Complex Functions (Power Series)

suspenc3
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Hi, Power Series' were not covered in my cal II class, so I don't know how to solve these. Is there a certain way to solve these?

Find the Radiusof convergence and open disk of convergence of the power series:

\frac{n^2}{2n+1}(z+6+2i)^n

I don't know how to latex the summation but it is there, n=0 - inf.

THanks
 
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See the formula for r here:

http://en.wikipedia.org/wiki/Radius_of_convergence

Note that is you have a function f(z), and you find its power series expansion about a point a, then its radius of convergence is just the distance from a to the nearest singularity of f. You can use this fact to find the radius of convergence of the power series of a real-valued function f(x) by finding the nearest singularity of the corresponding complex valued function f(z).
 
I should have written "Note that if you have an holomorphic function f(z),..."

I need to refresh my memory about all this...
 
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So you just take the lim as n->infinity of |\frac{C_n}{C_n+1}|?

or \frac{\frac{n^2}{2n+1}}{\frac{n^2}{2n+1}+1}?
 
That's probably the easiest way.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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