What Determines the Radius of Convergence in Complex Power Series?

[Ted123]

Homework Statement

[PLAIN]http://img153.imageshack.us/img153/4822/radiusm.jpg

Homework Equations

The Attempt at a Solution

Using the ratio test:

\left | \frac{e^{i(n+1)^2 \theta} \theta^{n+1} z^{(n+1)^2}}{e^{in^2 \theta} \theta ^n z^{n^2}} \right |

= | \theta ||e^{2n\theta i}||e^{i\theta}||z^{2n+1}|

= | \theta ||e^{2n\theta i}||z^{2n+1}|

since |e^{i\theta}|=1 whenever \theta\in\mathbb{R}

I know the radius of convergence R=1 but how do I deduce this by finding the limit as n\to \infty ?

 

[vela]

You can simplify a bit more since |e^2nθi|=1 as well.

Hint: Let z = re^iθ. Then |z^m| = |r^me^imθ| = |r^m||e^imθ| = r^m.

So you need |\theta| |z^{2n+1}| < 1 for the series to converge. This obviously holds if θ=0, so look at the case when θ≠0. Using the hint above, solve for |z| and then take the limit at n→∞.

 

[Ted123]

How did you get

So you need |\theta| |z^{2n+1}| < 1 for the series to converge.

to follow from your calculation in your hint?

For |z|>1 there's an easy reason why the series doesn't converge since the terms don't tend to 0.

 

[vela]

No, you need to start from |\theta| |z^{2n+1}| &lt; 1 and eventually get to |z|<1. What you wrote isn't the same nor is it true when the series converges (take, for example, when z=0).

 

[vela]

How did you get

to follow from your calculation in your hint?

Sorry, it didn't follow from the hint. It was supposed to follow from what you said in your first post. It's the condition for convergence from the ratio test.

For |z|&gt;1 there's an easy reason why the series doesn't converge since the terms don't tend to 0.