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So I have a physics background, not math, but i'm trying to understand at least conceptually WHY laurent series are computed in the way they are. For example, take the series expansion of

$$\frac{z}{(z-1)(z-2)}.$$

In this example they break it up in partial fractions:

$$\frac{2}{z-2}-\frac{1}{z-1}.$$

But now here's where the 'magic' happens. To find Laurent series from, say, $1 < |z| < 2$, they 'factor out' a $\frac{1}{z}$ from the $\frac{2}{z-2}$ term, then proceed to expand in powers of $\frac{1}{z}$, i.e. a normal Taylor series about $\frac{1}{z}$. My question is why do they do this seemingly trivial factorization? What about factoring $\frac{1}{z}$ means this gives you a valid series in that region? Nobody seems to explicitly say this--they just do it.

For the region $|z| > 2$, they factor out $\frac{1}{z}$ from the $\frac{1}{z-1}$ term and $\frac{2}{z}$ from the other term. Again, why does this work? Algebraically aren't these all the same function?

Michael Albanese
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user115418
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1 Answers1

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There's one expansion for $\frac{1}{z-2}$ in $|z| < 2$ and another for $|z| > 2$. In both cases, one uses the geometric series $\frac{1}{1-w}=1+w+w^{2}+w^{3}+\cdots$ as the standard model, which converges for $|w| < 1$. If you want to expand $\frac{1}{z-2}$ in $|z| > 2$, then you'll have $\frac{2}{|z|} < 1$, which gives the clue as to how to rewrite the fraction: $$ \frac{1}{z-2}=\frac{1}{z(1-\frac{2}{z})}=\frac{1}{z}\frac{1}{1-\frac{2}{z}} = \frac{1}{z}\left[1+\frac{2}{z}+\frac{2^{2}}{z^{2}}+\frac{2^{3}}{z^{3}}+\cdots\right],\;\; 2 < |z|. $$

Disintegrating By Parts
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