# Finding Region of convergence for complex series

1. Apr 9, 2012

### andrie55

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

Find the region of convergence for the following complex series and draw the region

Ʃ(i+z)^(2n-1)/2^(2n+1)

3. The attempt at a solution

Here is my hand written working, sorry i could figure out how to use the symbols

http://i43.tinypic.com/10xsnrc.jpg

Thank you

I am sorry I just remembered that links to images are not permitted. How do I delete this thread?

Last edited: Apr 9, 2012
2. Apr 9, 2012

### Dick

I think the figure is all right. Now you just want to figure out what region |(z+i)^2/4|<1 corresponds to. Simplify it a little.

3. Apr 9, 2012

### andrie55

Thats the problem mate. I just don't have a clue on what to do after this stage. I have look a various text books and online without success.

4. Apr 9, 2012

### micromass

Staff Emeritus
You want to get it in the form $|z-c|<r$ for some c and r.

Hint:

1) |cz|=c|z| for positive real c
2) |z²|=|z|²

5. Apr 9, 2012

### andrie55

To be honest I dont have a clue now as in most cases we have only worked out if the series in convergence or divergent. This is new to me. Are they any textbooks or websites u recommend for learning this stuff?

Anyway thanx for your help guys

6. Apr 9, 2012

### Dick

You may be thinking this is much harder than it actually is. You've already done the hard part. Do you know |z+i|=|z-(-i)|<r is an open disk of radius r around the point z=(-i)?

7. Apr 10, 2012

### andrie55

No I dont know that. So in this case c=-i?

8. Apr 10, 2012

### Dick

Sure, |z-c|<r is a disk centered on c. So if c=(-i) then you get |z+i|<r. Now what's r?

9. Apr 10, 2012

### andrie55

Since its convergence r is equals to 1. Unless i have to multiply by the denominator of 4?

10. Apr 10, 2012

### Dick

Work it out. If |(z+i)^2/4|<1 then |(z-(-i)|<? No, it's not 1.

11. Apr 10, 2012

### andrie55

cross multiplication makes it |(z+i)^2|<4
then square root of 4 =2
so |(z-(-i)|< 2

12. Apr 10, 2012

### Dick

Yes. The radius of convergence is 2.

13. Apr 10, 2012

### andrie55

Oh thank you very much I have learnt something very important. So now I just need to plot xy plane with i along the y axis and numbers (radius) along the x axis. The circle with initiate from point -i with radius of 2. So how will i know that the radius is 2 along the y axis since its a imaginary axis with no real numbers?

14. Apr 10, 2012

### Dick

You are making this sound more complicated than it is. If y is your imaginary axis then the point -i is the point (0,-1). Use that as a center and draw a circle of radius 2 around it.