Complex Series

**mohammadfawaz** · March 6th 2011, 07:04 AM

Hello!
Suppose that $\{z_n\}$ is a sequence of complex numbers satisfying $Re\{z_n\}\geq0$ and that the two series $\sum_{n=1}^\infty z_n$ and $\sum_{n=1}^\infty z_n^2$ are convergent. Prove that the series $\sum_{n=1}^\infty |z_n|^2$ is also convergent.

Well, assuming $z_n=x_n+iy_n$ , we can deduce from the given that the following series converge:
$\sum_{n=1}^\infty x_n$ , $\sum_{n=1}^\infty |x_n|$ ,
$\sum_{n=1}^\infty y_n$ ,
$\sum_{n=1}^\infty x_n^2-y_n^2$ ,
$\sum_{n=1}^\infty 2x_n y_n$ .

The sequence $\sum_{n=1}^\infty |z_n|^2$ can be written as:
$\sum_{n=1}^\infty x_n^2+y_n^2$ or $\sum_{n=1}^\infty |x_n^2-y_n^2+2ix_ny_n|$

I am stuck here for now. Any ideas?

Thanx

Mohammad

**Opalg** · March 6th 2011, 11:47 AM

Originally Posted by mohammadfawaz

Suppose that $\{z_n\}$ is a sequence of complex numbers satisfying $Re\{z_n\}\geq0$ and that the two series $\sum_{n=1}^\infty z_n$ and $\sum_{n=1}^\infty z_n^2$ are convergent. Prove that the series $\sum_{n=1}^\infty |z_n|^2$ is also convergent.

If the series of positive terms $\sum x_n$ converges, then so does $\sum x_n^2$ (comparison test, because $x_n<1$ for all sufficiently large n). If you also know that $\sum(x_n^2-y_n^2)$ converges, then (by subtraction) so does $\sum y_n^2$ . Hence so does $\sum(x_n^2+y_n^2).$

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