Under these conditions can I say that $\sum (a_{n}^{2} + b_{n}^{2})$ converges? I know that $\sum a_{n}$, $\sum b_{n}$, $\sum (a_{n}^2 - b_{n}^2)$ and $\sum 2a_{n}b_{n}$ converges. With that can I say that $\sum a_{n}^{2}$ and $\sum b_{n}^{2}$ converges?
PS. $a_{n}$ and $b_{n}$ are from $\sum z_{n}$(converge) where $z_{n} = a_{n} +ib_{n}$ and the other series are from the hipothesis that $\sum z_{n}^2$ converges and $a_{n} \geq 0$

The answer is no.  For example, take
$$
a_n = \frac{(-1)^n}{\sqrt{n+1}}, \quad b_n = \frac{(-1)^{\lfloor n/2\rfloor}}{\sqrt{n+1}}
$$
where $n=0,1,2,3,\dots$  Or, if you prefer, set
$$
z_n =  \frac{(-1)^n}{\sqrt{n+1}} + i\frac{(-1)^{\lfloor n/2\rfloor}}{\sqrt{n+1}}
$$
