# Need help

1. Feb 10, 2007

### racland

Anyone knows anything about the Radius on convergence in the complex plane (Complex Analysis)

2. Feb 10, 2007

Yes. Somebody knows something about the radius of convergence in the complex plane.

3. Feb 10, 2007

### ranger

:rofl: :rofl:

4. Feb 11, 2007

### HallsofIvy

Staff Emeritus
Radius of convergence in the complex plane is exactly the same as radius of convergence in the real numbers. Exactly what is your question?

5. Feb 13, 2007

### racland

Great

The problem states:
Find the Radius of Convergence of the following Power Series:
(a) Sumation as n goes from zero to infinity of Z^n!
(b) Sumation as N goes from zero to infinity of (n + 2^n)Z^n

For (a) I think the radius of convergence is 1 but I'm a bit unsure of that...

6. Feb 14, 2007

### HallsofIvy

Staff Emeritus
As I said- same thing as on the real line (except now it really is a radius!). Apply the "ratio" test: a series $\Sigma a_n$ converges absolutely if the limit ration
$$\lim_{n\rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right|$$
is less than 1.
(Diverges if that limit is greater than one, may converge absolutely or conditionally or diverge if it is equal to 1).

In particular, for Zn, we have |Zn+1/Zn|= |Z|. That series converges absolutely for |Z|< 1. (It obviously diverges for Z= 1 or -1 and diverges for |Z|> 1)

Try (n+ 2n)Zn yourself.

7. Feb 14, 2007

### StatusX

Is that $z^{n!}$? If so, you can prove a general result that if a_n is any increasing sequence of natural numbers, then $z^{a_n}$ converges iff |z|<1. This is the case HallsofIvy did if a_n=n, and yours if a_n=n!. The general proof follows from the result for a_n=n (which is the smallest increasing sequence of natural numbers).

Last edited: Feb 14, 2007