Proof that $\frac {1} {2\pi i} \oint \frac {{\rm d}z} {P(z)} $ over a closed curve is zero. 
Prove that $$\frac {1} {2\pi i} \oint \frac {{\rm d}z} {P(z)} $$ over a closed  smooth simple curve that contains all of the roots of the polynomial $P(z)$ is either zero if $n \geq 2$ or equals $\frac{1} {a_0} $ if it's degree is $1$.

Proof: $n\geq 2$ $$\frac {1} {P(z)=(z-z_0)(z-z_1)\dots(z-z_n)}$$ So after partial decomposition of the fraction I'll have something like $$\frac {A} {z-z_0} +\frac{B} {z-z_1}+\cdots$$ where $A,B \in \mathbb{C}$ (Is this fact true that $A,B$ wont be polynomials and just coefficients I can see it why in an intuitive way but cannot write a full proof of it for the decomposition trying to get a linear system or something) then use Cauchy's theorem and integrate all of these on  different circles with their center at each root  and argue that  the integral is the same as the sum of those circle integrals. Now since $\frac {1}{z-z_n}$ will be analytic inside the circle when I'm integrating on circles that do not contain $z_n$  some  of those integrals  will be zero. and now since my curves are of the form $\gamma(t)=z_n+e^{it}$ , $t \in [0,2\pi]$ the rest of the integrals  will look like $$\int_0^{2\pi}  \frac{Aie^{ti}} {e^{it}}\,{\rm d}t $$  and all of them will be $A_i$. 
Is my proof correct? Only part I cannot show is that I won't have any problem with decomposing the fraction. And I'm also using the assumption that after the decomposition I'll have as numerators simple complex numbers not not functions of any kind.
 A: I think you are over-complicating it a bit, it is not necessary to recover the coefficients of the partial fraction decomposition, it is enough to exploit the residue theorem and the ML lemma. If the degree of $P$ is $n\geq 2$ and all its zeroes lies inside $|z|<R$,
$$\forall M\geq R,\qquad \oint_{|z|=R}\frac{dz}{P(z)}=\oint_{|z|=M}\frac{dz}{P(z)} $$
since both integrals equal $2\pi i$ times the sum of the residues of $f(z)=\frac{1}{P(z)}$ at its poles. But if $|z|$ is large we have $\left|P(z)\right|\geq C |z|^n $, hence
$$\left|\oint_{|z|=M}\frac{dz}{P(z)}\right|\leq \frac{2\pi M}{CM^n} $$
that goes to zero as $M\to +\infty$, and the only non-trivial part of the claim is proved.
A: Let $P(z)=\sum_{k=1}^na_kz^k$ be a polynomial of order $n>1$.  Assume that all of the zeros, $z_k$, $k=1,\dots,n$, are distinct.  Then, we can write $P(z)=a_n\prod_{k=1}^n(z-z_k)$.  
Next, using partial fraction expansion, we can write the reciprocal of $P(z)$ as
$$\begin{align}
\frac{1}{P(z)}&=\sum_{k=1}^n\frac{A_k}{z-z_k} \tag 1
\end{align}$$
where $A_k$ are the residues.  Now, multiplying $(1)$ by $a_n\prod_{j=1}^n(z-z_j)$ yields
$$\begin{align}
\frac{a_n\prod_{j=1}^n(z-z_j)}{P(z)}=a_n\sum_{k=1}^n A_k \frac{\prod_{j=1}^n(z-z_j)}{(z-z_k)} \tag 2
\end{align}$$ 
The left-hand side of $(2)$ is equal to $1$, whereas the right-hand side is a polynomial of order $n-1$.  Therefore, the coefficients of the polynomial on the right-hand side must be zero for all but the zeroth order term.  In particular, the coefficient on the leading term is given by $a_n\sum_{k=1}^n A_k$.  Since, this must be zero, we find that
$$\sum_{k=1}^n A_k=0$$
Inasmuch as $\oint_C \frac{1}{P(z)}\,dz=2\pi i \sum_{k=1}^n A_k $, then 
$$\frac{1}{2\pi i}\oint_C \frac{1}{P(z)}\,dz=0$$
as was to be shown!
