Finding Region of convergence for complex series

[andrie55]

Homework Statement

I am struggling to answer this question please help

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

Ʃ(i+z)^(2n-1)/2^(2n+1)

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

 

[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.

 

[andrie55]

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.

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

 

[micromass]

You want to get it in the form |z-c|&lt;r for some c and r.

Hint:

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

 

[andrie55]

You want to get it in the form |z-c|&lt;r for some c and r.

Hint:

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

To be honest I don't 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 thanks for your help guys

 

[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)?

 

[andrie55]

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)?

No I don't know that. So in this case c=-i?

 

[Dick]

No I don't know that. So in this case c=-i?

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

 

[andrie55]

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

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

 

[Dick]

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

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

 

[andrie55]

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

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

 

[Dick]

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

Yes. The radius of convergence is 2.

 

[andrie55]

Yes. The radius of convergence is 2.

Oh thank you very much I have learned 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?

 

[Dick]

Oh thank you very much I have learned 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?

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.