Laurent Series $f(z)=\frac{1}{(z+1)(z-2)}$ with conditions

Question

I want to find the laurent series of the function $$f(z)=\frac{1}{(z+1)(z-2)}$$ for the following conditions:

• $|z|<1$

• $2<|z|<\infty$

I have found the series for first condition, by partial fractions as

$$f(z)=\frac13(\frac{1}{z-2}-\frac{1}{z+1})=-\frac12+\frac z4-\frac{3z^8}{8}+\cdots$$

But how to find the series for the second condition $2<|z|<\infty$, as in the second condition function is holomorphic everywhere?

We can use also use the long general procedure using contour integration, but that I know to be used when laurent series is to be found out at a particular point $z=z_0$, but how to perform it for intervals ?

Another doubt

And sometimes I get confused when we can say taylor's series will be same as laurent series?

Answer 1

I'll do one part. If $|z|>2,$ then $|2/z|<1,$ so we can apply the geometric series to $$\frac{1}{z-2}=\frac{1}{z}\frac{1}{1-\frac{2}{z}},$$ getting that $$\frac{1}{z-2}=\sum\limits_{n=0}^\infty \frac{2^n}{z^{n+1}}.$$ Can you do the other piece?

EDIT: In the comments below, I've discussed OP's confusion on the distinction between Laurent series and Taylor series. The summary is this: the Laurent series coefficients are given by integrals around a contour contained in the given annulus, and in this problem, these contours encircle a singularity at $z=2.$ Hence, we will have negatively-indexed terms. A Laurent series will be a Taylor series if there are no terms with negative indices. which is not the case here.

Answer 2

For $|z|<1$:

$$ \begin{align} \frac1{(z+1)(z-2)}&=\frac1{2(z+1)(z/2-1)}\\ &=-\frac12\left(\sum_{n\ge 0}(-1)^n x^n\right)\left(\sum_{m\ge 0}\left(\frac12\right)^m x^m\right)\\ &=-\frac12\sum_{k\ge 0}c_k x^k \end{align} $$ where $$ c_k=\sum_{j=0}^k(-1)^j\frac1{2^{k-j}}=\frac1{2^k}\sum_{j=0}^k (-2)^j=\frac{1-(-2)^{k+1}}{2^k(1-(-2))}=\frac13\cdot(2^{-k}+(-1)^k2) $$

For $|z|>2$: $$ \begin{align} \frac1{(z+1)(z-2)}&=\frac1{z^2(1+z^{-1})(1-2z^{-1})}\\ &=z^{-2}\left(\sum_{n\ge 0}(-1)^nz^{-n}\right)\left(\sum_{m\ge 0}2^m z^{-m}\right)\\ &=\sum_{k\ge 0}d_k z^{-k-2} \end{align} $$ when this time $$ d_k=\sum_{j=0}^k(-1)^j 2^{k-j}=2^k\sum_{j=0}^k\left(-\frac12\right)^j =2^k\frac{1-(-1/2)^{k+1}}{1-(-1/2)}=\frac{2^{k+1}+(-1)^k}{3} $$