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.