# Complex Functions (Power Series)

1. Oct 9, 2006

### suspenc3

Hi, Power Series' were not covered in my cal II class, so I dont 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 dont know how to latex the summation but it is there, n=0 - inf.

THanks

2. Oct 9, 2006

### quasar987

See the formula for r here:

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).

3. Oct 9, 2006

### quasar987

I should have written "Note that if you have an holomorphic function f(z),..."

I need to refresh my memory about all this... :grumpy:

Last edited: Oct 9, 2006
4. Oct 9, 2006

### suspenc3

So you just take the lim as n->infinity of $$|\frac{C_n}{C_n+1}|$$?

or $$\frac{\frac{n^2}{2n+1}}{\frac{n^2}{2n+1}+1}$$?

5. Oct 9, 2006

### quasar987

That's probably the easiest way.