How to find the Laurent Series
I'm going to try to give nearly cookbook style instructions on how to find most of the Laurent series that one would be asked to find in an introductory Complex Analysis course.
This means that we have a function \(f:\Bbb C\to \Bbb C\) and a point \(z=z_0\). We would like to write f as a series $$ f(z) = \sum_{-\infty}^\infty c_n (z-z_0)^n $$ And so what that "really" means is that we need to determine the values of the \(c_n\), since that's the only part of the above equation that we don't know.
1) Maybe there's nothing to do???
I mean look. The goal is to get the function into that form up there. If the function is ... already in that form ... well then you just don't do nothing. I mean, maybe you eat some popcorn or whatever, but you don't do any calculations.
For example if your function is \( f(z)=z^2 \) and you are expanding this at z=0 (i.e. \(z_0 = 0\)), well then f is already in the form that we want. You would just have \(c_2 = 1\) and for every \(n\ne 2\) we have \(c_n=0\). The series then looks like this:
$$ \sum_{-\infty}^\infty c_n(z-z_0)^n = \cdots + 0\cdot z^{-1} + 0 + 0 \cdot z + (1)z^2 + 0 \cdot z^3 + \cdots $$ $$ = z^2 $$
So in this example there's really just nothing you need to do. It's given to you, already done.
Here is another function which is already expanded at z = 0.
$$ \frac 1 z $$
In this case \(c_{-1} = 1\) and every other coefficient is zero.
---
Here’s a function already expanded at z = -1 just to show some variety:
$$ \frac 1 {z+1} + z+1 $$
So again the point here is: Before you go doing stuff, remember what you’re trying to obtain in the first place! Maybe there’s nothing to do at all, depending on what you are working on.
2) Just Do It (The Formula)
Ok but maybe life isn’t so easy on you. For example suppose you are given \(f(z) = e^z\) and you’re asked to expand this at z = 1. Well in this case you’re gonna see at the end that this isn’t all that “Laurenty” because this is really just going to end up being a Taylor series. But for now let’s pretend we don’t know that.
So maybe the first thing you try is, you reach for the formula. That is to say, the formula for the coefficients.
$$ a_n = \frac{1}{2\pi i}\oint_\Gamma \frac{f(z)}{(z-z_0)^{n+1}} \ dz $$
where \(\Gamma\) is any “nice” curve that encloses the point \(z_0\). In this example that means
$$ a_n = \frac{1}{2\pi i} \oint_\Gamma \frac {e^z}{(z-1)^{n+1}} \ dz $$
It will turn out that which \(\Gamma\) we choose just doesn’t matter, because I’m going to use Cauchy’s integral formula. So I could tell you that I arbitrarily pick \(\Gamma\) to be the unit circle centered at z = 1 but I could pick any other suitable curve and get the same results. I’ll just leave it as \(\Gamma\). You can imagine it’s your favorite suitable curve.
We will divide this into two cases, either \(n \ge 0\) or \(n < 0\). In the former case, I can use Cauchy’s integral formula
$$ \oint_\Gamma \frac{f(z)}{(z-z_0)^{n+1}} \ dz = \frac{2\pi i}{n!} f^{(n)}(z_0) $$
which in our example becomes \(\frac{2\pi i}{n!} (e^z)^{(n)}|_{z=1} \). The nth derivative of \(e^z\) is just \(e^z\) which is what makes it such a convenient “easy” example. Therefore this becomes just
$$ \frac{2\pi i}{n!} e^z | _{z=1} = \frac{2\pi i}{n!} e^1 $$
We can now say what the coefficient is, and it agrees with what we already knew from Taylor:
$$ a_n = \frac{1}{2\pi i} \left( \frac{2\pi i}{n!} e^1 \right) = \frac 1 {n!} e^1 $$
This still leaves us to determine what happens when \(n < 0\) but there the situation is even simpler. We must evaluate
$$ a_n = \frac{1}{2\pi i}\oint_\Gamma \frac{e^z}{(z-1)^{n+1}} \ dz $$
but for a strictly negative n, we can just regard this as \( e^z \) times \((z-1)^m\) where \( m = -(n+1) \ge 0\). Well in that case we just have a product of two functions which are analytic on and interior to \(\Gamma\)! This means that we now live in a beautiful paradise where the integral vanishes. I.e.
$$ a_n = \frac{1}{2\pi i} \cdot 0 = 0 $$
So we’re done! We now know all the coefficients. In the end we can summarize what we found by writing it in the convenient notation
$$ e^z = \sum_{n=0}^\infty \frac{e^1}{n!}(z-1)^n $$
Please note that you should not actually have to go through all this, since you should already know the expansion of \(e^z\). I chose this example just to be an especially easy demonstration of using the formula. But if you are asked to find the expansion of this particular function, you should already be familiar enough with Taylor expansions to cut to this result much faster using the more basic Taylor formula.
You might wonder: But will it always be this easy? What do I do if the negative values of n do not vanish like this?
To answer that: I cannot think of an example where the best way to handle tough calculations is by using the formula. If the formula is giving you tough calculations for the coefficients then you might find a better strategy by doing something other than the formula.
3) Use what you (are supposed to) know already
Many calculations of the coefficients simply use already know Taylor series, and just manipulate them through composition, addition, distribution, differentiation, and integration. In particularly unpleasant settings we may also need to multiply using “convolution”.
For instance, take \(e^{1/z}\) expanded at z = 0. If you think over the first two strategies you’ll see that they don’t apply. It’s not already in the form that we want, and it’s not easy to use the formula.
But we already know the expansion of \(e^z\) expanded at z = 0! It’s
$$ \sum_{n=0}^\infty \frac{1}{n!} z^n $$
So we may just compose (or “plug in”) 1 / z to get
$$ \sum_{n=0}^\infty \frac{1}{n!} (1/z)^n $$
This tells you, for instance, that \(c_{0} = 1\) by taking \(n = 0\). It also tells you \(c_{-1} = 1\) by taking \(n=1\) and it tells you \( c_{-2} = \frac 1 2 \) by taking \(n=2\) and so on.
Sometimes the Laurent series can be found just that easily!
---
Here’s another example: Expand \( z + e^{1/z} \) at z = 0. Well we just use the expansion for \(e^{1/z}\) just as above. We also need the expansion of z at the point. But that’s trivial because (a la the first strategy above where we don’t do nothin) it’s already in the desired form. Therefore we just put the terms together.
$$ z+ \sum_{n=0}^\infty \frac{1}{n!} (1/z)^n $$
is the Laurent series. We could try to somehow “simplify” this, but really I don’t think any other way of writing it is any “simpler”. This right here already shows you what all of the coefficients are! And that’s all we really wanted.
---
We can take another example, this time using differentiation. So for instance, suppose we want to find the Laurent expansion for \(\frac{1}{(1-z)^2} \) at the point \(z = 0\). We should recognize that this looks eerily similar to the function \(\frac 1 {1-z} \) which famously has as its expansion the geometric series
$$ \frac 1 {1-z} = \sum_{n=0}^\infty z^n $$
so long as we take \(|z|<1\). If you stare at this long enough, maybe you will recognize that \(\frac 1 {(1-z)^2}\) is the derivative of \(\frac 1 {1-z}\)! It’s probably not immediately obvious, but it kind of makes sense to notice this. After all the power has gone from -1 to -2, so that kind of seems like a derivative.
(Note: Of course we could also have squared the function, but that would have required us to think about multiplying. When it comes to multiplying two Laurent series, this can become extremely painful, so we would like to avoid it where possible.)
Using this idea, since \(\frac 1 {1-z} = \sum_{n=0}^\infty z^n\) then if we take the derivative on both sides, we get
$$ \frac{1}{(1-z)^2} = \sum_{n=1}^\infty nz^{n-1} $$
In fact this idea may and probably should already be familiar from Taylor series. In fact, everything that I’ve done in this example really just is a Taylor series, since there are no negative exponents. But we could also go the other direction, do a similar trick with integration, and find a Laurent series for \(-\ln (1-z)\) since this is just the antiderivative of \(\frac 1 {1-z}\).
Anyway, I won’t spend more time on this since I really just wanted to communicate this idea: One strategy for finding Laurent series is to take expansions that you already know, and manipulate them in certain ways.
4) Adjust a geometric series to fit your needs
Up to now I haven’t really been talking at all about the domain of convergence. For most of these the domain has been the complex plane, perhaps with the exception of a point. For geometric series we take a disk, but these subtleties haven’t been worth emphasizing. But now let us worry about them, in the face of a couple problems.
Take the example function \(\frac 1 {z-z^2} \). And let’s think about a few different points at which we could expand this. it will be important to notice that this has poles at z = 0 and z = 1. Perhaps unexpectedly, it will be easier to carry out the expansion at the poles than it will be at any point not at the poles.
To begin with consider expanding at z = 0. This is made one step extra easy because in some sense, some part of the function is already expanded there! When we factor the denominator we get
$$ \frac 1 z \cdot \frac 1 {1-z} $$
so that the left factor is already in the form that we want. We only need to concern ourselves with the right factor. For this, we are already chomping at the bit to apply the geometric series. That’s excellent but we will need to keep in mind that this is only the right thing to do, IF we are trying to find the expansion which is valid in \(|z|<1\). In that case, the following is the right way to get the expansion.
$$ \frac 1 z \sum_{n=0}^\infty z^n = \sum_{n=0}^\infty z^{n-1} $$
But what if we wanted an expansion valid in \(|z|>1\)? Clearly the problem factor is the right factor, since it’s the one that put the limitation on us that we now want to unsaddle. To correct it we can actually factor out a factor of 1 / z from it!
$$ \frac 1 z \cdot \frac{1}{1-z} = \frac 1 z \cdot \frac 1 z \cdot \frac{1}{1/z-1} $$
Now it looks almost kinda like it’s ready for a geometric series? But the 1-… part isn’t quite right, it’s flipped. Ok, fine, then we also factor out a -1 to flip it around like how we need it to be.
$$ -\frac 1 z \cdot \frac 1 z \cdot \frac{1}{1-1/z} $$
If we now apply the geometric series to this it will be valid where \(|1/z|<1\) which is the same as \(1< |z|\) … which is … precisely what we wanted this whole time! So the Laurent expansion is
$$ -\frac 1 {z^2}\sum_{n=0}^\infty (1/z)^n = -\sum_{n=0}^\infty (1/z)^{n-2} $$
---
If we were to expand this same function, this time at the point z = 1, we would go through a very similar process. The main difference is that the right factor would now be the one which is already in the form that we want, and we would have to instead fiddle with the factor \(\frac 1 z\) to do some kind of geometric series thing to it. This time we would need to introduce a 1-… so that it fits the geometric series formula. We can accomplish that by writing
$$ \frac 1 z = \frac 1 {z-1+1} = \frac{1}{1+(z-1)} $$
This is getting better. It’s not a 1-… on bottom, but if we make the + and -- then we get what we want.
$$ \frac 1 {1-(-[z-1])} $$
To this we may now apply the geometric series formula, valid where \(|-[z-1]| < 1\). That is to say, valid where \(|z-1|<1\). In this region we obtain
$$ \sum_{n=0}^\infty (-[z-1])^n $$
and if we now want to put together the Laurent series, it’s
$$ \frac{1}z \cdot \frac{1}{1-z} = \sum_{n=0}^\infty (-[z-1])^n \cdot \left(-\frac 1 {z-1}\right) $$
$$ = -\sum_{n=0}^\infty (-1)^n(z-1)^{n-1} $$
---
We could again consider, what if we didn’t want to do the expansion valid in \(|z-1|<1\) but instead valid in \(|z-1|>1\). Again we we do the same as above but this time factor out a certain factor from the denominator. This blog post is already massive and I’m getting tired of writing it, so I won’t detail this much more. It suffices to just remember the bigger point that I’m trying to make here: To get these Laurent series that we want, the strategy I’m broadly describing is to perform certain manipulations which lend themselves to geometric series.
---
The same is again true for when the expansion is not taken at any pole, like for instance if we were to try to perform the expansion at z = i. A good way to approach this is to again factor the polynomial, but then we need to manipulate both factors since neither of them is already in a form that we want for this expansion.
It is especially a good idea to first NOT have to deal with multiplication. As pointed out earlier, multiplication is a bit of hell when it comes to series, so it is best if we can avoid it. Since we’re dealing with polynomials, we can change out of multiplication and into addition by another well-known trick: Partial fraction decomposition.
Oh man I’m tired.
5) Other stuff. I guess.
Long as this blog post is, it definitely doesn’t encompass everything about finding the Laurent series of a given function at a given point. There are things I made a nod to in this post, but didn’t talk about. But there are also many techniques and results about finding Laurent series whose existence I haven’t even acknowledged. So this is not a complete guide, but just a guide to help a new student to organize their thoughts and understanding of the early techniques of finding Laurent series.
The end.