How do I write the Laurent series expansion of $\frac{1}{z-3}$ for $|z-3|>5$.
I'm trying to rewrite this into the geometric series form of $\frac{1}{1-r}$. I rewrote it as $-\frac{1}{3}(\frac{1}{1-\frac{z}{3}})$, but I don't think this is correct. I'm pretty sure that because the disk is centered around 3 I need to keep a $z-3$ in the "r" spot.
I made a mistake last night as honestly as possible, I hope below formulas can help you $$|z_1-z_2+z_2|\leq|z_1-z_2|+|z_2|\\|z_1-z_2|\geq |z_1|-|z_2|\\|z-2|\geq |z_-3|-|1|\\|z-2|+1\geq |z-3|>5 \\ \to |z-2|>4$$ you need soomething converge to $\frac1{z-3}$ if we look at $\frac r{1-r}=r+r^2+r^3+...$ we have to find $r$ such that series converge to $\frac1{z-3}$so $$\frac{r}{1-r}=\frac{1}{z-3}\to rz-3r=1-r\\r(z-3+1)=1\to r=\frac{1}{z-3+1}=\frac{1}{z-2}$$ now lets look at $|z-2|>4$ so $\frac{1}{|z-2|}<\frac 14$ convergence condition
finally we have $$\frac1{z-3}=\frac{r}{1-r}=\frac{\frac{1}{z-2}}{1-\frac{1}{z-2}}=(\frac{1}{z-2})+(\frac{1}{z-2})^2+(\frac{1}{z-2})^3+...$$