# Laurent series of $\frac{1}{z^3}+\frac{1}{z-2i}+\frac{1}{(z+1)^2}$

I'm having some conceptual trouble understanding how should the Laurent series be computed depending on the region I want it to converge on.

In this particular case, I want to get the Laurent series of $f(z) = \frac{1}{z^3}+\frac{1}{z-2i}+\frac{1}{(z+1)^2}$ around $z=0$ for the regions:

• $1 < |z| < 2$
• $|z| > 2$

My attempt:

We are going to obtain the Laurent series for each summand. The first one is always $1/z^3$ for both domains, so nothing to do there.

For $\frac{1}{z-2i} = \frac{1}{-2i}\frac{1}{1-\frac{z}{2i}}$ we can expand in the first domain as a geometric series, so we find that $\frac{1}{z-2i} = \frac{i}{2}\sum_{n=0}^\infty (\frac{z}{2i})^n$.

In the second region... I'm at a loss.

Similarly for the third summand. I do not know how to deal with it.

• $\frac {1}{(1+z)^2} = -\frac {d}{dz} \frac {1}{1+z}$ – Doug M Dec 20 '16 at 21:34

$$\frac1{(z+1)^2}=\frac d{dz}\frac{-1}{z+1}=\frac d{dz}\frac{-1/z}{1+1/z}=\frac d{dz}\frac{-1}z\sum_{n=0}^\infty\frac{(-1)^n}{z^n}$$
$$\frac1{z-2i}=\frac{1/z}{1-2i/z}=\frac1z\sum_{n=0}^\infty\frac{(-2i)^n}{z^n}$$
which converges for $|z|>2$.