gives you the calculation of the desired residues which states : 

If $c$ is a pole of order $n$, then the residue of $f$ around $z = c$ can be found by the formula:
  $$\text{Res}(f,c)=\frac{1}{(n-1)!}\lim_{z\to c}(z-c)^n f^{(n-1)}(z)$$
  where the "power" $(n-1)$ denotes the derivative of such order with respect to $z$.

Given this, can you now find the desired residue ? (look down below for infinity)
Generally, this is what you have to do in order to answer such question (and in the future calculate integrals involving complex integration and residues). Just keep in mind to always find correctly all of the singularities and their orders and applying the correct formula.
Here, I will list the other $2$ possible cases for the residues : 

The case of a simple pole :
At a simple pole $c$, the residue of $f$ is given by:
  $$\text{Res}(f,c)=\lim_{z\to c}(z-c) f(z)$$
  It may be that the function f can be expressed as a quotient of two functions, $f(z)=g(z)/h(z)$, where $g$ and $h$ are holomorphic functions in a neighbourhood of $c$, with $h(c) = 0$ and $h'(c) ≠ 0$. In such a case, L'Hôpital's rule can be used to simplify the above formula to:
  $$\text{Res}(f,c)=\frac{g(c)}{h'(c)}$$
Residue at infinity :
  $$\text{Res}(f(z),\infty) = -\text{Res}\bigg(\frac{1}{z^2}f\bigg(\frac{1}{z}\bigg),0\bigg)$$
  If the condition :
  $$\lim_{|z| \to \infty}f(z) = 0$$
  is met, then the residue at infinity can be computed using the following formula :
  $$\text{Res}(f,\infty)= -\lim_{|z|\to \infty}zf(z)$$
  If instead :
  $$\lim_{|z| \to \infty}f(z) = c \neq 0$$
  then the residue at infinity is :
  $$\text{Res}(f,\infty)= -\lim_{|z|\to \infty}z^2f'(z)$$

