Find Laurent expansion of $\frac{1}{z}$ in the annulus $1<|z-1|<5$.

Question

I tried this

$$\frac{1}{z}=\frac{1}{z-1+1}=\frac{1}{1-(-(z-1))}=\sum_{k=0}^{\infty}(-1)^k(z-1)^k.$$

However, this geometric series is only valid for $|z-1|<1.$ How can I fix this?

Answer 1

$$\frac1z=\frac1{z-1+1}=\frac1{z-1}\cdot\frac1{1+\frac1{z-1}}=\frac1{z-1}\sum_{n=0}^\infty\frac{(-1)^n}{(z-1)^n}$$

and the above is true for

$$\left|\frac1{z-1}\right|<1\implies|z-1|>1$$