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

Question

How can I compute the Laurent series of $$f(z)=\frac{1}{(z-1)(z-2)}$$ on the circular ring $A(0;1,2)=B(0,2)\setminus \overline{B(0,1)}$?

I tried to take the Cauchy product of geometric series but these series don't converge in the area.

Answer 1

So you want $\;1<|z|<2\;$:

$$\frac1{(z-1)(z-2)}=\frac1{z-2}-\frac1{z-1}=\frac12\cdot\frac1{1-\frac z2}-\frac1z\cdot\frac1{1-\frac1z}$$

Now use the well known developments of geometric series with ratio less than one, and observe that both

$$\left|\frac z2\right|<1\;,\;\;\;\left|\frac1z\right|<1$$

Answer 2

First rewrite $f$ as $$ -\frac{1}{2}\frac{1}{1-\frac{z}{2}}+\frac{1}{1-z}$$ Now, the first you can rewrite using geometric series as long as $|\frac{z}{2}|<1\iff|z|<2,$ so in particular in your domain $A$.

The second would converge only in $|z|<1.$ The trick is to rewrite it as (which is geometrically sort of doing circle inversion https://en.wikipedia.org/wiki/Inversive_geometry#Circle_inversion) $$-\frac{1}{z}\frac{1}{1-\frac{1}{z}},$$ so that now its GS again converges if $|\frac{1}{z}|<1\iff1<|z|$. I.e. in particular in your domain $A$.

Now sum the two and you are done

Answer 3

The constraints are $1 < \lvert z \rvert < 2$, which can be rewritten $\lvert z/2 \rvert < 1$ and $\lvert 1/z \rvert < 1$.

Using partial fraction decomposition, we obtain $$f(z) = \frac{-1}{2}\frac{1}{1-z/2} - \frac{1}{z}\frac{1}{1-1/z}.$$

Therefore $$f(z) = \frac{-1}{2}\sum_{i=0}^\infty \left(\frac{z}{2}\right)^i - \sum_{i=-1}^{-\infty} z^{i}.$$