Comparing Complex Homework: Solving for |z|<1 & |z|>1

[Ted123]

Homework Statement

[PLAIN]http://img225.imageshack.us/img225/7501/complexh.jpg

Homework Equations

The Attempt at a Solution

Can I say that, for both $|z|<1$ and $|z|>1$ ,

$\displaystyle \left | \frac{1}{n^2} \left ( \frac{1}{1+z^n} \right ) \right | \leq \frac{1}{n^2}$ .

So, since $\displaystyle \sum^{\infty}_{n=1} \frac{1}{n^2}$ converges, $\displaystyle \sum^{\infty}_{n=1} \frac{1}{n^2} \left ( \frac{1}{1+z^n} \right )$ converges absolutely in both cases (a) and (b) by the comparison test?

 

[losiu99]

Well, it's not true that
$$\left|\frac{1}{1+z^n}\right|\leq 1$$
for $$|z|\neq 1$$. However,
$$|1+z^n|\geq |1-|z|^n|$$
so, for $$|z|<1$$ we have
$$\left|\frac{1}{1+z^n}\right|\leq\frac{1}{1-|z|^n}\leq\frac{1}{1-|z|}$$
and for $$|z|>1$$ we have
$$\left|\frac{1}{1+z^n}\right|\leq\frac{1}{|z|^n-1}\leq\frac{1}{|z|-1}$$
Then you can use the rest of your argument.

 

[Ted123]

Well, it's not true that
$$\left|\frac{1}{1+z^n}\right|\leq 1$$
for $$|z|\neq 1$$. However,
$$|1+z^n|\geq |1-|z|^n|$$
so, for $$|z|<1$$ we have
$$\left|\frac{1}{1+z^n}\right|\leq\frac{1}{1-|z|^n}\leq\frac{1}{1-|z|}$$
and for $$|z|>1$$ we have
$$\left|\frac{1}{1+z^n}\right|\leq\frac{1}{|z|^n-1}\leq\frac{1}{|z|-1}$$
Then you can use the rest of your argument.

So how does my inequality: $\left | \frac{1}{n^2} \left ( \frac{1}{1+z^n} \right ) \right | = \frac{1}{n^2} \left | \left ( \frac{1}{1+z^n} \right ) \right |\leq \frac{1}{n^2}$

follow from the upper bounds of your last 2 inequalities?

So the following is true:

$\frac{1}{1-|z|} \leq 1$ for $|z|<1$

and

$\frac{1}{|z|-1}\leq 1$ for $|z|>1$

and I can multiply both sides of the inequlity by $\frac{1}{n^2}$

 

[losiu99]

No, these two inequalities are not true. However, for $$|z|<1$$
$$\left|\frac{1}{n^2}\left(\frac{1}{1+z^n}\right)\right|\leq\frac{1}{n^2}\frac{1}{1-|z|}=\frac{C}{n^2}$$
and so the series are convergent, by comparison test. The same for $$|z|>1$$. Your upper bounds were slightly wrong, but the idea of comparing with $$\frac{1}{n^2}$$ was ok.