Write $\frac {1}{1+z^2}$ as a power series centered at $z_0=1$ I'm trying to solve a question where I need to write $\frac {1}{1+z^2} $ as a power series centered at $z_0=1$
I'm not allowed to use taylor expansion. So my first thought was to rewrite the function in a form where I can apply the basic identity:
$$ \frac{1}{1-x} = \sum_{n=0}^\infty x^n $$
So let's rewrite the original function $f(z) = \frac {1}{1+z^2} $ and use above identity:
$$\frac {1}{1+z^2} = \frac {1}{1-(-z^2)} $$
But observe that I need to center it at $z_0=1$ so I rewrote it to:
$$\frac {1}{1-(-z^2)}=\frac {1}{2z-(-(z-1)^2)} $$
I can factor out $\frac {1}{2z}$ and use the basic identity to get :
$$\frac {1}{2z}\frac {1}{1-\frac{(z-1)^2}{2z}} = \frac {1}{2z}\sum_{n=0}^\infty \left(\frac{-(z-1)^2}{2z}\right)^n =\frac {1}{2z}\sum_{n=0}^\infty \frac{(-1)^n(z-1)^{2n}}{(2z)^n} = \sum_{n=0}^\infty \frac{(-1)^n(z-1)^{2n}}{(2z)^{n+1}}$$
The problem is if I ask wolfram alpha: power series$ \frac{1}{1+z^2} $centered at z = 1 it will give me a different answer that doesn't match the sum I found.
Can anyone help me figure this one out? What am I doing wrong?

$$\frac1{1+z^2}=\frac1{(z-1)^2+2(z-1)+2}\\
=\frac12\left(1+(z-1)+\frac12(z-1)^2\right)^{-1}\\
=\frac12\left(1-((z-1)+\frac12(z-1)^2)+((z-1)+\frac12(z-1)^2)^2-...\right)\\
=\frac12-\frac12(z-1)+\frac14(z-1)^2+0(z-1)^3+...$$
