Laurent Series of $\frac{1}{z^2(z-2)}$

Question

Expand Laurent series $f(z)=\frac{1}{z^2(z-2)}$ at $|z-2|>2$

Before starting, why isn't it Taylor series? we do not have singularities.

Answer 1

We can write

$$f(z) = \frac 1{z^2(z-2)} = \frac{1}{z-2}\frac d{dz} \left( -\frac 1z \right) = \frac{1}{z-2}\frac d{dz} \left( \frac {-1}{2+(z-2)} \right)\\ = -\frac{1}{2(z-2)}\frac d{dz} \left( \frac 1{1+\frac{z-2}2} \right) $$

so it is enough to find Laurent series of $\frac 1{1+\frac{z-2}2}$. Since $|z-2| > 2$, it follows that $\left|\frac{2}{z-2}\right|<1$ and thus

$$\frac 1{1+\frac{z-2}2} = \frac{\frac{2}{z-2}}{\frac{2}{z-2}+1} = \frac{2}{z-2}\sum_{n=0}^\infty \left( -\frac{2}{z-2}\right)^n,$$

where we used that appropriate geometric series converges.

Differentiate it and multiply by $-\frac{1}{2(z-2)}$ to obtain Laurent series of $f$ on the annulus $|z-2|>2$.