Find the Laurent series of the function and give the residue

Question

I have $f(z)={z^2\over {z^2-1}}$. I want to find the Laurent series of $f$ and the residue at the point $z_0=1$.

Can I say that $f$ has a pole of order 2 at $z=1?$ Or is that only used when the denominator looks like $(z-z_0)^m$.

I let $H(z)=z^2/z+1$, then take the 1st and 2nd derivative. I get $H(1)=1/2, H'(1)=3/4$ and $H''(1)=1/4$. Would the residue of the function at $z=1$ just be $1/2$?

And I do not know how to get the Laurent series.

Answer 1

$$f(z)={1\over{z-1}}{z^2\over {z+1}}={1\over{z-1}}{(z-1+1)^2\over {z-1+2}}={1\over 2}{1\over{z-1}}{((z-1)+1)^2\over{1+(z-1)/2}}=\cdots$$