Finding the Laurent series of $f(z)=\frac{1}{(z-1)^2}+\frac{1}{z-2}$?

Can anyone help me out with finding the Laurent series of
$f(z)=\dfrac{1}{(z-1)^2}+\dfrac{1}{z-2}$ in $\{z \in \Bbb C: 2<|z-4|<3\}$?

Solutions Collecting From Web of "Finding the Laurent series of $f(z)=\frac{1}{(z-1)^2}+\frac{1}{z-2}$?"

A related technique. Here is how you advance.

$$ f(z)=\dfrac{1}{(z-1)^2}+\dfrac{1}{z-2} =\dfrac{1}{((z-4)+3)^2}+\dfrac{1}{(z-4)+2}$$

$$ = \dfrac{1}{9\left(\frac{(z-4)}{3}+1\right)^2}+\dfrac{1}{(z-4)(1+\frac{2}{z-4})} $$

$$ = \dfrac{1}{9\left(w+1\right)^2}+\dfrac{1}{(z-4)(1+t)}, $$


$$ w= \frac{(z-4)}{3}\,\quad t= \frac{2}{z-4}. $$

Now, recalling the geometric series, we have, the power series of $g(w)=\frac{1}{(1+w)(1+w)}$ will converge for $|w|<1$ which implies
$|z-4| <3 $. On the other hand, the power series $h(t)=\frac{1}{1+t}$ will converge for $|t|<1$ which gives $ |z-4|>2 $.