I am trying to find the Laurent series for $$\frac{1}{z^3(z-5)^2}$$, in the region $0<|z-5|<5$. I am not sure how to go about solving this because the region $0<|z-5|<5$ is surrounding the point $z=5$.
Another example that I encountered is $$\frac{1+z}{z(z-4)^3}$$ for $0<|z-4|<4$
How to find the Laurent series for functions of this type?
On
$$\begin{align}\frac1{z^3}&=\frac12\frac{\partial^2}{\partial z^2}\frac1 z\\&=\frac12\frac{\partial^2}{\partial z^2}\frac1{5-5+z}\\&=\frac12\frac{\partial^2}{\partial z^2}\frac{\frac15}{1-\frac{5-z}5}\\&=\frac1{10}\frac{\partial^2}{\partial z^2}\sum_{n=0}^\infty\frac{(5-z)^n}{5^n}\\&=\frac1{10}\sum_{n=2}^\infty\frac{n(n-1)(5-z)^{n-2}}{5^n}\\&=\frac1{10}\sum_{n=2}^\infty\frac{n(n-1)(-1)^n}{5^n}(z-5)^{n-2}\end{align}$$
Divide this by $(z-5)^2$ to get your function:
$$f(z)=\frac1{10}\sum_{n=2}^\infty\frac{n(n-1)(-1)^n}{5^n}(z-5)^{n-4}$$
I would advocate using the change of coordinates $w=z-5$, so that one is expanding in powers of $w$. The function is now $$\frac1{w^2(w+5)^3}.$$ We are assuming that $|w|<5$, so then $$\frac1{(w+5)^3}=\frac{(1+w/5)^{-3}}{5^3}.$$ We can now expand $(1+w/5)^{-3}$ by the binomial theorem, as $|w/5|<1$.