Find the Laurent series expansion of $f(z)=\frac{1}{z(z-1)(z-2)}$ for $1<|z|<2$

So I started of by doing partial fractions:


First of all I have some important conceptual doubts. The first term, $\frac{1}{2z}$ appears to be holomorphic in $1<|z|<2$ this should imply that it admits a taylor series expansion. However, when I try to expand $\frac{1}{2z}$ around $0$ which is the center of my disk, I don't know where to evaluate the $f^{(n)}(z)$ terms which appear in the expansion.

For the second term, I neither know what to do (the same for the third). Can anyone help me out?

  • 2
    $\begingroup$ Just first note that it won't admit a Taylor series expansion around the origin if it's not holomorphic at the origin. $\endgroup$ – Chill2Macht May 14 '16 at 17:36

The $z \mapsto { 1\over z}$ is analytic in the annulus $1<|z|<2$ and does possess a Taylor expansion based on any point in the annulus. Since it is not analytic at $z=0$, it does not have a Taylor series expansion around the origin.

However, the term $z \mapsto {1 \over 2} { 1\over z}$ is a perfectly fine term in the context of expanding about zero in the above annulus.

For the term $z \mapsto {1 \over z-1}$, note that for $|z| >1$ we can write ${1 \over z-1} = {1 \over z} {1 \over 1-{1 \over z}}$ which we can write as ${1 \over z} \sum_{k=0}^\infty {1 \over z^k}$ from which we can read off the Laurent expansion.

The third term is similar, but remember that $|z|<2$ so the expansion is slightly different: ${1 \over z-2} = - {1 \over 2} { 1 \over 1-{z \over 2}}$, and, as above, we can read off the Laurent expansion (in fact, a Taylor series in this case).


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