Decompose into simple fractions over $\mathbb{C}$

Question

Decompose into simple fractions over $\mathbb{C}$: $$f(x) = \frac{1}{(x^2-1)^n}$$

I know how to decompose some fractions by $\mathbb{R}$, but I have no idea how analyze this over $\mathbb{C}$ and solve dependency from $n$.

Answer 1

$x^2-1 = (x+1)(x-1)$, so you want a partial fraction decomposition of the form $$ \dfrac{1}{(x+1)^n (x-1)^n} = \sum_{j=1}^n \frac{a_j}{(x+1)^j} + \sum_{j=1}^n \frac{b_j}{(x-1)^j}$$ If $ t = x+1$, we have by the binomial series $$ \frac{1}{t^n(t-2)^n} = (-1)^n \sum_{k=0}^\infty {k+n-1 \choose k} t^{k-n} 2^{-n-k} $$ so that $$a_j =(-1)^n {2n-1-j \choose n-j} 2^{j-2n},\ j=1 \ldots n$$ and similarly $$b_j = (-1)^{n+j} {2n-1-j \choose n-j} 2^{j-2n},\ j=1 \ldots n$$

Answer 2

If, for instance, $n=3$, you can do it by writing $\frac1{(x^2-1)^3}$ as$$\frac{A_1}{x+1}+\frac{A_2}{x-1}+\frac{A_3}{(x+1)^2}+\frac{A_4}{(x-1)^2}+\frac{A_5}{(x+1)^3}+\frac{A_6}{(x-1)^3},$$This will lead you to the system$$\left\{\begin{array}{l}-A_1+A_2-A_3-A_4-A_5+A_6=1\\A_1+A_2+2A_3-2A_4+3A_5+3A_6=0\\2A_1-2A_2-3A_5+3A_6=0\\-2A_1-2A_2-2A_3+2A_4+A_5+A_6=0\\-A_1+A_2+A_3+A_4=0\\A_1+A_2=0.\end{array}\right.$$Its solution is$$A_1=-\frac3{16},\ A_2=\frac3{16},\ A_3=-\frac3{16},\,A_4=-\frac3{16},\ A_5=\frac18,\text{ and }A_6=-\frac18.$$