# Homework Help: Radius of convergence question(complex function)

1. Jul 10, 2010

### nhrock3

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 10, 2010

### Gib Z

More precisely, your teacher was using the theorem that if it is known a power series diverges for some number P, then it diverges for all complex numbers greater in magnitude than P, |z|>|P|. So for the first example in your post, since the function is not analytic at z=1, the radius of convergence is certainly 1 or smaller, definitely not bigger.

3. Jul 10, 2010

### nhrock3

so why the other example dont work by that logic
?

4. Jul 10, 2010

### Gib Z

Because you are wrong and the radius of convergence for that is indeed |z|< 1.

It could never possibly be |z|>1, the radius of convergence is always in the shape of a ball from some center, not an annulus. By the comparison test, if the power series converges for some complex value P, it converges for all |z|<|P|

5. Jul 10, 2010

### vela

Staff Emeritus
It depends on how you expand the function actually. For example, you can write

$$\frac{1}{1-z} = 1+z+z^2+\cdots$$

which converges for |z|<1, but you could also write it as

$$\frac{1}{1-z} = \frac{1}{z(1/z-1)} = -\frac{1}{z}\left(1+\frac{1}{z}+\frac{1}{z^2}+\cdots\right) = -\frac{1}{z}-\frac{1}{z^2}-\frac{1}{z^3}-\cdots$$

This series converges for |z|>1.

You didn't tell us what the actual series were in each example, so talking about a radius of convergence doesn't really make sense. All of the functions are analytic everywhere except at the poles.

6. Jul 10, 2010

### Gib Z

That's a Laurent Series, not a Power Series though. The concept of Radius of Convergence is specific to Power series.

7. Jul 10, 2010

### nhrock3

the original says the develop exaple2+example3 by 1<|z|<2

8. Jul 10, 2010

### Gib Z

It sound's like then you are being asked to find Laurent Series for those functions valid in that annulus. Radius of Convergence refers only to Power Series.

9. Jul 10, 2010

### nhrock3

yess

10. Jul 10, 2010

### Gib Z

Okay then...you should start a new thread with all your working, and this time, with the complete entire original problem. You didn't say anything Laurent series in the original post.