Partial fraction of complex polynomials : a linear algebra proof I will show a proof I think I devised which I will explain in details. 
I will prove that 

Suppose that $(P,Q) \in (\mathbb{C}[X])^2$ such that 
  $$  \deg(P)<\deg(Q), \deg(Q)>1\\
\forall x \in \mathbb{C}, Q(x)=c\prod_{i=1}^n(x-\alpha_i)^{m_i},$$
where $\deg(Q) = \sum_{i=1}^nm_i$, and $\alpha_i$ denotes the $i^\text{th}$ root of $Q$ whose multiplicty is $m_i$. Then, there exists a unique set of values $\{a_{i,j}\}_{\substack{1\leq i\leq n \\1\leq j \leq m_i}}$
  such that
  $$\frac{P(x)}{Q(x)}=\sum_{i=1}^n\sum_{j=1}^{m_i}\frac{a_{i,j}}{(x-\alpha_i)^j}.$$

Here is the proof outline 
Define $$E=\left\{\frac{P}{Q}\,\middle|\, P\in \mathbb{C}[X] \wedge \deg(P)<\deg(Q)\right\}\\
F=\left\{\sum_{i=1}^n\sum_{j=1}^{m_i}\frac{a_{i,j}}{(x-\alpha_i)^j}\,\middle|\, a_{i,j}\in{\mathbb C}\mbox{ for all }i,j\right\}.$$
First, we prove that $E$ and $F$ are both vector spaces with equal dimension $d=\deg(Q)$. Then, proving that $F\subseteq E$ implies that $E=F$, (dimension theory on finite vector spaces) then  the uniqueness of vector decomposition in a base yields the result. 
Thus, the proof steps are the following :


*

*Prove that  E=$\DeclareMathOperator{\spann}{span} \spann{(\frac{1=x^0}{Q(X)}},\dots,\frac{x^{d-1}}{Q(x)})$, and therefore $E$ is a vector space (done, will post soon) 

*Prove that the elements of the set are linearly independent, and thus, $\dim{E}=d$ (done as well)

*Do the same work with $ F=\spann{(\frac{1}{(x-\alpha_1)},\dots,\frac{1}{(x-\alpha_1)^{m_1}},\dots,\frac{1}{(x-\alpha_n)^{m_n}})}$ (need help to do this rigourously)

*Show that $F\subset E$ (help needed)


Can anyone help me to write those steps? 
Thanks! 
Hint : To write $\spann$ in math put the following at the top of your answer: 
$\DeclareMathOperator{\spann}{span}$

then use \spann{something} in math mode
 A: Step 3: It is easy to see that $F$ is spanned by the set $\left\{\frac{1}{(x-\alpha_i)^j}\right\}_{\substack{j=1,2\ldots,m_i\\i=1,2,\ldots,n}}$. Next, we show that the elements of the set are linearly independent by looking for a non-trivial solution for the following equation:
\begin{align}
\sum_{i=1}^n\sum_{j=1}^{m_i}\frac{a_{ij}}{(x-\alpha_i)^j}=0&\implies  \sum_{i=1}^n\sum_{j=1}^{m_i}\frac{a_{ij}(x-\alpha_i)^{m_i-j}}{(x-\alpha_i)^{m_i}}=0\\
&\implies   \sum_{i=1}^n\sum_{j=1}^{m_i}\frac{a_{ij}(x-\alpha_i)^{m_i-j}\prod_{k=1,k\neq i}^{n}(x-\alpha_k)^{m_k}}{\prod_{k=1}^n(x-\alpha_k)^k}=0\\
&\implies   \sum_{i=1}^n\sum_{j=1}^{m_i}a_{ij}(x-\alpha_i)^{m_i-j}\prod_{k=1,k\neq i}^{n}(x-\alpha_k)^{m_k}=0.
\end{align}
Substituting $x=\alpha_l$ in the above expression gives $a_{lm_l}=0$, for all $l=1,2,\ldots,n$. Thus, we get
\begin{equation}
 \sum_{i=1}^n\sum_{j=1}^{m_i-1}a_{ij}(x-\alpha_i)^{m_i-j}\prod_{k=1,k\neq i}^{n}(x-\alpha_k)^{m_k-1}=0.
\end{equation}
After repeated substitution of $x=\alpha_l$ in the above simplified expression, we obtain that $a_{ij}=0$, for $j=1,2\ldots,m_i$ and $i=1,2,\ldots,n$. Thus, the elements of the set are linearly independent.
Step 4: Using a simplication similar to that in step 3, we get that any element of $F$ can be expresed as the ratio of two polynomials: $$    \frac{\sum_{i=1}^n\sum_{j=1}^{m_i}a_{ij}(x-\alpha_i)^{m_i-j}\prod_{k=1,k\neq i}^{n}(x-\alpha_k)^{m_k}}{\prod_{k=1}^n(x-\alpha_k)^k}= \frac1c  \frac{\sum_{i=1}^n\sum_{j=1}^{m_i}a_{ij}(x-\alpha_i)^{m_i-j}\prod_{k=1,k\neq i}^{n}(x-\alpha_k)^{m_k}}{Q(x)}$$
Clearly, the degree of the numerator at most $d-1$ and thus, it is less than the degree of denominator.  Hence, $F\subseteq E$.
