Find the Laurent series of the function$$ f(z) = \frac{z+1}{z(z-4)} $$ in the annulus $0<|z-4|<4$.
My approach: $$ \begin{aligned} f(z) &= \frac{z+1}{z(z-4)} =\left(\frac{-1}{4}\frac{1}{z}+\frac{5}{4}\frac{1}{z-4}\right)\\ &=\frac{-1}{4}\left(\frac{1}{z}+\frac{5}{4-z}\right) = \frac{-1}{4}\left(\frac{1}{z}+\frac{5}{4}\frac{1}{1-z/4}\right). \end{aligned} $$ By the region, I'm pretty sure the second term in the expression would just be $\sum(\frac{z}{4})^n$. I am not sure how to deal with $1/z$ in the bracket cause it seems we cannot determine whether $z$ is in the region or not. Any thoughts? Thanks
The region is centered at $z=4$, so the series should be as well. The series in the question is centered at $z=0$.
Partial fractions is a good way to go, but the $(z+1)$ can, and should, be included in the partial fractions. $$ \begin{align} \frac{z+1}{z(z-4)} &=\frac14\left(\frac5{z-4}-\frac1z\right)\tag1\\[3pt] &=\frac14\left(\frac5{z-4}-\frac1{4+(z-4)}\right)\tag2\\ &=\frac14\left(\frac5{z-4}-\frac{1/4}{1+\frac{z-4}4}\right)\tag3\\ &=\frac{5/4}{z-4}-\sum_{k=0}^\infty\left(-\frac14\right)^{k+2}(z-4)^k\tag4 \end{align} $$ Explanation:
$(1)$: partial fractions
$(2)$: center the second term at $z=4$
$(3)$: write the second term as $\frac1{1+x}$
$(4)$: apply the series for $\frac1{1+x}$