Verify that the function $f(z) = \frac {z}{(z − 1)(z + 2)}$ is analytic in the annulus $4 < |z − 1 − 4i| < 5$ and find the Laurent series expansion of f centered at $1 + 4i$ for this annulus.
I understand that the poles are $z=1$ and $z=-2$. Because these values are not in the annulus, the function is analytic on said annulus.
I have a hard time with Laurent series expansion, though.
Hint:
Observe that
$$\frac z{(z-1)(z+2)}=\frac13\left(\frac1{z-1}+\frac2{z+2}\right)=\frac13\left(\frac1{4i+z-1-4i}+\frac2{3+4i+z-1-4i}\right)=$$
$$\frac13\left(\frac1{z-1-4i}\frac1{1+\frac{4i}{z-1-4i}}+\frac2{3+4i}\frac1{1+\frac{z-1-4i}{3+4i}}\right)$$
Finally (for me, not for you...), observe that we both have
$$\left|\frac{4i}{z-1-4i}\right|<1\;,\;\;\;\left|\frac{z-1-4i}{3+4i}\right|<1\;\;\;\;\text{(why?)}$$
Now use the development for geometric series with constant quotient $\;r\;,\;\;|r|<1\;$ ...