Let's consider the annulus $1\le|z|\le2$ and $$f(z)=\frac{2z+3}{z^2+3z+2}$$ Find its Laurent series.
First I simplify the expresion: $$f(z)=\frac{2z+3}{(z+1)(z+2)}=(2z+3)\left(\frac1{z+1}-\frac1{z+2}\right) \\= \frac{2z+3}{z+1}-\frac{2z+3}{z+2}=\frac1{z+1}+2-\left(2-\frac1{z+2}\right) \\=\frac1{z+1} +\frac1{z+2}=f_1(z)+f_2(z)$$
I say the series representation of $f$ diverges in $1\le|z|\le2$, since both $f_1,f_2$ are geometric series, and those diverge when $|z|\ge1$.
Is this correct?
Hint: $\dfrac1{z+1}=\dfrac{\frac1{z}}{1-(-\frac1{z})}$, and $\dfrac1{z+2}=\dfrac{\frac1{2}}{1-(-\frac{z}{2})}$