Consider the function $f(z) = \dfrac{5z-3}{(z+1)(z-3)}$ Consider the function $f(z) = \dfrac{5z-3}{(z+1)(z-3)}$
i) Find the Laurent series of $f(z)$ for the annular domain $1 < |z| < 3$ 
ii) Using part i) or otherwise, find the value of the contour integral $$\int_{C}\dfrac{5z-3}{z^{5}(z+1)(z-3)}dz$$
where $C$ denotes the circle $|z|= 2$ oriented in the counter clockwise direction.
I just started my journey with Residues, and i am confused with the answer they gave for part b). 
For part a), i have computed the answer to be $$f(z) = \sum_{n=0}^{\infty}\dfrac{(-1)^{2}\cdot2}{z^{n+1}} +\sum_{n=0}^{\infty}\dfrac{-1}{3^n}\cdot z^n$$ 
valid for $1<|z|<3$ and i checked it to be correct.
However, for part b, they just gave me $$\int_{C}\dfrac{5z-3}{z^{5}(z+1)(z-3)}dz=\int_{C}\dfrac{f(z)}{z^5}dz=2\pi ia_4 = 2\pi i\dfrac{-1}{3^4}$$
I am completely lost! I only know how to use the slower method. Please help
 A: This looks like homework, so I'll give you hints:
i) Every rational polynomial function which is proper (greater degree in the denominator) admits a partial fraction decomposition with only constants in the numerator. Could these terms have Laurent expansions which are familiar?
ii) Where are the poles enclosed by the contour? (this should be easy) What can you say about the value of the integral based on the knowing the poles? (you mentioned residues --a promising avenue). How might I get the residue of a function whose Laurent series is known? That is, does knowledge of the Laurent series give me a way to calculate the residue? 
Edit: Since OP has stated it's not a homework question, I'm adding more detail at his request. 
Given that you have already calculated the Laurent series, I believe you understand the partial fraction decomposition you need for i). For ii) you have 
$$
\int_C \frac{f(z)}{z^5}dz = \int_C \frac{1}{z^5}\left\{ \sum_{n = 0}^\infty \frac{2}{z^{n+1}}-\sum_{n = 0}^\infty \frac{1}{3^n}z^n\right\} dz.
$$
Doug M. has already mentioned the fact that $\oint z^n dz = 0$ whenever $n \neq -1$. This is because every value of $z^n$ except for $n = -1$ has an anti derivative which is single-valued. In this case Cauchy's theorem is merely a restatement of the fact that a line integral about a closed path in a conservative field is zero. 
Since the first sum is a (uniformly convergent on $1< |z| < 3$) series in $z^n$ where $ n < -1$ for each term, the integral of the series vanishes.
We are left with
$$
\int_C \frac{f(z)}{z^5}dz = -\int_C \sum_n\frac{1}{3^n} z^{n-5} dz.
$$
The integrand is 
$$
\frac{1}{z^5}+\frac{1}{3\cdot z^4}+\frac{1}{3^2\cdot z^3}+\cdots+\frac{1}{3^4\cdot z}+\cdots
$$
and is also uniformly convergent on the annulus, thus all integrals of $z^{n (\neq -1)}$ vanish. The anti-derivative of  $1/z$ however is $\log z$ which also includes the multi-valued argument function. That's why it isn't zero. 
You are left with the integral 
$$
\int_c \frac{1}{z}dz = 2\pi i,
$$
the equality following because the winding number of a circular path is 1. This gives you
$$
\int_C \frac{f(z)}{z^5}dz = -\frac{1}{3^4}\int_c \frac{1}{z}dz = -\frac{2\pi i}{3^4}. 
$$
What you have just done is give a basic argument (though not a proof) for the residue theorem.
A: For any contour that encloses the origin.
$\oint z^n dz = 0$ for all $n$ except $n = -1$
If you can decompose function to a Laurant series, then when you integrate, the $z^{-1}$ term is going to be all important.
$z^{-5} (\sum_{n=0}^{\infty}\dfrac{(-1)^{2}\cdot2}{z^{n+1}} +\sum_{n=0}^{\infty}\dfrac{-1}{3^n}\cdot z^n)$
The $z^{-1}$ term is  $\frac {-1}{3^4} z^{-1}$
