I would appreciate a nudge in the right direction for the following questions, as I haven't really had any exposure to finding Laurent Series expansions for more complicated functions

Find the Laurent Series expansion for:

$$1)\frac{z-7}{z^2+z-2}$$ $ 1<|z|<2$ , $|z|>2$ , $0<|z-1|<1$

For this question I tried to follow the standard route for this question by using partial fraction decomposition and then attempting to put each fraction in the form of $\frac{1}{1-z}$ but I feel like I'm having an issue when I come to this stage:

By partial fraction decomposition we have: $$\frac{z-7}{(z-1)(z+2)}=\frac{3}{z+2} - \frac{2}{z-1}$$ And attempting to put $\frac{3}{z+2}$ and $-\frac{2}{z-1}$ in the form $\frac{1}{1-z}$ has so far yielded an embrassing $$\frac{3}{z+2}=\frac{3}{2+z}=\frac{3}{2}\frac{1}{1+\frac{z}{2}}$$ At this point I think it's as simple as doing some algebraic manipulation but I can't even seem to do this? And once I've done that is this still just a simple "plug and play" type scenario with Taylor Series?

$$2)\frac{1-cosz}{(z-2\pi)^3}$$ I don't even know where to start for this. I thought perhaps partial fraction decomposition again but I don't see how this would help (if even possible) since we would still struggle to get the form of $\frac{1}{1-z}$

$3)$Let $f(z)= \sqrt {z^2-3z+2}$, $1<|z|<2$. Does the Laurent series expansion for $f(z)$ exist? Justify your answer.

This is completely beyond me as I can't imagine where I would begin to check if the expansion exists. I'm only used to (as you can probably tell) calculating Laurent Series of the form $\frac{1}{1-z}$

Additionally I'm not entirely sure how the conditions i.e $1<|z|<2$ etc affect the question or how the Laurent Series expansion is calculated, so if someone could briefly explain that I'd appreciate it.



1 Answer 1


For context, the correct expansion is $$\frac{1}{1-w} = \sum_{n=0}^\infty w^n\qquad\text{when $|w|<1$.}$$ Which means it's critical to manipulate the expression so that $|w|$ is in fact smaller than 1.

1) The partial fraction expansion is the correct first step: $$\frac{z-7}{z^2+z-2} = \frac{3}{z+2} - \frac{2}{z-1}.$$ Where to go from here depends on the domain you're interested in. On the annulus $1<|z|<2$, we want $$\frac{3}{z+2} = \frac{3}{2}\cdot\frac{1}{1+\frac{z}{2}} = \frac{3}{2}\cdot\frac{1}{1-\left(-\frac{z}{2}\right)} =\sum_{n\ge0}(-1)^n\frac{3}{2^{n+1}}z^n\qquad \text{because $\displaystyle{\left|-\frac{z}{2}\right|<1}$,}$$ but $$-\frac{2}{z-1} = \frac{2}{z}\cdot\frac{1}{1-\frac{1}{z}} = \sum_{n<0} 2z^n\qquad\text{because $\displaystyle{\left|\frac{1}{z}\right|<1}$.}$$

The Laurent series for the original expression on the annulus is just the sum of these.

On the complement of the disk, you want to do something similar, but keep in mind that now the first fraction has to be expanded in powers of $2/z$ rather than $z/2$.

On the punctured disk $0<\left|z-1\right|<1$, first rewrite the partial fractions in terms of $t=z-1$, expand about $t$, then re-express in terms of $z$. For the purposes of the first fraction, $t$ is small. The second is just $2/t$; not much else to be done there. (Note this expansion will actually work out to $0<\left|z-1\right|<3$.)

2) Here the question is to find (a?) Laurent series for $$\frac{1-\cos z}{(z-2\pi)^3}.$$ I can only give general advice here since you don't specify a domain, but:

  • There are well-know Taylor expansions (and therefore Laurent expansions) for $\cos z$ and $\sin z$ centered at $z=0$. If the center is somewhere else, there are always sum-of-angle formulas.
  • There is a general rule for multiplying Laurent series: $$\left(\sum_n a_nz^n\right)\cdot\left(\sum_n b_nz^n\right) = \sum_n c_nz^n\qquad\text{where $\displaystyle{c_n = \sum_{i+j=n}a_ib_j}$.}$$ (Note that this only necessarily works if both series terminate in the negative direction or both terminate in the positive direction; we want that sum to be finite or there's a whole bunch of issues that come up.)

3) Is $\sqrt{z^2-3z+2}$ even well-defined on the annulus? If not, it definitely doesn't have a Laurent series.

  • $\begingroup$ This is a phenomenal explanation for the first question - thank you so much for this! I guess the different domains meant that I am required to calculate the Laurent Series expansion for each domain. My only question is shouldn't one of the sums for the Laurent Series (the principle part, I think it is called) have $z^{-n}$? As for the second one I forgot to say the domain is $|z-2\pi|>0$. Does this mean I use the Laurent Series for cos, then take 1-cosz multiplied by $\frac{1}{(z-2\pi)^3}$? Finally how would I go about showing 3) is not well defined on the annulus? Many thanks!! $\endgroup$
    – Overclock
    Mar 8, 2018 at 20:59
  • 1
    $\begingroup$ * Yes, you'll get different expansions for different domains. * In general you'll have terms of negative degree, yes, and they show up in the series for $2/(z-1)$. (Notice I'm taking $n<0$ for that one). * For 2), that's correct. You'll want the Laurent series about $z=2\pi$ rather than $z=0$, like in the last part of 1). Since $\cos z$ is $2\pi$-periodic that one will be pretty simple. (cont.) $\endgroup$
    – Chad Groft
    Mar 8, 2018 at 22:25
  • 1
    $\begingroup$ For 3), here's what happens. Every complex number except for 0 has two complex square roots that are additive inverses of each other (1 has $\pm 1$, $-1$ has $\pm i$, $i$ has $\pm (1+i)/\sqrt{2}$, etc.) If you start at some number $z$, choose a square root for it, and continuously vary $z$ (staying away from 0), the square root will vary continuously as well. But if you move $z$ in a loop around 0, you won't end up with the same square root at the end: it'll have picked up a minus sign. That's what happens with 3). Not sure how much rigor you need. $\endgroup$
    – Chad Groft
    Mar 8, 2018 at 23:02
  • 1
    $\begingroup$ For contrast: if instead you were interested in the region $|z|>2$, then every loop would go around $z=1$ and $z=2$ the same number of times, and would pick up the same number of minus signs from each, and they'd all cancel out. So there would be a well-defined square-root function (two in fact, one for each choice of square root at a convenient starting point), and each would have a Laurent series. $\endgroup$
    – Chad Groft
    Mar 8, 2018 at 23:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .