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|&lt;1 and |z|&gt;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|&lt;1 we have
\left|\frac{1}{1+z^n}\right|\leq\frac{1}{1-|z|^n}\leq\frac{1}{1-|z|}
and for |z|&gt;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|&lt;1 we have
\left|\frac{1}{1+z^n}\right|\leq\frac{1}{1-|z|^n}\leq\frac{1}{1-|z|}
and for |z|&gt;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|&lt;1

and

\frac{1}{|z|-1}\leq 1 for |z|&gt;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|&lt;1
<br /> \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}<br />
and so the series are convergent, by comparison test. The same for |z|&gt;1. Your upper bounds were slightly wrong, but the idea of comparing with \frac{1}{n^2} was ok.