I need to find the Laurent series expansion of
$$f(z)=\frac{1}{(z-4)(z-1)}$$
about $1<\vert z \vert <4.$
So after partial fraction decomp I get
$$\frac{1}{(z-4)(z-1)}=\frac{1}{3}\bigg(\frac{1}{z-4}-\frac{1}{z-1} \bigg).$$
And I can rewrite this as
$$\frac{1}{3}\bigg(-\frac{1}{4}\bigg( \frac{1}{1-\frac{z}{4}}\bigg) +\frac{1}{z}\frac{1}{1-\frac{1}{z}}\bigg)$$
So I get
$$\frac{1}{3}\bigg(-\sum_{n=0}^\infty (\frac{z^n}{4^{n+1}})+\sum_{n=-1}^{-\infty}z^n\bigg).$$
Did I do this last step correctly? And for the integral I get $\frac{2\pi i}{3}$. Domain of integral is circle center at $4$ of radius $1$.
Yes your solution is correct.
Last step $$\frac{1}{3}\bigg(-\sum_{n=0}^\infty (\frac{z^n}{4^{n+1}})+\sum_{n=1}^{\infty}\frac{1}{z^n}\bigg).$$ here the co-effient of $\frac{1}{z}$ is $\frac{1}{3}$.
Using Residue theorem the value of integral is: $2\pi i$ $\times$ (co-efficient of $\frac{1}{z}$)
which is $\frac{2\pi i}{3}$.