Rewriting the quotient of taylor series (to calculate residue)

Question

To calculate the residue of a complex integral, I look at the taylor expansion of $ \frac{\cos(z)}{\sin(z)}$, and want to find the coefficient in front of the $z^{-1}$ term.

I have been given the solution, which is: $$ \frac{\cos(z)}{\sin(z)} = \frac{1 - \frac{1}{2}z^2 + \ldots}{z(1 - \frac{1}{6}z^2 + \ldots)} =^{*} \frac{1}{z}(1 - \frac{1}{2}z^2)(1 + \frac{1}{6}z^2) + \ldots = \frac{1}{z} - \frac{1}{3}z + \ldots$$ so the residue is 1.

I can't, however, figure out how he performs the step I've marked with an asterisk (from fraction to no fraction, so to speak). Can anybody help me out?

Thanks in advance!

Answer 1

Note that for $|w|<1$, $$\frac{1}{1-w}=\sum_{n\geq o}w^n=1+w+o(w).$$ Hence $$\frac{1}{1 - \frac{1}{6}z^2+o(z^2)}=1+\frac{1}{6}z^2+o(z^2).$$