$f(z)=\frac{z}{(z-1)(z+4)}$ and I need the Laurent series for $0<|z-1|<5$ and $5<|z-1|$.
I think I have $0<|z-1|<5$ figured by decomposing $f$ into $\frac{1}{5(z-1)}+\frac{4}{5(z+4)}$ which results in coefficients for the Laurent series being $\frac{1}{5}$ when $n=-1$ and $\frac{4(-1)^n}{5^{n+2}}$ for $n=0$ and $n>0$.
For $5<|z-1|,$ $\frac{4}{5(z+4)}=\frac{4}{5(z-1)}[1-\frac{5}{z-1}+\left(\frac{5}{(z-1)}\right)^2-\ldots]$. I am uncertain of how to add $\frac{1}{5(z-1)}$ to the expansion of $\frac{4}{5(z+4)}$. Any thoughts would be helpful.
for $|z-1|<5$ we have $\frac{|z-1|}{5}<1$ so $$f(z)=\frac{1}{(z-1)+5}+\frac{1}{z-1}\times\frac{1}{(z-1)+5}$$ therefore $$f(z)=\frac{1}{5}\sum\limits_{n=0}^{\infty}(-1)^n(\frac{z-1}{5})^n+\frac{1}{5(z-1)}\sum\limits_{n=0}^{\infty}(-1)^n(\frac{z-1}{5})^n.$$ Also, for $|z-1|>5$ we have $\frac{5}{|z-1|}<1$ so we have that $$f(z)=\frac{1}{z-1}\times\frac{1}{1+\frac {5}{(z-1)}}+\frac{1}{(z-1)^2}\times\frac{1}{1+\frac {5}{(z-1)}}$$ so that $$f(z)=\frac{1}{z-1}\sum\limits_{n=0}^{\infty}(-1)^n(\frac{5}{{z-1}})^n+\frac{1}{(z-1)^2}\sum\limits_{n=0}^{\infty}(-1)^n(\frac{5}{z-1})^n.$$