<s>[INST] How to expand $(x+y)^{-n}$? How to expand $(x+y)^{-n}$ by binomial theorem, where $n$ is a positive integer? Is there any limitation for $x$ and $y$? If it can be expanded, how to compute the  coefficients? Many thanks in advance.
 [/INST] This has a singularity along the line $y = -x$. The expansion in terms of powers of $y$ and $x$ will differ depending on whether you are above or below this line.
No binomial theorem, but note that
$$
\sum_{n = 0}^\infty z^n = \frac{1}{1-z}
$$
whenever $|z| < 1$.
If we relabel $y = -w$, and look at $(x - w)^{-1}$, we get
$$
\frac{1}{x-w} = \frac{1}{x}\frac{1}{1-\frac{w}{x}}
$$
and
$$
\frac{1}{x-w} = -\frac{1}{w}\frac{1}{1-\frac{x}{w}}
$$
which converge separately for $x > w$ and $w > x$ in each case (above and below the line x = w respectively).
But then $(x - w)^{-1}$ can be written as
$$
\sum_{n = 0}^{\infty} x^{-n -1} w^n 
$$
and
$$
- \sum_{n = 0}^{\infty} x^n w^{-n-1} 
$$
in each region.
Now, 
$$
\partial_x^{n} (x - w)^{-1} = (-1)^{n} n! (x - w)^{-n-1}
$$
where I have written $\partial_x$ for the partial derivative with respect to $x$. 
So
$$
(x - w)^{-n-1} = \frac{(-1)^n}{n!} \partial_x^n \left(\sum_{k > 0} x^{-k-1} w^k\right) \\
= \frac{(-1)^n}{n!} \sum_{k = 0}^\infty (-k-1)(-k-2)\cdots (-k - n) x^{-k-1-n}w^k \\
= (-1)^n \sum_{k = 0}^\infty (-1)^n \left(\begin{array}{c} k+n \\ k \end{array}\right) x^{-k-1-n}w^k \\
= x^{-n} \sum_{k = 0}^\infty \left(\begin{array}{c} k+n \\ k \end{array}\right) x^{-k-1}w^k
$$
for the first case, and similarly,
$$
(x - w)^{-n-1} = \frac{(-1)^{n+1}}{n!} \partial_x^n \left(\sum_{k > 0} x^k w^{-k-1}\right) \\
= \frac{(-1)^{n+1}}{n!}\sum_{k \geq n} (k)(k-1)\cdots(k-n+1)x^{k-n} w^{-k-1} \\
= (-1)^{n+1} \sum_{k \geq n} \left(\begin{array}{c} k \\ n \end{array}\right)x^{k-n} w^{-k-1}\\
= (-1)^{n+1}w^{-n}\sum_{k = 0}^\infty \left(\begin{array}{c} k+n \\ n \end{array}\right)x^k w^{-k-1}
$$
for the second.
Note that this works for any values $x, y$ $(= -w)$ in $\mathbb{C}$.
</s>