# Homework Help: Laurent series question

1. Jul 10, 2010

### nhrock3

find the laurent series of $$f(x)=\frac{-2}{z-1}$$+$$\frac{3}{z+2}$$
for
1<|z|<2

i was by my teacher that the radius of convergence
is what smaller then the number which makes the denominator 0.
if
$$f(x)=\frac{1}{1-z}$$
then
because 1-1=0
so
it is analitical on
|z|<1

so if i apply the same logic
$$f(x)=\frac{-2}{z-1}$$

1 still makes denominator 0
and
it is analitical on
|z|<1
it is analitical on
|z|>1

for
$$f(x)=\frac{3}{z+2}$$
-2 makes denominator 0
so |z|<-2 (but its illogical because |z| is a positive numbe)

where is my mistake?

2. Jul 11, 2010

### HallsofIvy

I doubt that your teacher said "the radius of convergence is what smaller then the number which makes the denominator 0." I suspect that he/she said something like "the radius of convergence is not larger than the absolute value of the number which makes the denominator 0". And that only applies to series centered on z= 0. Generally, the radius of convergence of a Laurent series about a point runs to the nearest pole.

To find the Laurent series between z= 1 and z= 2, I recommend using the middle point- z= 3/2. That way, the "radius of convergence" will be 1/2 and so cover the entire region between 1 and 2. And, in my opinion, the simplest way to do that is to change the variable- let y= z- 3/2. Then z-1= z- 3/2+ 1/2= y+ 1/2 and z- 2= z- 3/2- 1/2= y- 1/2.

$$\frac{-2}{z- 1}+ \frac{3}{z- 2}= \frac{-2}{y+ 1/2}+ \frac{3}{y- 1/2}$$

To do each of those, write them as limits of geometric series. For example,
$$\frac{-2}{y+ 1/2}= \frac{-4}{2y+1}= \frac{-4}{1- (-2y)}$$
which is the sum of a geometric series with first term -4 and common ration -2y. Of course, such a series will converge for -1< -2y< 1 or -1/2< y= z-3/2< 1/2, 1< z< 2, exactly what you want.

3. Jul 11, 2010

### nhrock3

i know that when they say that |z| is greater then sothing then
we develop it b negative powers

why they said that
$$\frac{-2}{z- 1}$$
is analytical in |z|>1
??
you said
"not larger than the absolute value of the number which makes the denominator 0"
so its supposed to be |z|<1

why its not so
??

4. Jul 11, 2010

### vela

Staff Emeritus
What's the definition of an analytic function? It doesn't seem you know what that term means.

5. Jul 11, 2010

### nhrock3

analityc function means that it is defined in every point of the region
no singular point in the given region
for example:
5<|z|<6
from -5 to 6 the is no singular point for the function
it is analitical from -5 to 6

6. Jul 11, 2010

### vela

Staff Emeritus
To be a bit more precise, we say a function f(z) is analytic at a point z0 if it can be represented by a power series in a neighborhood about that point:

$$f(z) = \sum_{n=0}^\infty a_n (z-z_0)^n$$

For example, take f(z)=1/(1-z). We can come up with a power series for f(z) expanded about z0=0, namely

$$\frac{1}{1-z} = \sum_{n=0}^\infty z^n$$

so f(z) is analytic at z=0. Similarly, we can also find a power series for f(z) about the point z=3,

$$\frac{1}{1-z} = \frac{1}{-2-(z-3)} = -\frac{1}{2(1+(z-3)/2)} = -\frac{1}{2} \sum_{n=0}^\infty (-1)^n\left(\frac{z-3}{2}\right)^n$$

so f(z) is analytic at z=3. In fact, you can find a power series about every point in the complex plane except for z=1, so this function is analytic everywhere except z=1.

Note that asking if a function is analytic at some point z is not the same as asking if a particular series for the function is valid at z. The fact that the first series above doesn't converge at z=3 says nothing about whether f(z) is analytic there.

So say you have a function f(z) and you expand it in a power series about the point z0. This expansion is good for some neighborhood of z0, so you'd like to know how big this neighborhood actually can be. The answer is that the series is only going to be valid up until you run into a singularity. The distance between z0 and the singularity is the radius of convergence of the series. This is the idea that your professor was trying to get across to you.

For example, the function f(z)=1/(1-z) is singular at z1=1. The first series was for f(z) expanded about z0=0, so its radius of convergence is |z1-z0|=1. The second series was for f(z) expanded about z0=3, so the radius of convergence for that series is 2, the distance from z0=3 to the singularity at z=1.

Unlike a power series, a Laurent series includes negative powers of (z-z0). This allows you to express f(z) as a series about z0 that converges in a region where the power series won't. Again, if we consider f(z)=1/(1-z) expanded about z0=0, we have

$$\frac{1}{1-z} = \sum_{n=0}^\infty z^n$$

for the region |z|<1, and

$$\frac{1}{1-z} = -\frac{1}{z}\sum_{n=0}^\infty \left(\frac{1}{z}\right)^n$$

for the region |z|>1. The different regions will have different Laurent series.

Determining the regions of convergence is similar to finding the radius of convergence of a power series: you look at where the singularities are. For example, take the function

$$g(z)=\frac{1}{(z-1)(z+4)}$$

It has singularities at z=1 and z=-4. Say we want to expand about the point z0=0. The closest singularity is at a distance of 1, and the other one is at a distance of 4. So you'll have three regions to worry about: |z|<1, 1<|z|<4, and |z|>4.

Suppose you want to find series about the point z0=2. The closest singularity is z=1, which is at a distance of 1 from z0=2, and the other singularity is at a distance of 6 from z0=2. So, in this case, the regions are: |z-2|<1, 1<|z-2|<6, and |z-2|>6.

Last edited: Jul 11, 2010
7. Jul 12, 2010

### nhrock3

why this series cannot be developed
??

Last edited by a moderator: May 4, 2017
8. Jul 12, 2010

### vela

Staff Emeritus
You tell me.

9. Jul 12, 2010

### nhrock3

we have from -4 till 2 our point from which we develop is -2
so our distances are
2 and 4
or areas are
|z+2|<2
2<|z+2|<4
|z+2|>4

1<|z+2|<4

i dont have such area
so it cannot be developed becaus there is no such area
?

10. Jul 12, 2010

### vela

Staff Emeritus
Right. You can't have just one Laurent series the converges everywhere in that region.