Hello all at StackExchange,

I'm having some trouble understanding computing the Laurent series for different domains.

Here's my approach to finding the Laurent series for $\dfrac{3}{(z+1)(z-2)}$ for the following domains.

1) $|z| < 1$

2) $ 1 < |z| < 2$

3) $|z| > 2$

So, using partial fraction decomposition, I obtain $\dfrac{1}{z-2} - \dfrac{1}{z+1}$.

The idea is to obtain series for both fractions, which is understandable (Taylor series and power series of complex functions is understood), but I am not understanding $\bf{why}$ the fractions become modified as follows (the algebra is understood, not the idea).

1) We change $\dfrac{1}{z-2} = -\dfrac{1}{2}\left[\dfrac{1}{1-\frac{z}{2}}\right]$ and leave $\dfrac{1}{z+1} = \dfrac{1}{z-(-1)}$.

2) We still have $\dfrac{1}{z-2} = -\dfrac{1}{2}\left[\dfrac{1}{1-\frac{z}{2}}\right]$ but instead $\dfrac{1}{z+1} = \dfrac{1}{z}\left[\dfrac{1}{1+\frac{1}{z}}\right]$.

3) We change $\dfrac{1}{z-2} = \dfrac{1}{z}\left[\dfrac{1}{1-\frac{2}{z}}\right]$ but keep $\dfrac{1}{z+1} = \dfrac{1}{z}\left[\dfrac{1}{1+\frac{1}{z}}\right]$.

Do the fractions change because of analyticity in the domains, and in particular, why do we factor a term of $\dfrac{1}{z}$ for some of them? Much help would be appreciated, and regards.

  • $\begingroup$ I believe for part 1) you should have: $\dfrac{1}{1-(-z)}$ because the sum of the geometric series needs to be of the form shown by blf below, where the $1$ in the denominator is positive. $\endgroup$ – andraiamatrix Dec 17 '13 at 5:04

I think that the reason is that you need to use the identity $$\frac{1}{1-w} = \sum_{n=0}^\infty w^n$$ which holds only for $|w| < 1$. Notice that if $|z|<1$, then $\frac{1}{z+1} = \sum_{n=0}^\infty (-z)^n$ is valid, but not for $|z| > 1$. That's why in the other two cases, you need to write it as $$\frac{1}{z} \cdot \frac{1}{1+\frac 1 z} = \frac{1}{z} \sum_{n=0}^\infty \left(-\frac 1 z\right)^{n}$$ in order to use the identity, since now $\left|\frac 1z\right| < 1$.

I might be mistaken, so please correct me if you find a mistake in what I have said.

  • $\begingroup$ This is the reason that I learned in class. After factoring $1/z$ then the geometric series will converge only for $|1/z|<1$ which can be rearranged to give $1<|z|$ $\endgroup$ – andraiamatrix Dec 17 '13 at 4:41

I think blf gave the answer you probably want. But I'll add some extra reasoning:

For 1) with the factored fractions you've given, the corresponding geometric series you find for $-\frac{1}{2}\left[\frac{1}{1-\frac{z}{2}}\right]$ will be valid in $|z|<2$ and the series for $\frac{1}{1-(-z)}$ will be valid in $|z|<1$ so together they are both valid in the smaller of the two circles (which is your domain $|z|<1$).

For 2) you have your series representation for $\frac{1}{z-2}$ is already valid inside $|z|<2$ but the series for $\frac{1}{z+1}$ that you found in part 1) is only valid for $|z|<1$ and you need it valid in $|z|>1$ so that is the series that you will need to alter.

For 3) your series for $\frac{1}{z+1}$ from part 2) will be valid in $|z|>1$ so it can be left alone. But the series representation for $\frac{1}{z-2}$ is still only valid inside the circle $|z|<2$ so you will need to alter it to make it valid in $|z|>2$ so that they will be both be valid in the larger of the two circles which is your domain $|z|>2$.


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