# Homework Help: Laurent series

1. Oct 12, 2009

### oddiseas

1. The problem statement, all variables and given/known data

Laurent expansion for 2/z+4 -1/z+2

I can derive the laurent expansion, but i would like a better understanding, so if anyone could tell me if my understanding is right or wrong,

2. Relevant equations
I know that the region we want is an annulus.

But i am trying to figure this out, so i can have a picture in my head, which seems to work for me. Is this logic correct:
1) we have a circle cenered at -2, and if we keep the magnitude of z greater than 2 than we ensure that -1/z+2 is entirely analytic, and outside of this circle and we dont have any problems.(But really an open disk is analytic but not simply connected, so why cant we just use |z|<2.)

2)We have another circle of radius 4, but we place the singularity on the boundary of the circle, thus if we keep the magnitude of z<4 we ensure that we dont run into the singularity.

QUESTION: why when deriving the laurent series in books if we have:
|z|>2 for the Laurent series to be defined they write 1/z+2=1/z(1+2/z).
It seems to me that they do this because z>2, which meens it wont include z=o and therefore we dont have a problem with 2/z at z=0, thus simplifyibg the calculation.Is this the actual reason?

But if this is so for 2/z+4 we have |z|<4 and thus we may run into the singularity at zero
so we cant express it this way.and we use 1/2(z/2+1). Is this a correct analysis? and if not then why do we do it?

In addition why cant i keep the equation as it originally appeared as 2/z+4 -1/z+2 and derive the expression using the binomial theorm.Why do we change it? is the only reason to simplify the calculation?

3. The attempt at a solution

2. Oct 12, 2009

### Office_Shredder

Staff Emeritus

I obviously can't see the steps taken in the book, but for 1/z(1+2/z) this is commonly done because 1/(1+2/z) has a known Taylor series expansion when |z|>2, i.e. |2/z|<1

A lot of times when trying to find the Laurent series of a function, it turns out that you really have a bunch of holomorphic functions that are in denominators, or maybe with funny terms, and it helps to write those out as Taylor series and then perform some algebra to get a single Laurent series

3. Oct 12, 2009

### n!kofeyn

One thing that is good to know is that for problems such as this, there are multiple annuli that the given function has a Laurent expansion in. For example, the function you gave has Laurent expansions for |z|<2, 2<|z|<4, and |z|>4. This means that there are 3 different Laurent expansions for each of these annuli.

Now, to find the Laurent expansions, it boils down to manipulating your function so that you can take advantage of the geometric series to find your Laurent expansions. I'll do the case of |z|<2.
$$\frac{2}{z+4} - \frac{1}{z+2} = \frac{2}{4(1-(-z/4))} - \frac{1}{2(1-(-z/2))} = \frac{1}{2} \sum_{n=0}^\infty \left(-\frac{z}{4} \right)^n - \frac{1}{2} \sum_{n=0}^\infty \left(-\frac{z}{2} \right)^n$$
I used the fact that the two geometric series converge for |-z/4|<1, which is equivalent to |z|<4, and |-z/2|<1, which is equivalent to |z|<2. This means I have found the Laurent expansion for this function in the annulus (actually circular region) |z|<2. (You can of course simplify it a little more from where I left off.)

You do the same thing, for the other two annuli, but just make note of what values the geometric series converge for. This will determine how you manipulate the functions so that you can use the geometric series.

My last comment is that each of these annuli (and thus the Laurent expansions) are centered at z=0. You mentioned there is a circle centered at z=-2, which is incorrect. Notice that we removed the circles centered at zero with radii 2 and 4, which removed the two singularities z=-2,-4.

4. Oct 12, 2009

### oddiseas

according to the definition in my book, the laurent series is used when we specifically want a valid series expansion of a function that contains a singular point.It says in this case we cant avoid the singularity, by expanding around an alternate region , so we use the laurent expansion with an annular region "centered" on the singularity. You say that it is centered at zero. But the book specifically states that it is centered at the singularity.

5. Oct 12, 2009

### n!kofeyn

What is the book and page number?

6. Oct 12, 2009

### oddiseas

Its called complex variables by K.A stroud and its on page 306

7. Oct 12, 2009

### n!kofeyn

Well I was hoping I either had the book or could find the page on Google Books, but I don't and couldn't. It just depends on what the problem is asking. If the problem is asking you to find the Laurent series expansion about z=0, then the three annuli listed above are the correct ones, and the method I explained is correct. This is most likely the case. The Laurent series is similar to the power series as it can be centered at different points. It doesn't have to be centered at the singularity. The annulus just can't contain that singularity.

An example is the function $$\frac{1}{(z-1)(z-2)}$$.
Centering the Laurent series at z=0, it can be broken up similar to the problem above, and then you can utilize the geometric series to find its Laurent expansion. But let's say we want to center it at z=1 and to have it be convergent in the annulus 0<|z-1|<1. Then
$$\frac{1}{(z-1)(z-2)} = \frac{1}{z-1} \cdot \frac{-1}{1-(z-1)} = -\frac{1}{z-1} \sum_{n=0}^\infty (z-1)^n = -\sum_{n=0}^\infty (z-1)^{n-1}$$
Note that the geometric series converges for |z-1|<1.

Also, I made a small error above because I mentioned there was a Laurent expansion for |z|<2. It should be 0<|z|<2 since the Laurent series is only for an annulus.

8. Oct 13, 2009

### HallsofIvy

A Laurent series doesn't have to be "centered on a singularity" but if it isn't it will be a Taylor's series, not a "Laurent series" because it will not contain any negative powers. A "singularity" is any point at which the function is not analytic. NOT centered on a singularity means centered on a point at which the function is analytic and, by definition, the function is equal to its Taylor's series around that point.

9. Oct 14, 2009

### oddiseas

ok, thanks to all, that makes sense now

10. May 20, 2011

### santoshrana21

i m just a beginner in complex variable. So can any genius help me to get concept how to find laurent expansion and residue. Please describe in simple procedure.