Find Laurent series for $\frac{(z+1)}{z(z-4)^3}$ in $0<|z-4|<4$.
First we perform partial fraction and we get: $\frac{A}{z}+\frac{B}{z-4}+\frac{C}{(z-4)^2}+\frac{D}{(z-4)^3}$.
My first question is: for the last three terms, do we just leave them like that? Because they all seem to have the desired form.
My second question is: How do we deal with the first term? Do we just let $w=z-4$, then our region becomes $0<|w|<4$, and we have $\frac{A}{z}=\frac{A}{w+4}=A\frac{1}{4}\frac{1}{1-(-\frac{w}{4})^2}=A\frac{1}{4}\sum_{j=1}^{\infty}(-\frac{w}{4})$? But do we have to use geometric series to do the trick? What if I want to have a hard time and find the Taylor series of this thing?
Thank you.
Note that $$\frac{1}{z}=\frac{1}{z-4+4}=\frac{1}{4}\frac{1}{\frac{z-4}4+1}$$
By hypothesis $|z-4|<4$, so indeed you want to expand this in positive powers of $z-4$,
$$\frac 1 z=\frac 1 4\sum_{\nu \geqslant 0}(-1)^\nu \left(\frac{z-4}4\right)^{\nu}$$
This should be also evident from the fact that $1/z$ is holomorphic in $B_4(4)$ so represents the regular part of the Laurent separation of your function (you've already found the principal part), and hence only positive terms will appear.