Taylor Series Expansion for $\frac1{(1+z^2)}$

Question

Given $f(z) = \frac{1}{1+z^2}$, I want to find the Taylor series of $f(z)$ about $z_0 = 0$. Intuitively, and based on the formation of a standard power series I have $f(z) = \sum (-1)^n(z^2)^n$. What I need now is to find the power series for $(\frac{1}{1+z^2})^2$ and $(\frac{1}{1+z^2})^3$. I assume I'll need to use partial fractions here but I'm a little stuck at the start. Tips would be appreciated.

Answer 1

$$1/(1+z^2) = \sum_{n=0}^\infty z^{2n} (-1)^n$$

$$1/(1+z^2)^2 = \sum_{n=0}^\infty (z+1)z^{2n} (-1)^n$$

Answer 2

Hint: the derivative of $$ x\to \frac1{1+x^2} $$ is$$x\to -2x\times \frac{1}{(1+x^2)^2} $$

and the derivative of $$ x\to \frac1{(1+x^2)^2} $$ is$$x\to -2\times 2x\frac{1}{(1+x^2)^3} $$

Answer 3

Let $$\frac{1}{(1+z^2)^2}=\sum_{n=0}^{\infty}a_nz^n$$ then:

$$(1+z^2)\frac{1}{(1+z^2)^2}=(1+z^2)\sum_{n=0}^{\infty}a_nz^n=\sum_{n=0}^{\infty}a_nz^n+a_nz^{n+2}$$

You can write it also as (let $a_{-1}=a_{-2}=0$):

$$\sum_{n=0}^{\infty}a_nz^n+a_nz^{n+2}=\sum_{n=0}^{\infty}(a_n+a_{n-2})z^n$$

But you know that

$$\frac{1}{(1+z^2)}=\sum_{n=0}^{\infty}a_nz^n+a_nz^{n+2}=\sum_{n=0}^{\infty}b_nz^n$$

So finally you have $b_n=a_{n}+a_{n-2}$.