# Complex Series

• Mar 6th 2011, 07:04 AM
Complex Series
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

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 $\sum(x_n^2+y_n^2).$