Finding the Laurent series of a given function

Question

How do I find the Laurent series expansion for the function

$\frac{z}{(z+1)(z+2)}$

which converges for $1 < |z| < 2$ and diverges elsewhere.

I have done the partial fractions for $f(z) = \frac{2}{z+2} - \frac{1}{z+1}$.

Answer 1

It is not true that the series converges in $1<|z| < 2$. The function is perfectly well defined in $z = 0$.

So the function behaves well in the disk $|z| < 1$ and you can find it's Laurent series (which is equal to the taylor series, since the function is holomorphic there)

To do that you just need to find the laurent series of $\frac 2{z+2}$ and $\frac 1{z+1}$ and sum.

You need to remember that $\frac 1{1-z} = \sum_{n=0}^\infty z^n$

Spoiler for $\frac 1{z+1}$

Recognize that $\frac 1{z+1} = \frac{1}{1 - (-z)} = \sum_{n=0}^\infty (-1)^nz^n$

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In $-1$ we find a pole, so we can't fo forward. But we can write the laurent expansion around $z = -1$ and we'll find that this converges for $|z+1| < 1$ (that is, in the disk centered in $-1$ of radius $1$)

You can't do the same argument for $z= -2$, also.

Answer 2

As you found already, $f(z) = \frac{2}{2+z} - \frac{1}{1+z}$
Let's look at each fraction separately:

$-\frac{1}{1+z}$ is analytic in $|z| \gt 1$ and since $|\frac{1}{z}| \lt 1$ we have: $$ -\frac{1}{1+z} = -\frac{1}{z}\cdot\frac{1}{1-(-\frac{1}{z})} = -\frac{1}{z}(1-\frac{1}{z}+\frac{1}{z^2}-\dots) = -\frac{1}{z}+\frac{1}{z^2}-\frac{1}{z^3}+\dots $$

$\frac{2}{2+z}$ is analytic in $|z|\lt2$, we have $|\frac{z}{2}| \lt 1$ and we can represent it with it's Taylor series: $$ \frac{2}{2+z} = \frac{1}{1-(-\frac{z}{2})} = 1 -\frac{z}{2}+\frac{z^2}{4}-\dots $$

Therefore the Laurent series: $$ \dots -\frac{1}{z^3}+\frac{1}{z^2}-\frac{1}{z}+1-\frac{z}{2}+\frac{z^2}{4}-\dots $$ Converges in $1 \lt |z| \lt 2$