Prove that $\sum_{j=0}^{\infty}z_{j}$ converges absolutely iff $\sum_{j=0}^{\infty}x_{j}$ and $\sum_{j=0}^{\infty}y_{j}$ converges absolutely, where $z_{j} = x_{j} + iy_{j}$.
My first approach
Let us prove the implication $(\Rightarrow)$ first.
Let $\varepsilon > 0$. Then it exists a natural $n_{0}\geq 0$ s.t. \begin{align*} n > m\geq n_{0} \Rightarrow \sum_{j=m}^{n}|z_{j}| < \varepsilon \Rightarrow \sum_{j=m}^{n}|x_{j}| \leq \sum_{j=m}^{n}|z_{j}| < \varepsilon \end{align*} which implies that $\sum_{j=0}^{\infty}|x_{j}|$ converges because it is a Cauchy sequence.
Similar argument applies to $\sum_{j=0}^{\infty}y_{j}$.
Conversely, let us prove the implication $(\Leftarrow)$.
Let $\varepsilon > 0$. Then it exists $n_{0}$ s.t. \begin{align*} n > m \geq n_{0} \Rightarrow \sum_{j=m}^{n}|x_{j}| < \varepsilon/2\,\,\wedge\,\,\sum_{j=m}^{n}|y_{j}| < \varepsilon/2 \Rightarrow \sum_{j=m}^{n}|z_{j}| \leq \sum_{j=m}^{n}|x_{j}| + \sum_{j=m}^{n}|y_{j}| < \varepsilon \end{align*} which proves that $\sum_{j=0}^{\infty}|z_{j}|$ converges absolutely due to the same reason as before.
My second approach
Let us assume that $\sum_{j=0}^{\infty}|z_{j}|$ converges.
Then it is bounded. Hence the sequence $\sum_{j=0}^{n}|x_{j}|\leq \sum_{j=0}^{n}|z_{j}|\leq M$ is increasing and bounded too.
Consequently, it converges. Similar reasoning applies to $\sum_{j=0}^{n}|y_{j}|$.
The converse implication results from the same reasoning. Indeed, one has that \begin{align*} \sum_{j=0}^{n}|z_{j}| \leq \sum_{j=0}^{n}|x_{j}| + \sum_{j=0}^{n}|y_{j}| \leq M_{1} + M_{2} := M \end{align*} whence we conclude that the series $\sum_{j=0}^{\infty}z_{j}$ converges absolutely, and we are done.
Could someone please tell me if the wording of my proofs are correct?
Is there another way to prove it differently from the proposed solutions?
Any contribution is appreciated.