I am trying to study complex analysis and I've come across this $$ \begin{align} f(z)= \frac{1}{1+z^2} \end{align} $$ I need to determine the Laurent series expansion for the annulus $1<|z-2i|<3$
I think that I need to write $f$ in the form of $\frac{A}{1-\frac{1}{1-(z-2i)}}$ and then use the geometric series but I don't know how.
I see that $$ f(z)= \frac{1}{(i-z)(i+z)} = \frac{1}{2i}\frac{1}{i-z}+\frac{1}{2i}\frac{1}{i+z} $$ but I don't know what to do next. I also understand that the Laurent series will depend on the annulus and in our case there: $|z|<1$ , $1<|z|<3$ ( which is the one we're interested ) and $|z|>3$.
Any assistance or guidance with this will be greatly appreciated. Thank you for your time!
Let $z=2i+w$, then $$f(w)=\frac{1}{2i}\frac{1}{-i-w}+\frac{1}{2i}\frac{1}{3i+w}$$
$|w|>1$ so the first term can be written as $\frac{1/w}{-i/w-1}$, with a geometric series in $i/w$.
Also $|w|<3$ so the second term can be written as a geometric series from $\frac13\frac{1}{i+w/3}$.